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BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a)

Odisha State Board BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 9 Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a)

Question 1.
ନିମ୍ନଲିଖୂତ ମନୋମିଆଲଗୁଡ଼ିକୁ ସାନରୁ ବଡ଼ ଘାତାଙ୍କ କ୍ରମରେ ସଜାଇ ଲେଖ ।
1.4y3, √2y2, -51, 7y8, -8y4, \frac{11}{13}y9, √3y
ସମାଧାନ:
ମନୋମିଆଲଗୁଡ଼ିକୁ ସାନରୁ ବଡ଼ ଘାତାଙ୍କ କ୍ରମରେ ସଜାଇ ଲେଖିଲେ ହେବ –
-51, √3y, √2y2, 1.4y3, -8y4, 7y8, ଓ \frac{11}{13}y9 .

Question 2.
ନିମ୍ନରେ ପ୍ରଦତ୍ତ ମନୋମିଆଲ୍‌ଗୁଡ଼ିକ ମଧ୍ୟରୁ ସଦୃଶ ମନୋମିଆଲ୍‌ଗୁଡ଼ିକୁ ବାଛି ପୃଥକ୍ ଭାବେ ଲେଖ ।
12x2, -3x, \frac{1}{\sqrt{2}} x3, -5x2, \frac{x}{7}, 15, √3x3, 10x4, \frac{8}{11}
ସମାଧାନ:
(12x2, -5x2), (-3x, \frac{x}{7}), (\frac{1}{\sqrt{2}} x3, √3x3), (15, \frac{8}{11})
ସଦୃଶ ମନୋମିଆଲ୍‌ଗୁଡ଼ିକୁ ବନ୍ଧନୀ ମଧ୍ଯରେ ରଖାଯାଇଛି ।

Question 3.
ନିମ୍ନସ୍ଥ ପ୍ରତ୍ୟେକ ପ୍ରକାର ପଲିନୋମିଆଲ୍‌ରୁ ଦୁଇଟି ଲେଖାଏଁ ଉଦାହରଣ ଦିଅ ।

(i) ଣୂନଘାତା ପଲିନୋମିଆଲ୍
ସମାଧାନ:
2, –\frac{1}{4}

(ii) ଏକପଦ ବିଶିଷ୍ଟ ଦ୍ୱିଘାତୀ ପଲିନୋମିଆଲ୍
ସମାଧାନ:
5x2, -3x2

(iii) ଦୁଇପଦ ବିଶିଷ୍ଟ ତ୍ରିଘାତୀ ପଲିନୋମିଆଲ୍
ସମାଧାନ:
2x3 – 3x2, \frac{4}{5} x3 + 5x

(iv) ତିନିପଦ ବିଶିଷ୍ଟ ଦ୍ୱିଘାତୀ ପଲିନୋମିଆଲ୍
ସମାଧାନ:
x2 – 3x + 2, 3x2 – 4x – 5
(ଦତ୍ତ ଉଦାହରଣ ବ୍ୟତୀତ ଅନ୍ୟ ଉଦାହରଣ ଦିଆଯାଇପାରେ ।)

BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a)

Question 4.
ଯାେଗ କର

(i) 2y3 – 3y – 4, 2 – y3 + 5y
ସମାଧାନ:
(2y3 – 3y –  4) + (2 – y3 + 5y)
= (2y2 – 3y –  4) + (-y3 + 5y + 2) (ପ୍ରତ୍ୟେକ ପଲିନୋମିଆଲ୍‌କୁ ଘାତାଙ୍କ କ୍ରମରେ ସଜାଇବା)
= (2y3 –  y3) + (-3y + 5y) + (- 4 + 2) (ସଦୃଶ ପଦଗୁଡ଼ିକୁ ଏକାଠି କରି)
= y3 + 2y – 2
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = y3 + 2y – 2

(ii) 3x4 – 2x3 – 5 + x – 5x2, 3x3 + 2x2 – x4 – x + 1
ସମାଧାନ:
(3x4 – 2x3 – 5 + x – 5x2) + (3x3 + 2x2 – x4 – x + 1)
= (3x4 – 2x3 – 5x2 + x – 5) + (-x4 + 3x3 + 2x2 – x + 1) (ଘାତାଙ୍କ କ୍ରମରେ ସଜାଇ)
= (3x4 – x4) + (-2x3 + 3x3) + (-5x2 + 2x2) + (x – x) + (-5 + 1) (ସଦୃଶ ପଦଗୁଡ଼ିକୁ ଏକାଠି କରି)
= 2x4 + x3 – 3x2 + 0 – 4 = 2x4 + x3 – 3x2 – 4
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = 2x4 + x3 – 3x2 – 4

(iii) \frac{3}{4} x2\frac{4}{5} x – 3, \frac{1}{4} x2 + \frac{4}{5} x + 2
ସମାଧାନ:
\left(\frac{3}{4} x^2-\frac{4}{5} x-3\right)+\left(\frac{1}{4} x^2+\frac{4}{5} x+2\right)
= \left(\frac{3}{4} x^2+\frac{1}{4} x^2\right)+\left(-\frac{4}{5} x+\frac{4}{5} x\right) + (-3 + 2) (ସଦୃଶ ପଦଗୁଡ଼ିକୁ ଏକାଠି କରି)
= \frac{4}{4} x2 + 0.x – 1 = x2 – 1
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = x2 – 1

(iv) 2.1x3 + 3.2x2 + 5 – 3x, 1.9x3 – 1.2x2 + 2x – 1
ସମାଧାନ:
(2.1x3 + 3.2x2 + 5 – 3x) + (1.9x3 – 1.2x2 + 2x – 1)
= (2.1x3 + 3.2x2 – 3x + 5) + (1.9x3 – 1.2x2 + 2x – 1)
= (2.1x3 + 1.9x3) + (3.2x2 – 1.2x2) + (-3x + 2x) + (5 – 1)
= 4x3 + 2x2 – x + 4
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = 4x3 + 2x2 – x + 4

(v) \frac{1}{2}z3\frac{3}{2} z2 + 6z, \frac{1}{2} z2\frac{1}{2} z3 – 3z – 1, z3 + 2z2 + 3z – 4
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a)

(vi) 8x – 3xy + 2xyz, 2xy – 5x + 3xyz, xy – 3x + 4xyz
ସମାଧାନ:
(8x – 3xy + 2xyz) + (2xy – 5x + 3xyz) + (xy – 3x + 4xyz)
= (2xyz – 3xy + 8x) + (3xyz + 2xy – 5x) + (4xyz + xy – 3x)
= (2xyz + 3xyz + 4xyz) + (- 3xy + 2xy + xy) + (8x- 5x – 3x)
= 9xyz + 0 + 0 = 9xyz
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = 9xyz

(vii) 5x2 – 2xy + y2, 4xy – 2y2 – 3x2, 4y2 – xy – x2
ସମାଧାନ:
(5x2 – 2xy + y2) + (4xy – 2y2 – 3x2) + (4y2 – xy + x2)
= (5x2 – 2xy + y2) + (-3x2 + 4xy – 2y2) + (-x2 – xy + 4y2)
= (5x2 – 3x2 – x2) + (-2xy + 4xy – xy) + (y2 – 2y2 + 4y2)
= x2 + xy + 3y2
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = x2 + xy + 3y2

Question 5.
ବିୟୋଗ କର :

(i) 6x3 – 13x2 + 14 ରୁ -x3 + 2x – 7x2 + 11
ସମାଧାନ:
(6x3 – 13x2 + 14) – (-x3 + 2x – 7x2 + 11)
= (6x3 – 13x2 + 14) – (-x3 – 7x2 + 2x + 11) (ଘାତାଙ୍କ କ୍ରମରେ ଲେଖିଲେ)
= (6x3 – 13x2 + 14) + {- (-x3 – 7x2 + 2x + 11)} [ a – b = a + (- b)]
= (6x3 – 13x2 + 14) + (x3 + 7x2 – 2x – 11)
= 6x3 + x3 – 13x2 + 7x2 – 2x + 14 – 11 (ସଦୃଶ ପଦଗୁଡ଼ିକୁ ଏକାଠି କରି ସଜାଇ ଲେଖିଲେ)
= 7x3 – 6x2 – 2x + 3
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = 7x3 – 6x2 – 2x + 3
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 1

(ii) t4 – 11 + 2t2 – t3 ରୁ 2t3 – 8t2 – 10
ସମାଧାନ:
(t4 – 11 + 2t2 – t3) – (2t3 – 8t2 – 10)
= (t4 – t3 + 2t2 – 11) – (2t3 – 8t2 – 10)
= (t4 – t3 + 2t2 – 11) + {-(2t3 – 8t2 – 10)}
= (t4 – t3 + 2t2 – 11) + (-2t3 + 8t2 + 10)
= t4 – t3 – 2t3 + 2t2 + 8t2 – 11 + 10 = t4 – 3t3 + 10t2 – 1
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = t4 – 3t3 + 10t2 – 1

(iii) \frac{12}{13} y2\frac{5}{13} y3 – 15 ରୁ – \frac{1}{13} y2 + \frac{8}{13} y3 + 20
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 2

(iv) 2.5x3 – 7 – 3.5x2 ରୁ 2.5x2 + 1.5x3 + 8 – 2x
ସମାଧାନ:
(2.5x3 – 7 – 3.5x2) – (2.5x2 + 1.5x3 + 8 – 2x)
= (2.5x3 – 3.5x2 – 7) – (1.5x3 + 2.5x2 – 2x + 8)
= (2.5x3 – 3.5x2 – 7) + {-(1.5x3 + 2.5x2 – 2x + 8)}
= (2.5x3 – 3.5x2 – 7) + (-1.5x3 – 2.5x2 + 2x – 8)
= 2.5x3 – 1.5x3 – 3.5x2 – 2.5x2 + 2x – 7 – 8
= x3 – 6x2 + 2x – 15
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = x3 – 6x2 + 2x – 15

(v) x2 – 2xy + 3y2 ରୁ 2x2 – xy – 2y2
ସମାଧାନ:
(x2 – 2xy + 3y2) – (2x2 – xy – 2y2)
= (x2 – 2xy + 3y2) + {-(2x2 – xy – 2y2)}
= (x2 – 2xy + 3y2) + (-2x2 + xy + 2y2)
= x2 – 2x2 – 2xy + xy + 3y2 + 2y2 = -x2 – xy + 5y2
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = -x2 – xy + 5y2

(vi) 2x2 – 3xy – 4xy2 ରୁ x2 – xy – 2xy2
ସମାଧାନ:
(2x2 – 3xy- 4xy2) – (x2 – xy – 2xy2)
= (2x2 – 3xy – 4xy2) + {-(x2 – xy – 2xy2)}
= (2x2 – 3xy – 4xy2) + (-x2 + xy + 2xy2)
= 2x2 – x2 – 3xy + xy – 4xy2 + 2xy2 = x2 – 2xy – 2xy2
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = x2 – 2xy – 2xy2

(vii) a – 3b + 2c ରୁ 3b – 7c + 2a
ସମାଧାନ:
(a – 3b + 2c) – (3b – 7c + 2a) = (a – 3b + 2c) – (2a + 3b – 7c)
= (a – 3b + 2c) + {-(2a + 3b – 7c)} = (a – 3b + 2c) + (-2a – 3b + 7c)
= a – 2a – 3b – 3b + 2c + 7c = -a – 6b + 9c
∴ ନିର୍ଣ୍ଣେୟ ଯୋଗଫଳ = -a – 6b + 9c

(viii) \frac{1}{2} a + \frac{2}{3} b – \frac{3}{2} c ରୁ a – \frac{1}{3} b + \frac{1}{2} c
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 3

Question 6.
ନିମ୍ନରେ ଦତ୍ତ ପଲିନୋମିଆଲ୍‌ଗୁଡ଼ିକର ଗୁଣଫଳ ସ୍ଥିର କରି ଗୁଣଫଳର ଘାତ ନିରୂପଣ କର ।

(i) 2x2 – 3x + 5 ଓ x2 + 8x + 2
ସମାଧାନ:
(2x2 – 3x + 5)(x2 + 5x + 2)
= 2x2(x2 + 5x + 2) -3x(x2 + 5x + 2) + 5(x2 + 5x + 2) ( ବଣ୍ଠନ ନିୟମ)
= 2x4 + 10x3 + 4x2 – 3x3 – 15x2 – 6x + 5x2 + 25x + 10 (ପୁନଃବଣ୍ଟନ ନିୟମ)
= 2x4 + (10x3 – 3x3) + (4x2 – 15x2 + 5x2) + (-6x + 25x) + 10 (ସଦୃଶ ପଦଗୁଡ଼ିକୁ ଏକାଠି ରଖୁ)
= 2x4 + 7x3 – 6x2 + 19x + 10
ଗୁଣଫଳର ଘାତ = 4 ।
∴ ଯଦି p(x) ଓ q(x) ଦୁଇଟି ପଲିନୋମିଆଲ୍ ହୁଏ, ତେବେ {p(x) × g(x)} ର ଘାତ = p(x)ର ଘାତ + q(x)ର ଘାତ ।
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 4

(ii) y3 – 5y2 + 11y ଓ y5 – 20y4 + 17
ସମାଧାନ:
(y3 – 5y2 + 11y)(y5 – 20y4 + 17)
= y3(y5 – 20y4 + 17) – 5y2(y5 – 20y4 + 17) + 11y(y5 – 20y4 + 17)
=y8 – 20y7 + 17y3 – 5y7 + 100y6 – 85y2 + 11y6 – 220y5 + 187y
=y8 + (-20y7 – 5y7) + (100y6 + 11y6) – 220y5 + 17y3 – 85y2 + 187y
= y8 – 25y7 + 111y6 – 220y5 + 17y3 – 85y2 + 187y
ଗୁଣଫଳର ଘାତ = 8 ।

(iii) (2x + 3) ଓ 5x2 – 7x + 8
ସମାଧାନ:
(2x + 3)(5x2 – 7x + 8)
= 2x(5x2 – 7x + 8) + 3(5x2 – 7x + 8) = 10x3 – 14x2 + 16x + 15x2 – 21x + 24
= 10x3 + (-14x2 + 15x2) + (16x – 21x) + 24 = 10x3 + x2 – 5x + 24
ଗୁଣଫଳର ଘାତ = 3 ।

BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a)

(iv) (x – 1), (7x – 9) ଓ 3x3 – 14x2 + 8
ସମାଧାନ:
(x – 1)(7x – 9)(3x3 – 14x2 + 8)
= {(x – 1)(7x – 9)} (3x3 – 14x2 + 8) = {x(7x – 9) – 1(7x – 9)} (3x3 – 14x2 + 8)
= (7x2 – 9x – 7x + 9)(3x3 – 14x2 + 8) = (7x2 – 16x + 9)(3x3 – 14x2 + 8)
= 7x2 (3x3 – 14x2 + 8) – 16x(3x3 – 14x2 + 8) + 9(3x3 – 14x2 + 8)
= 21x5 – 98x4 + 56x2 – 48x4 + 224x3 – 128x + 27x3 – 126x2 + 72
= 21x5 + (- 98x4 – 48x4) + (224x3 + 27x3) + (56x2 – 126x2) – 128x + 72
= 21x5 – 146x4 + 251x3 – 70x2 – 128x + 72
ଗୁଣଫଳର ଘାତ = 5 ।

(v) (x2 + y2) ଓ (x4 – x2y2 + y4)
ସମାଧାନ:
(x2 + y2)(x4 – x2y2 + y4)
= x2 (x4 – x2y2 + y4) + y2 (x4 – x2y2 + y4) = x6 – x4y2 + x7y4 + x4y2 – x2y4 + y6
= x6 + (-xy + x4y2) + (x2y4 – x2y4) + y6 = x6 + 0 + 0 + y6 = x6 + y6
ଗୁଣଫଳର ଘାତ = 6 ।

(vi) (2x + 3y), (2x – 3y) ଓ (4x2 + 9y2)
ସମାଧାନ:
(2x + 3y)(2x – 3y)(4x2 + 9y2) = {(2x + 3y)(2x – 3y)} (4x2 + 9y2)
= {2x(2x – 3y) + 3y(2x – 3y)} (4x2 + 9y2)
= (4x2 – 6xy + 6xy – 9y2)(4x2 + 9y2) = (4x2 – 9y2)(4x2 + 9y2)
= 4x2 (4x2 + 9y2) – 9y2 (4x2 + 9y2) = 16x4 + 36x2y2 – 36x2y2 – 81y4
= 16x4 + 0 – 81y4 = 16x4 – 81y4
ଗୁଣଫଳର ଘାତ = 4 ।

Question 7.
ଭାଗଫଳ ଓ ଭାଗଶେଷ ନିରୂପଣ କର ।

(i) (x3 – 1) + (x – 1)
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 5

(ii) (-81y2 + 64) + (8 – 9y)
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 6

(iii) (2x3 – 7x2 – x + 2) + (x2 – 3x – 2)
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 7

(iv) (x3 – 14x2 + 37x – 26) + (x – 2)
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 8

(v) (t3 – 6t2 + 11t – 6) + (t2 – 5t + 6)
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 9

(vi) (8a2 – 34ab + 21b2) + (4a + 3b)
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 10

(vii) (16xy2 – 21x2y + 9X3 – 4y3) + (x – y)
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 11

(viii) (x4 + x2y2 + y4) + (x2 – xy + y2)
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a) 12

ଭାଜ୍ୟ = ଭାଜକ x ଭାଗଫଳ + ଭାଗଶେଷ Euclidian Algorithmର ପ୍ରୟୋଗରେ ଭାରକ୍ରିୟାର ସତ୍ୟତା ନିରୂପଣ କରାଯାଇଥାଏ ।

BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(a)

Question 8.
ଯଦି p(x) = 3x3 – 6x2 + 2 ଏବଂ q(x) = 2x2 – 5x + 1 ତେବେ

(i) 2p(x) – 5q(x)
ସମାଧାନ:
ଯଦି p(x) = 3x3 – 6x2 + 2 ଏବଂ q(x) = 2x2 – 5x + 1
2p(x) – 5q(x) = 2(3x3 – 6x2 + 2) – 5(2x2 – 5x + 1)
= 6x3 – 12x2 + 4 – 10x2 + 25x – 5
= 6x3 – 12x2 – 10x2 + 25x + 4 – 5 = 6x3 – 22x2 + 25x – 1

(ii) 4p(x) + 3q(x) ବମାନ ସ୍ଥିର କରା
ସମାଧାନ:
ଯଦି p(x) = 3x3 – 6x2 + 2 ଏବଂ q(x) = 2x2 – 5x + 1
4p(x) + 3q(x) = 4(3x3 – 6x2 + 2) + 3(2x2 – 5x + 1)
= 12x3 – 24x2 + 8 + 6x2 – 15x + 3
= 12x3 – 24x2 + 6x2 – 15x + 8 + 3 = 12x3 – 18x2 – 15x + 11

Question 9.
ଯଦି p(x) = 2x3 + 3x + 5 ଏବଂ q(x) = x2 + 4x + 1 ଓ r(x) = x – 1 ହୁଏ ତେବେ ଦର୍ଣ।ଅ ଯେ,

(i) p(x) × q(x) = q(x) × p(x)
ସମାଧାନ:
ଯଦି p(x) = 2x3 + 3x + 5 ଏବଂ q(x) = x2 + 4x + 1 ଓ r(x) = x – 1
ବାମପାର୍ଶ୍ଵ = p(x) × q(x) = (2x3 + 3x + 5)(x2 + 4x + 1)
= 2x3 (x2 + 4x + 1) + 3x(x2 + 4x + 1) + 5(x2 + 4x + 1)
= 2x5 + 8x4 + 2x3 + 3x3 + 12x2 + 3x + 5x2 + 20x + 5
= 2x5 + 8x4 + 5x3 + 12x2 + 5x2 + 3x + 20x + 5
= 2x5 + 8x4 + 5x3 + 17x2 + 23x + 5
ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ = q(x) × p(x) = (x2 + 4x + 1)(2x3 + 3x + 5)
= x2 (2x3 + 3x + 5) + 4x (2x3 + 3x + 5) + 1(2x3 + 3x + 5)
= 2x5 + 3x3 + 5x2 + 8x4 + 12x2 + 20x + 2x3 + 3x + 5
= 2x5 + 8x4 + 3x3 + 2x3 + 5x2 + 12x2 + 20x + 3x + 5
= 2x5 + 8x4 + 5x3 + 17x2 + 23x + 5 (ପଲିନୋମିଆଲ୍ ଦ୍ଵୟର ଗୁଣନ କ୍ରମବିନିମୟୀ ।)
ବାମପାର୍ଶ୍ଵ = ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ (ପ୍ରମାଣିତ)

(ii) p(x) × {(q(x) + r(x)} = p(x) . q(x) + p(x) . r(x)
ସମାଧାନ:
ବାମପାର୍ଶ୍ଵ = P(x) × {q(x) + r(x)}
= (2x3 + 3x + 5) x {(x2 + 4x + 1) + (x – 1)} = (2x3 + 3x + 5) + (x2 + 4x + 1 + x – 1)
= (2x3 + 3x + 5)(x2 + 5x) = 2x3 (x2 + 5x) + 3x(x2 + 5x) + 5(x2 + 5x)
= 2x5 + 10x4 + 3x3 + 15x2 + 5x2 + 25x = 2x5 + 10x4 + 3x3 + 20x2 + 25x
ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ = P(x) . q(x) + p(x) . r(x)
= (2x3 + 3x + 5)(x2 + 4x + 1) + (2x3 + 3x + 5)(x – 1)
= 2x3 (x2 + 4x + 1) + 3x(x2 + 4x + 1) + 5(x2 + 4x + 1)+ 2x3 (x – 1) + 3x(x – 1) + 5(x – 1)
= 2x5 + 8x4 + 2x3 + 3x3 + 12x2 + 3x + 5x2 + 20x + 5 + 2x4 – 2x3 + 3x2 – 3x + 5x – 5
= 2x5 + 8x4 + 2x4 + 2x3 + 3x3 – 2x3 + 12x2 + 5x2 + 3x2 + 3x + 20x – 3x + 5x + 5 – 5
= 2x5 + 10x4 + 3x3 + 20x2 + 25x
ବାମପାର୍ଶ୍ଵ = ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ (ପ୍ରମାଣିତ)
(ପଲିନୋମିଆଲ୍ ଦ୍ଵୟର ଗୁଣନ କ୍ରମବିନିମୟୀ ।)

Question 10.
ସରଳ କର :

(i) (x2 – 3x + 5) + (2x2 – x – 2) – (3x2 + 7x – 3)
ସମାଧାନ:
(x2 – 3x + 5) + (2x2 – x – 2) – (3x2 + 7x – 3)
= x2 – 3x + 5 + 2x2 – x – 2 – 3x2 – 7x + 3
= x2 + 2x2 – 3x2 – 3x – x – 7x + 5 – 2 + 3
= 0 – 11x + 6 = -11x + 6

(ii) (x2 – xy + 2y2) – (2x2 + 4xy + 3y2) + (4x2 – 2xy – y2)
ସମାଧାନ:
(x2 – xy + 2y2) – (2x2 + 4xy + 3y2) + (4x2 – 2xy – y2)
= x2 – xy + 2y2 – 2x2 – 4xy – 3y2 + 4x2 – 2xy – y2
= x2 – 2x2 + 4x2 – xy – 4xy – 2xy + 2y2 – 3y2 – y2
= 3x2 – 7xy – 2y2

(iii) (x + b + c) (a – b + c) – (a + b – c) (a – b – c)
ସମାଧାନ:
(a + b + c)(a – b + c) – (a + b – c)(a – b – c)
= [a(a – b + c) + b(a – b + c) + c(a – b + c)] – [a(a – b – c) + b(a – b – c) – c(a – b – c)]
= (a2 – ab + ca + ab – b2 + bc + ca –  bc + c2) – (a2 – ab – ac + ab – b2 – bc – ca + bc + c2)
= (a2 – b2 + c2 – ab + ab + ca + ca + bc – bc) – (a2 – b2 + c2 – ab + ab – ac – ac – bc + bc)
= (a2 – b2 + c2 + 2ca) – (a2 – b2 + c2 – 2ca) = a2 – b2 + c2 + 2ca – a2 + b2 – c2 + 2ca
= a2 – a2 – b2 + b2 + c2 – c2 + 2ca + 2ca = 4ca

The Thief and the Tiger Question Answer Class 8 English Chapter 2 BSE Odisha

Odisha State Board BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger Textbook Exercise Questions and Answers.

Class 8th English Chapter 2 The Thief and the Tiger Question Answers BSE Odisha

The Thief and the Tiger Class 8 Questions and Answers

Session – 1
Pre-Reading (ପ୍ରାକ୍-ପଠନ):

→ Man rides a horse or an elephant or a donkey. Does he ever ride a lion or a tiger or a bear? Why? See the picture below. What do you see? A man is riding a tiger. Is it possible? When? Where? Let’s read the story and see how.
Pre reading
ଲୋକଟି ଗୋଟିଏ ଘୋଡ଼ା, କିମ୍ବା ହାତୀ କିମ୍ବା ଗଧ ଉପରେ ଚଢ଼ିଛି । ସେ କେବେ ସିଂହ, କିମ୍ବା ବାଘ କିମ୍ବା ଭାଲୁ ଉପରେ ଚଢ଼ିଥାଏ କି ? କାହିଁକି ? ତଳେ ଥ‌ିବା ଚିତ୍ରଟିକୁ ଦେଖ । କ’ଣ ଦେଖୁଛ ? ଗୋଟିଏ ଲୋକ ବାଘ ଉପରେ ଚଢ଼ିଛି କେତେବେଳେ ? କେଉଁଠି ? ଆସ ଗପଟି ପଢ଼ି ସବୁକଥା ଜାଣିବା ।

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

II. While Reading

Text

  • SGP – I
  • Read paragraphs 1-3 silently and answer the questions that follow.
    (୧ମରୁ ୩ୟ ଅନୁଚ୍ଛେଦ ପର୍ଯ୍ୟନ୍ତ ପାଠକରି ନୀରବରେ ପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

1. Once …………………………………………… in the stable.
ଥରେ ଜଣେ ରାଜା ଥିଲେ । ସେ ତାଙ୍କର ବଳବାନ ଓ ଦ୍ରୁତଗାମୀ ଅଶୁମାନଙ୍କ ପାଇଁ ଖୁବ୍ ପ୍ରସିଦ୍ଧ ଥିଲେ । ସେ ସବୁଠାରୁ ଦକ୍ଷ ଶକ୍ତିଶାଳୀ ଘୋଡ଼ାମାନଙ୍କୁ ଆଣି ତାଙ୍କ ଘୋଡ଼ାଶାଳରେ ରଖୁଥିଲେ । ଦିନେ ଗୋଟିଏ ଚୋର ତାଙ୍କର ଗୋଟିଏ ଘୋଡ଼ାକୁ ଚୋରି କରିନେବାକୁ ଇଚ୍ଛା ପ୍ରକାଶ କଲା । ଗୋଟିଏ ବାଘ ଚୋରଟିର ଏ ଯୋଜନା ବିଷୟରେ ଜାଣିବାକୁ ପାଇଲା । ସେ ସେହି ଚୋରଟିର ମାଂସ ଖାଇବାକୁ ଚିନ୍ତାକଲା । ଏଣୁ ସେଇ ରାତିରେ ବାଘଟି ରାଜାଙ୍କ ଘୋଡ଼ାଶାଳରେ ଏକ ନିରାପଦ ସ୍ଥାନରେ ଛପିରହିଲା । ବାଘଟି ଘୋଡ଼ାମାନଙ୍କ ସହ ନୀରବରେ ଶାନ୍ତ ଅବସ୍ଥାରେ ଛିଡ଼ା ହୋଇ ରହିଲା । ସତେଯେମିତି ସେ ଘୋଡ଼ାଶାଳର ଅନ୍ୟ ଘୋଡ଼ାମାନଙ୍କ ପରି ଗୋଟିଏ ଘୋଡ଼ା ।

2. After sometime …………………………………………………… rode on it.
କିଛି ସମୟ ପରେ ଚୋରଟି ଘୋଡ଼ାଶାଳରେ ପ୍ରବେଶକଲା । ଘୋଡ଼ାଶାଳଟି ଅନ୍ଧାର ଥିଲା । ସେ ପ୍ରତ୍ୟେକ ଘୋଡ଼ାର ପିଠିରେ ହାତମାରି କେଉଁଟି ସବୁଠୁ ଶକ୍ତିଶାଳୀ ଘୋଡ଼ା ତାହା ପରୀକ୍ଷା କରିବାକୁ ଲାଗିଲା । ଶେଷରେ ସେ ବାଘର ପିଠିରେ ହାତ ପକାଇଲା । ଏଇଟି ସବୁଠୁ ଭଲ ଘୋଡ଼ା ବୋଲି ଚିନ୍ତାକଲା ଏବଂ ସେଇଟିକୁ ଘୋଡ଼ାଶାଳର ବାହାରକୁ ଆଣିଲା । ତା’ପରେ ସେ ବାଘର ପାଟିରେ ଲଗାମ ବାନ୍ଧିଲା ଏବଂ ତା’ଉପରେ ଚଢ଼ିଗଲା ।

3. The tiger ……………………………………… anything in the dark.
ବାଘର ଆଗରୁ ଏପରି କୌଣସି ଅଭିଜ୍ଞତା ନଥିଲା । ସେ ଚୋରକୁ ଅତି ଶକ୍ତିଶାଳୀ ବ୍ୟକ୍ତି ବୋଲି ଭାବିଲା । ସେ ଚୋରଟିକୁ ଭୟଙ୍କର ଭାବରେ ଡରିଯାଇଥିଲା । ଏଣୁ ଅତି ଦ୍ରୁତଗତିରେ ଦୌଡ଼ିବାକୁ ଲାଗିଲା । ଚୋରଟିର ମଧ୍ୟ ଏପରି ଏକ ପ୍ରାଣୀ ଉପରେ ଚଢ଼ିବାର ଅଭିଜ୍ଞତା ନଥିଲା । ସେ ଏହାକୁ ଅତି ଶକ୍ତିଶାଳୀ ଘୋଡ଼ା ବାଛି ଆଣିପାରିଛି ବୋଲି ଚିନ୍ତାକଲା । ସେ ଜାଣି ନଥିଲା ଯେ ସେ ଏକ ବାଘ ଉପରେ ଚଢ଼ିଛି କାରଣ ସ୍ଥାନଟି ସମ୍ପୂର୍ଣ୍ଣ ଅନ୍ଧକାର ଥିଲା ।

Word Meaning

famous : well-known to many people I renowned (ପ୍ରସିଦ୍ଧ, ବିଖ୍ୟାତ)
swift : fast moving (କ୍ଷିପ୍ର / ଶୀଘ୍ର ଗତି କରୁଥିବା)
stable : house for horses (ଘୋଡ଼ାଶାଳ)
steal : the act of taking something from someone unlawfully (ଚୋରିକରିବା)
plan : design / scheme (ଯୋଜନା)
flesh : meat (ମାଂସ)
hid : ଲୁଚାଇ

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

silently : without speaking / without making a sound (ନୀରବରେ )
touched: be in direct physical contact (ଛୁଇଁବା)
bridle: leather band put on the head of a horse to control its movement/reins
experience: the accumulation of knowledge or skills that results
from direct participation in events or activities (ଅନୁଭୂତି)
powerful : having great influence (କ୍ଷମତାଶାଳୀ / ବଳଶାଳୀ)
terribly : horribly / causing fear (ଭୟଙ୍କର ଭାବରେ)
imagine : fancy / think / suppose (କଳ୍ପନାକରିବା)

Comprehension Questions and Answers: (ସଂକ୍ଷିପ୍ତ ପ୍ରଶ୍ନୋତ୍ତର)

Question 1.
What was the king famous for?
(ରାଜା କେଉଁଥୂପାଇଁ ପ୍ରସିଦ୍ଧ ଥିଲେ ?)
Answer:
The king was famous for his strong and swift horses.

Question 2.
What did the thief plan to do?
(ଚୋରଟି କ’ଣ କରିବାପାଇଁ ଯୋଜନା କଲା ?)
Answer:
The thief wanted to steal a horse.

Question 3.
Why did the tiger hide in the stable?
(ବାଘଟି ଘୋଡ଼ାଶାଳରେ ଛପିରହିଥିଲା କାହିଁକି ?)
Answer:
The tiger came to know about the thief’s plan and thought of eating the thief s flesh.

Question 4.
Why did the thief touch the back of each horse?
(ଚୋରଟି କାହିଁକି ପ୍ରତି ଘୋଡ଼ା ପିଠିରେ ହାତ ମାରୁଥିଲା ?)
Answer:
The thief touched the back of each horse to steal the best one.

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Question 5.
Why did he think the tiger to be the best horse ?
(ସେ ବାଘକୁ କାହିଁକି ସର୍ବଶ୍ରେଷ୍ଠ ଘୋଡ଼ା ବୋଲି ଭାବୁଥିଲା ?)
Answer:
He thought the tiger was the best horse because he felt the tiger’s back was different from the other horses.

Question 6.
How did he ride on it?
(ସେ କିପରି ଏହା ଉପରେ ଚଢ଼ିଲା ?)
Answer:
He rode on the tiger putting a bridle on its mouth thinking it was a horse.

Question 7.
He did not know that he was riding a tiger. Why?
(ସେ ଜାଣି ନଥୁଲା ଯେ ସେ ଏକ ବାଘ ଉପରେ ଚଢ଼ିଥିଲା, ଏହାର କାରଣ କ’ଣ ?)
Answer:
He didn’t know that he was riding a tiger, because there was darkness inside the stable.

Question 8.
Where did the tiger run into?
(ବାଘ କେଉଁଠାକୁ ଦୌଡ଼ି ପଳାଇଲା ।)
Answer:
The tiger ran into the forest.

Session – 2

  • SGP – 2
  • Read paragraph 4 silently and answer the questions that follow.
    (ଚତୁର୍ଥ ଅନୁଚ୍ଛେଦଟି ପାଠକରି (ନୀରବରେ) ପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

4. The day…………………………………… as it could.
ରାତି ପାହିଲା । ଅନ୍ଧକାର ଦୂରୀଭୂତ ହେଲା । ସେତେବେଳେ କେବଳ ସେ ଜାଣିପାରିଲା ଯେ ସେ ଭୁଲବଶତଃ ଗୋଟିଏ ବାଘ ଉପରେ ଚଢ଼ିଛି । ସେ ଭୟଭୀତ ହୋଇପଡ଼ିଲା । ବାଘଟି ମଧ୍ୟ ଭୟଭୀତ ହୋଇ ଅଧିକ ଦ୍ରୁତଗତିରେ ଦୌଡୁଥାଏ । ଚୋରଟି କ’ଣ କରିବ, ସ୍ଥିର କରିପାରିଲା ନାହିଁ । ସେ ପ୍ରାୟ ଅଚେତ ହେବାଭଳି ଅନୁଭବ କଲା । ଦୌଡୁଥିବାବେଳେ ବାଘ ଗୋଟିଏ ଗଛଦେଇ ଦୌଡ଼ୁଥିଲା । ଏହି ସମୟରେ ଚୋରଟି ଗଛର ଡାଳକୁ ଧରି ଗଛ ଉପରେ ଚଢ଼ିଗଲା । ବାଘ ଖୁସି ଅନୁଭବ କଲା । କାରଣ ସେ ଲୋକଟି ପାଇଁ ସେ ଅତ୍ୟଧ୍ୱ ଭୟଭୀତ ହୋଇପଡ଼ିଥିଲା । ପାରୁପର୍ଯ୍ୟନ୍ତ ଦ୍ରୁତ ବେଗରେ ବାଘଟି ବଣମଧ୍ୟକୁ ଦୌଡ଼ିଲା ।

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Word Meaning

dawn : day break / beginning of the day (ପ୍ରଭାତ / ଉଷା)
disappear : vanish from sight (ଅନ୍ତର୍ଦ୍ଧାନ ହୋଇଯିବା / ଅଦୃଶ୍ୟ ହୋଇଯିବା)
frightened : scared / intense fear (ଭୟଭୀତ କରାଇବା)
fast : swift (ଦୃତ / ଦ୍ରୁତଗାମୀ)
faint : to feel weak and lose consciousness (ଅଚେତ ହୋଇଯିବା)
passed : to cross (ଅତିକ୍ରମ କରିବା)
branch : part of a tree (ଶାଖା)
climb : ascend (ଚଢ଼ିବା)

Comprehension Questions and Answers: (ସଂକ୍ଷିପ୍ତ ପ୍ରଶ୍ନୋତ୍ତର)

Question 1.
When did the thief come to know that he was riding a tiger?
(ଚୋରଟି କେତେବେଳେ ଜାଣିପାରିଲା ଯେ ସେ ଏକ ବାଘ ଉପରେ ଚଢ଼ିଛି ?)
Answer:
When the day dawned and darkness disappeared the thief came to know that he was riding a tiger.

Question 2.
How did he save himself?
(ସେ କିପରି ନିଜକୁ ରକ୍ଷାକଲା ?)
Answer:
When the tiger ran by a tree beside the road the thief caught hold of one of the branches of the tree and climbed up.

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Question 3.
Why was the tiger happy?
(ବାଘ କାହିଁକି ଖୁସି ହୋଇଗଲା ?)
Answer:
The tiger was terribly afraid of the rider. So when the rider climbed up the tree he was happy.

  • SGP – 3 (Text book page No. 34)
  • Read paragraphs 5 – 8 silently and answer the questions that follow.
    (୫ମ ଠାରୁ ଅଷ୍ଟମ ଅନୁଚ୍ଛେଦ ପର୍ଯ୍ୟନ୍ତ ନୀରବରେ ପାଠକରି ପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

5. After…………………………………………. dead body.
କିଛି ସମୟ ପରେ ଚୋରଟି ଗଛ ଉପରୁ ଓହ୍ଲାଇ ଗଛତଳେ ବିଶ୍ରାମ ନେଲା । ଅତ୍ୟଧ‌ିକ ଭୟ ଓ କ୍ଳାନ୍ତ ଅନୁଭବ କରିଥିବାରୁ ସେ ସେଠାରେ ଶୋଇପଡିଲା । ଗଭୀର ନିଦ୍ରା ଯୋଗୁଁ ସେ ଗୋଟିଏ ମୃତ ଶବଭଳି ଜଣାପଡୁଥିଲା । ସେହି ବାଟଦେଇ ଗୋଟିଏ ଗଧ୍ଵ ଯାଉଥିଲା । ସେ ଲୋକଟିକୁ ଏକ ମୃତ ଶବ ଭାବି ତା’ର ମାଂସ ଖାଇବାକୁ ଇଚ୍ଛାକଲା । ସେ ମନକୁମନ କହିଲା, ମୁଁ କି ଭାଗ୍ୟବାନ, ଏଇ ଶବଟି ମୋର ସପ୍ତାହେରୁ ଅଧିକ କାଳ ଖାଦ୍ୟ ହୋଇପାରିବ । କିନ୍ତୁ କେହିଜଣେ ଏହାକୁ ବଣଭିତରକୁ ଟାଣିନେବାପାଇଁ ମୋତେ ସାହାଯ୍ୟ କରିବା ଦରକାର ।

6. Thinking so, ………………………………….. with a rope”.
ଏଇକଥା ଚିନ୍ତା କରି ଗଧ୍ଵ ଆଉ ଅନ୍ୟ ଗୋଟିଏ ପ୍ରାଣୀର ସାହାଯ୍ୟ ପାଇଁ ବଣ ଭିତରକୁ ଗଲା ଏବଂ ସେଇ ବାଘଟି ସହିତ ତା’ର ସାକ୍ଷାତ ହେଲା । ସେ ବାଘକୁ କହିଲା, ‘ବାଘ ମହାଶୟ’ ଗୋଟିଏ ମଣିଷର ଶବକୁ ଟାଣି ଆଣିବାରେ ମୋତେ ସାହାଯ୍ୟ କରିବ କି ? ମୁଁ ଏହାର ଅଧା ତୁମକୁ ଦେବି । ‘ଗତ ରାତିରେ ଘଟିଥିବା ଘଟଣାର ଅନୁଭୂତିରୁ ବାଘ କହିଲା, ‘ତୁମେ ମୋତେ ଠକିଦେବ ନାହିଁ ତ ? ମୋତେ ଏକା ଛାଡ଼ିଦେଇ ତୁମେ ଦୌଡ଼ି ପଳାଇଯିବନି ତ ? ‘ଗଧ୍ଵ କହିଲା, ‘ଆମେ ଦୁହେଁ ଦିହିଁଙ୍କୁ ଗୋଟିଏ ଦଉଡ଼ିରେ ବାନ୍ଧିଦେବା, ତା’ହେଲେ କେହି କାହାକୁ ଛାଡ଼ି ପଳାଇ ଯାଇପାରିବା ନାହିଁ ।’’
SGP 3

7. The wolf ……………………………………. wolf died.
ଗଧ୍ଵ ଓ ବାଘ ଦୁହେଁ ଦୁହିଁଙ୍କୁ ଗୋଟିଏ ଦଉଡ଼ିରେ ବାନ୍ଧି ସନ୍ତର୍ପଣରେ ଶବ ପାଖକୁ ଚାଲିଲେ । ଚୋରଟି ନିଦରୁ ଉଠିଯାଇଥିଲା । ସେ ଏମାନଙ୍କ ଆସିବା ଜାଣିପାରି ଭୟରେ ଚିତ୍କାର କରି କହିଲା, ‘ହଇରେ ବାଘ ତୁ ପୁଣି ଆସିଛୁ ?’’ ଏବେ ବାଘଟି ସେ ଚୋରକୁ ପୁଣି ଦେଖ୍ ଭୟଭୀତ ହୋଇପଡ଼ିଲା ଏବଂ ଗଧୂକୁ ଘୋଷାରିନେଇ ଯେତେ ପାରେ ସେତେ ଜୋର୍‌ରେ ଦଉଡ଼ିବାକୁ ଲାଗିଲା । ବିଚରା ଗଧୂଟି ବାଟରେ ମରିଗଲା ।

8. Since …………………………………………………… that day.
ସେବେଠୁ ବାଘ ଆଉ ମଣିଷ ମାଂସ ନ ଖାଇବା ପାଇଁ ପ୍ରତିଜ୍ଞାକଲା । ଚୋରଟି ମଧ୍ୟ ରକ୍ଷା ପାଇଯାଇଥିବାରୁ ଖୁସି ହେଲା ଏବଂ ସେ ମଧ୍ୟ ଆଉ ଚୋରି ନ କରିବାପାଇଁ ପ୍ରତିଜ୍ଞାକଲା ।

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Word Meaning

tired: of strength or energy (କ୍ଳାନ୍ତ)
wolf: a wild carnivorous animal (ଗଧୂ)
luck : destiny/fate / fortune (ଭାଗ୍ୟ)
last : continue to stay (ଚାଲୁ ରଖୁବା | ତିଷ୍ଠିବା)
drag : to pull by force (ଘୋଷାରିଘୋଷାରି ଟାଣିନେବା)
suspicious: a doubtful condition, feeling that something is wrong (ସନ୍ଦେହପରାୟଣ )
cheat: betray/deceive (ଠକିବା)
leave: to give up/abandon (ଛାଡ଼ିବା)
alone : lonely (ଏକୁଟିଆ)
suggest : to give opinion (ମତାମତ ଦେବା)
tie : to tag/join (ବାନ୍ଧିବା / ସଂଯୁକ୍ତ କରିବା)
rope : twisted cord (ଦଉଡ଼ି / ରଜ୍ଜୁ)
awake: rise/get up (ଉଠିପଡ଼ିବା)
footsteps: found of feet/sound, of a person walking (ପାଦ ଶବ୍ଦ)
promised : to make a promise / assure (ପ୍ରତିଜ୍ଞାକରିବା)
desire: want/like to get (ଇଚ୍ଛା).
give up: to avoid/abandon (ଛାଡ଼ିଦେବା )
stealing: to take others without knowledge (ଚୋରି)

Comprehension Questions and Answers: (ସଂକ୍ଷିପ୍ତ ପ୍ରଶ୍ନୋତ୍ତର)

Question 1.
Why did the thief fall fast asleep?
(ଚୋରଟି ଗଭୀର ନିଦ୍ରାରେ ଶୋଇଁପଡ଼ିଲା କାହିଁକି ?)
Answer:
The thief fell fast asleep because he was afraid and tired.

Question 2.
Who saw him ? What was his plan ?
(କିଏ ତାକୁ ଦେଖୁଲା ? କ’ଣ ସେ ଚିନ୍ତାକଲା ?)
Answer:
A wolf saw him. He thought he as dead and planned to use him as his food for more than a week long.

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Question 3.
What did he want the tiger to do?
(ସେ ବାଘକୁ କଣ କରିବାପାଇଁ ଚାହିଁଲା ?)
Answer:
He wanted the tiger to help him to drag the dead body into the forest.

Question 4.
What did the tiger say?
(ବାଘ କ’ଣ କହିଲା ?)
Answer:
The tiger asked him if he was going to cheat him or not.

Question 5.
“Won’t you run away leaving me alone ?” Who said this?
(ତୁ ମୋତେ ଏକୁଟିଆ ଛାଡ଼ି ପଳାଇଯିବୁନି ତ ? ଏକଥା କିଏ କହିଲା ?)
Answer:
“Won’t you run away leaving me alone ?” The tiger said this to the wolf.

Question 6.
What did the wolf say?
(ଗଧୂଟି କ’ଣ କହିଲା ?)
Answer:
The wolf suggested to the tiger that they would tie each other with a rope so that no one of them could run away leaving another.

Question 7.
What awoke the thief?
(ଚୋରଟି କାହିଁକି ଉଠିପଡ଼ିଲା ?)
Answer:
The footsteps of the tiger and the wolf awoke the thief.

Question 8.
Why did the tiger run away?
(ବାଘ କାହିଁକି ଦୌଡ଼ି ପଳାଇଲା ?)
Answer:
The tiger ran away frightened of the thief.

Question 9.
How did the wolf die?
(ଗଧୂ ମଲା କିପରି ?)
Answer:
The wolf died being dragged by the tiger as they tied each other with a rope.

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Question 10.
What did the tiger promise?
(ବାଘ କ’ଣ ପ୍ରତିଜ୍ଞା କଲା ?)
Answer:
The tiger promised no to desire for human flesh any more.

Question 11.
What did the thief stop doing?
(ଚୋର ସେହିଦିନଠାରୁ କ’ଣ ନ କରିବାକୁ ସ୍ଥିରକଲା ?)
Answer:
The thief stopped stealing from that day.

Session – 3
III. Post-Reading (ପଢ଼ିବା ପରେ)

1. Visual Memory Development Technique (VMDT)
Whole Text : the tiger and the thief in the stable……
(ଘୋଡ଼ାଶାଳରେ ବାଘ ଏବଂ ଚୋର…
the tiger with the thief on its back ran for life
(ବାଘ ଚୋରକୁ ପିଠିରେ ବସାଇ ଜୀବନବିକଳରେ ପଳାଇଲା)
……….. the thief and the tiger save themselves
(ଚୋର ଏବଂ ବାଘ ଉଭୟେ ନିଜ ନିଜକୁ ରକ୍ଷାକଲେ ।
……………the wolf’s plan and he died.
(ଗଧୂର ଯୋଜନା ଏବଂ ତା’ର ମୃତ୍ୟୁ)

Part Text (Para-5): “What good luck !, This dead man will last me more than a week. But someone should help me drag the dead body”.
(କି ଭାଗ୍ୟ ! ଏହି ଶବଟି ମୋର ସପ୍ତାହେରୁ ଅଧିକକାଳ ଖାଦ୍ୟ ହୋଇ ପାରିବ । କିନ୍ତୁ ଏହାକୁ ଘୋଷାରିନେବାପାଇଁ ମୋତେ କେହିଜଣେ ସାହାଯ୍ୟ କରିବା ଦରକାର ।)

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

2. Comprehension Activity : (ବୋଧ ପରିମାପକ କାର୍ଯ୍ୟାବଳୀ)

(a) Choose the correct alternatives and complete each sentences.
(ଠିକ୍ ବିକଳ୍ପ ବାଛି ବାକ୍ୟଗୁଡ଼ିକ ପୂର୍ଣ କର ।)

Question 1.
The thief put a _______________ on the tiger’s mouth.
(A) saddle
(B) bridle
(C) chain
(D) rope
Answer:
(B) bridle

Question 2.
When the tiger and the thief saw each other, _______________.
(A) only the tiger was frightened
(B) only the thief was frightened
(C) both were frightened
(D) none was frightened
Answer:
(C) both were frightened

Question 3.
“What a good luck !“ said _______________.
(A) the king
(B) the tiger
(C) the thief
(D) the wolf
Answer:
(D) the wolf

Question 4.
The wolf’s final plan was to ________ the dead body.
(A) drag
(B) bury
(C) burn
(D) eat
Answer:
(D) eat

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Question 5.
Seeing the thief, the tiger ran for life _______________ the wolf.
(A) dragging
(B) carrying
(C) leading
(D) following
Answer:
(A) dragging

(b) Given below are some sentences. They are about what happened in the story. But they are not in the right order. Fill in the boxes with correct serial numbers to rearrange the sentences. (ତଳେ କେତେକ ବାକ୍ୟ ଲେଖାହୋଇଛି । ସେଗୁଡ଼ିକ ଗଳ୍ପରେ ଘଟିଯାଇଥିବା ଘଟଣା ସମ୍ପର୍କିତ । କିନ୍ତୁ ସେଗୁଡ଼ିକ ଠିକ୍ କ୍ରମ ଅନୁସାରେ ନାହାନ୍ତି । ବାମପଟେ ଥ‌ିବା ଖାଲି ଘରଗୁଡ଼ିକରେ ଠିକ୍ କ୍ରମ ନମ୍ବର ଲେଖ୍ ବାକ୍ୟଗୁଡ଼ିକ ସଜାଅ ।)

[ ] The wolf requested the tiger to drag the man.
[ ] The thief got up.
[ ] The tiger among the horses stood silently.
[ ] Once, a thief came inside the stable to steal a horse.
[ ] It was dark everywhere.
[ ] The thief fell asleep like a dead man.
[ ] He climbed up a tree.
[ ] He thought the tiger to be the best horse.
[ ] The tiger ran for life dragging the wolf.
[ ] At night a tiger entered the stable.
[ ] The wolf and the tiger tied each other with a rope.

Answer:
[ 8 ] The wolf requested the tiger to drag the man.
[ 10] The thief got up.
[ 3 ] The tiger among the horses stood silently.
[ 1 ]Once, a thief came inside the stable to steal a horse.
[ 4 ] It was dark everywhere.
[ 7 ] The thief fell asleep like a dead man.
[ 6 ] He climbed up a tree.
[ 5 ] He thought the tiger to be the best horse.
[ 11] I The tiger ran for life dragging the wolf.
[ 2 ] At night a tiger entered the stable.
[ 9 ] The wolf and the tiger tied each other with a rope.

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Session – 4

3. Listening : (ଶ୍ରବଣ)
On the chart below, the characters in the story are written in the boxes from left to right at the top. Some words related to the characters are given in the boxes from top to bottom at the left. Your teacher will read out the words one by one. Listen to him/her carefully and put a tick (✓) in the box on the word line below the character. One is done for you.
(ତଳ ଚାର୍ଟରେ ବିଷୟର ଚରିତ୍ରଗୁଡ଼ିକ ପୃଷ୍ଠାର ଅଗ୍ରଭାଗରେ ବାମରୁ ଡାହାଣକୁ ଲେଖାଯାଇଛି । ସେମାନଙ୍କ ଚରିତ୍ର ସହିତ ଖାପଖାଉଥ‌ିବା କେତେକ ଶବ୍ଦ ଉପରୁ ତଳକୁ ବାମପଟେ ଲେଖାଯାଇଛି । ତୁମ ଶିକ୍ଷକ ସେ ଶବ୍ଦଗୁଡ଼ିକୁ ଗୋଟିଏ ପରେ ଗୋଟିଏ ପାଠକରିବେ । ଯତ୍ନର ସହିତ ତାଙ୍କ ଶବ୍ଦଗୁଡ଼ିକ ଶୁଣ ଏବଂ ଶବ୍ଦ ଧାଡ଼ି ପାଖରେ ଥ‌ିବା ଖାଲି ଘରମାନଙ୍କରେ ସେମାନଙ୍କ ଚରିତ୍ରର ଗୁଣକୁ ଖାପଖାଉଥ‌ିବା ଶବ୍ଦ ପାଖରେ ଚିହ୍ନ ଦିଅ ।)
listening
Answer:

Characters thief horse tiger wolf
Words
Strong
flesh
drag
steal
swift
speed
Search
ride
stable
run
died
branch
Forest
bridle
slept


BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

4. Speaking : (କଥନ)
• Practise the following dialogues.
(ନିମ୍ନ ସଂଳାପଗୁଡ଼ିକ ଅଭ୍ୟାସ କର ।)
• Step : (କେଉଁ କେଉଁ ସ୍ତରଦେଇ ଏହି ସଂଳାପ ପାଠ କରାଯିବ ।)
• Rehearsal-teacher reads aloud, and students listen. The teacher reads aloud and students repeat after him/her dialogue by dialogue.
• Teacher vs Students.
• Students vs students (in two groups)
(ଶିକ୍ଷକ ପାଟି କରି ପଢ଼ିବେ – ଛାତ୍ରଛାତ୍ରୀମାନେ ଶୁଣିବେ)
ଶିକ୍ଷକ ପାଟି କରି ପାଠ କରିବାପରେ ଛାତ୍ରଛାତ୍ରୀ ସେଗୁଡ଼ିକୁ ଆଉଥରେ ଦୋହରାଇ କହିବେ ।
ଶିକ୍ଷକ –
କହିବେ –
ଛାତ୍ରଛାତ୍ରୀ କହିବେ – କହିବେ ।
ପରସ୍ପର ସହିତ – (ଦୁଇଟି ଦଳରେ ବିଭକ୍ତ ହୋଇ)
(They do this reading from the text.)

Wolf : Good Luck ! This dead man will last me more than a week.
But who will help me drag the dead body ?
Tiger: What are you looking for, Mr. Wolf?
Wolf: Mr. Tiger, will you help me drag this dead body?
Tiger: Why should I?
Wolf: I’ll give you half of it. ,
Tiger: Aren’t you going to cheat me?
Wolf: No, no, not at all. How can I?
Tiger: Won’t you run away leaving me alone?
Wolf: How can that be? We’ll tie each other with a rope.
Tiger: Good idea! That’ll do.

Session – 5

5. Vocabulary : (ଶବ୍ଦଜ୍ଞାନ)
Match who lives where. Write the serial numbers in brackets. One is done for you.
Vocabulary
Answer:
Vocabulary 1

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Session  – 6

6.  Writing : (ଲିଖନ)

a. In comprehension Activity No. 2 b you have rearranged the sentences of the story. Use the sentences serially and write the story in the space given below. (ସଂପ୍ରତି କାର୍ଯ୍ୟ ପ୍ରକ୍ରିୟା No. 2(b)ରେ ଗଛର ବାକ୍ୟଗୁଡ଼ିକ ଠିକ୍ କ୍ରମରେ ସଜାଯାଇଛି ।ବାକ୍ୟଗୁଡ଼ିକ କ୍ରମ ଅନୁସାରେ ଲେଖୁ ଗଳ୍ପଟି ପ୍ରସ୍ତୁତ କର ।)

The Thief And The Tiger
______________________________________________
______________________________________________
______________________________________________
______________________________________________
Answer:

  • Once a thief came inside the stable to steal a horse. At night a tiger entered the stable.
  • The tiger among the horses stood silently.
  • It was dark everywhere.
  • He thought the tiger to be the best horse.
  • He climbed up a tree.
  • The thief fell asleep like a dead man.
  • The wolf requested the tiger to drag the man.
  • The wolf and the tiger tied each other with a rope.
  • The thief got up.
  • The tiger ran for life dragging the wolf.

(b). Write answers to the following questions.

Question (i).
Where did the thief and the tiger hide? Why?
(ଚୋର ଏବଂ ବାଘ କେଉଁଠି ଲୁଚିଥିଲେ ? କାହିଁକି ?)
Answer:
They hid in the king’s stable.
The thief wanted to steal a horse.
The tiger thought of eating man’s flesh.

Question (ii).
Why did the thief think the tiger to be the best horse?
(ଚୋର ବାଘକୁ କାହିଁକି ସର୍ବଶ୍ରେଷ୍ଠ ଘୋଡ଼ା ବୋଲି ଭାବିଲା ?)
Answer:
He thought so because the back of the tiger gave him a smooth silky touch.

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Question (iii).
Why was the thief frightened when it was the day?
(ଦିନ ହୋଇଯିବା ପରେ ଚୋରଟି ଭୟଭୀତ ହୋଇପଡ଼ିଲା କାହିଁକି ?)
Answer:
The thief was frightened when it was the day because he could not see due to darkness inside the stable at night.

Question (iv).
How did he save himself?
(ସେ କିପରି ନିଜକୁ ରକ୍ଷାକଲା ?)
Answer:
When the tiger was running in fear with great speed passing under a tree the thief was caught hold of one branch of the tree and climbed up.

Question (v).
Where did the theif take rest? Why did he fall asleep?
(ଚୋର କେଉଁଠି ବିଶ୍ରାମ ନେଲା ? ସେ କାହିଁକି ଗଭୀର ନିଦ୍ରାରେ ଶୋଇପଡ଼ିଲା ?)
Answer:
The thief took rest under the tree.

Question (vi).
What was the wolf’s plan?
(ଗମ୍ଵାର ଯୋଜନା କ’ଣ ଥିଲା ?)
Answer:
The wolf planned to use the dead body as his food for more than a week long.

Question (vii).
What sort of help did the wolf want from the tiger? What was his offer to him?
(ଗଧୂଆ ବାଘ ନିକଟରୁ କିପରି ସାହାଯ୍ୟ ଆଶା କରିଥିଲା । ସେ ତାକୁ କେଉଁ ଉପହାର ଦେବାପାଇଁ ପ୍ରତିଶୃତି ଦେଲା ?)
Answer:
The wolf wanted the tiger to help him drag the body to the safest place in the forest.

Question (viii).
What was the tiger’s suspicion?
(ବାଘର ସନ୍ଦେହ କ’ଣ ଥିଲା ?)
Answer:
The tiger feared that the wolf would cheat him and run away leaving him alone.

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Question (ix).
Why did the two animals tie each other with a rope?
(କାହିଁକି ଦୁଇଜଣ ପ୍ରାଣୀ ପରସ୍ପରକୁ ଏକ ଦଉଡ଼ିରେ ବାନ୍ଧିଲେ ?)
Answer:
The two animals tied each other with a rope so that neither of them could cheat or run away living another.

Question (x).
What did the tiger do when the thief shouted at him?
(ମଣିଷର ବଡ଼ପାଟି ଶୁଣି ବାଘ କ’ଣ କଲା ?)
Answer:
The tiger ran as fast as he could getting frightened and seeing the thief.

Question (xi).
What did the tiger promise?
(ବାଘ କ’ଣ ପ୍ରତିଜ୍ଞାକଲା ?)
Answer:
The tiger promised not to desire for human flesh anymore.

Question (xii).
What did the thief stop doing?
(ଚୋର କେଉଁ ଅଭ୍ୟାସ ବନ୍ଦ କରିଦେଲା ?)
Answer:
The thief stopped stealing anymore.

Session – 7

7. Mental Talk : (ମାନସିକ ଆଳାପ)

  • The tiger was stronger than the thief, but not so clever. Fear made him weaker.
    (ବାଘ ମଣିଷଠାରୁ ବଳବାନ ଥିଲା; କିନ୍ତୁ ଚତୁର ନଥିଲା । ଭୟ ତାକୁ ଦୁର୍ବଳ କରିଦେଲା ।)
  • Mind power is mightier than muscle power.
    (ମନର ବଳ ଶାରୀରିକ ବଳଠାରୁ ଅଧୁକ ଶକ୍ତିଶାଳୀ)

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

Tail-Piece

Did you like the story, “The Thief and The Tiger”?
(ଚୋର ଏବଂ ବାଘ ଗଳ୍ପଟି ଭଲ ଲାଗିଲା କି ?)
Read a story here, is more interesting than this.
(ଆଉ ଗୋଟିଏ ଏହିପରି ଗପ ପଢ଼ । ଏହାଠାରୁ ଅଧିକ ମନୋରଞ୍ଜନ କି)

The Liger On A Tiger (ବାଘ ଏବଂ ଫାଘ)

One day…………………………………………………. they could.

ଦିନେ ଗୋଟିଏ ଲୋକର ଘୋଡ଼ାଛୁଆଟି ହଜିଯାଇଥିଲା । ସେ ତାକୁ ଖୋଜି ଖୋଜି ବଣଭିତରକୁ ପଶିଯାଇଥିଲା । ସେ ବହୁତ କ୍ଳାନ୍ତ ହୋଇପଡ଼ିଥିଲା ଏବଂ ଆଗକୁ ଯିବାକୁ ଅକ୍ଷମ ହୋଇପଡ଼ିଥିଲା । ସେ ଘରକୁ ଫେରିଆସିବାକୁ ଇଚ୍ଛାକଲା । ମାତ୍ର ଘରକୁ ଫେରିବା ବାଟ ନପାଇ ବାଉଳା ହୋଇଗଲା । ସେତେବେଳକୁ ରାତି ହୋଇଯାଇଥିଲା । ଘରକୁ ଫେରିଯିବା ଆଉ ସମ୍ଭବ ନଥିଲା । ସେ ଗୋଟିଏ କୁଡ଼ିଆ ଦେଖିବାକୁ ପାଇଲା । ଏହା ଗୋଟିଏ ବୁଢ଼ୀଲୋକର କୁଡ଼ିଆ ଥିଲା । ଲୋକଟି ସେଠାରେ ଆଶ୍ରୟ ନେବାକୁ ବୁଢ଼ୀକୁ ଅନୁରୋଧ କଲା । ତା’ଘରେ ମାତ୍ର ଦୁଇଟି କୋଠରି ଥିଲା । ସେ ଗୋଟିଏ କୋଠରିରେ ତାରି ନାତୁଣୀ ସହିତ ରହୁଥିଲା ଏବଂ ଅନ୍ୟ କୋଠରିଟିରେ ଘରର ଜିନିଷପତ୍ର ସବୁ ରହିଥିଲା । ସେହି ଭଣ୍ଡାରଘରେ ତାକୁ ରହିବାକୁ ବୁଢ଼ୀ ଅନୁମତି ଦେଲା ।

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

ନାତୁଣୀଟି ଘୋଡ଼ାଛୁଆ ବିଷୟରେ କିଛିକିଛି ଶୁଣିଥିଲା । ଏଣୁ ସେ ଲୋକଟି ପାଖକୁ ଯାଇ ଘୋଡ଼ାଛୁଆ ବିଷୟରେ ସବୁକଥା ଜାଣିବାକୁ ଚାହିଁଲା । କିନ୍ତୁ ତା’ର ବୁଢ଼ୀ ମା’ ତାକୁ କହିଲା, ‘ଆଦୌ ନୁହେଁ । ତୁ ସେଠାକୁ ଆଦୌ ଯିବା ଉଚିତ ନୁହେଁ । ସେଠାରେ ବାଘ ଫାଘ ତୋତେ ଟେକିନେଇଯିବେ ।’’ ପ୍ରକୃତରେ ‘ଫାଘ’ ବୋଲି କୌଣସି ପ୍ରାଣୀ ନଥିଲା । କିନ୍ତୁ ବୁଢ଼ୀ ଘର ପଛପଟକୁ ପ୍ରତିଦିନ ଆସୁଥ‌ିବା ବାଘଟି ବୁଢ଼ୀ ମୁହଁରୁ ଫାଘ ସମ୍ପର୍କରେ ଶୁଣିବାକୁ ପାଇ ବିଚଳିତ ହୋଇପଡ଼ିଲା । ସେ ଭାବିଲା ବୋଧହୁଏ ଫାଘଟି ତା’ଅପେକ୍ଷା ଅଧିକ ଭୟଙ୍କର ଏକ ପ୍ରାଣୀ । ସେ ଗୋଟିଏ ରାକ୍ଷସ କିମ୍ବା ଭୂତ । ସେ ଖୁବ୍ ଭୟଭୀତ ହୋଇପଡ଼ିଲା । ସେ ସେଠାରୁ ପଳାଇଯିବାକୁ ଉପାୟ ଖୋଜିଲା । ଟିକିଏ ପରେ ଲୋକଟି ବାହାରକୁ ଆସି ଏହା ଦେଖିଲା ।
Tail piece

ସେତେବେଳକୁ ସକାଳ ହୋଇ ଆସୁଥାଏ । ସେଇ ଜାଲୁଜାଲୁଆ ଅନ୍ଧାର ଭିତରେ ଲୋକଟି ବାଘଟିକୁ ଦେଖି ତା’ର ଘୋଡ଼ାଛୁଆ ବୋଲି ଭାବିଲା । ସେ ସଙ୍ଗେ ସଙ୍ଗେ ସେଠାକୁ ଛୁଟିଯାଇ ବାଘର ମୁହଁକୁ ଗୋଟିଏ କନାରେ ବାନ୍ଧିପକାଇଲା । ଏହାଫଳରେ ବାଘର କାନ, ନାକ, ମୁଣ୍ଡ ଓ ବେକ ଜଣାପଡ଼ିଲା ନାହିଁ । ସେ ସଙ୍ଗେ ସଙ୍ଗେ ବାଘର ପିଠିରେ ବସିଗଲା । ବାଘର କି ଭୟ ! ସେ ଲୋକଟିକୁ ଫାଘ ବୋଲି ଭାବି ଜୀବନ ବିକଳରେ ଦୌଡ଼ିବାକୁ ଲାଗିଲା । ଯେତେବେଳେ ରାତି ପାହିଲା, ଲୋକଟି ଦେଖିଲା ଯେ ସେ ଗୋଟିଏ ବାଘ ପିଠିରେ ବସିଛି । କ’ଣ କରିବ ? କେମିତି ସେ ବାଘ ପିଠିରୁ ଖସି ଜୀବନ ବଞ୍ଚାଇବ ! ଯେତେବେଳେ ବାଘଟି ଗୋଟିଏ ବରଗଛ ତଳ ଦେଇ ଦୌଡୁଥିଲା, ଲୋକଟି ସେହି ସମୟରେ ବରଗଛର ଏକ ଡାଳକୁ ଧରି ଗଛ ଉପରେ ଚଢ଼ିଗଲା ଏବଂ କହିଲା, ‘ହେ ଭଗବାନ ! ତୁମକୁ ଧନ୍ୟବାଦ !

ଯାହାହେଉ ମୁଁ ବଞ୍ଚିଯାଇଛି ।’’ ବାଘଟି ମଧ୍ୟ ଫାଘ କବଳରୁ ମୁକ୍ତି ପାଇ ନିଜକୁ ଧନ୍ୟ ମନେକଲା । ବାଘଟି ଦୌଡ଼ି ନ ପଳାଇ ଗଛତଳେ ନିଶ୍ଵାସ ମାରିଲା ଏବଂ ଅନ୍ୟ ବାଘମାନଙ୍କୁ ଚିତ୍କାର କରି ଗଛ ଉପରେ ଥ‌ିବା ଭୟଙ୍କର ପ୍ରାଣୀ ଫାଘ ବିଷୟରେ ଜଣାଇଲା । ସବୁ ବାଘମାନେ ଆସି କହିଲେ, ‘ଖବର କ’ଣ ? କିଏ ତୁମର ମୁହଁ ବାନ୍ଧି ପକାଇଛି ?’’ ବାଘଟି ଦୀର୍ଘଶ୍ଵାସ ମାରି କହିଲା, ‘‘ଭାଇମାନେ ! ମୁଁ ମୃତ୍ୟୁମୁଖରୁ ବଞ୍ଚ୍ ଫେରିଆସିଛି । ମୋତେ ଗୋଟିଏ ‘ଫାଘ’ ଧରିନେଇଥିଲା । ମୁଁ ତାକୁ ପୂଜା ଅର୍ପଣ କରିବାକୁ ପ୍ରତିଜ୍ଞା କରିବାରୁ ସେ ମୋତେ ଛାଡ଼ିଦେଇଛି । ମୁଁ ଯଦି ତାକୁ ପୂଜା ଅର୍ପଣ ନଦିଏ; ତେବେ ସେ ମୋତେ ପୁଣି ଧରିନେଇଯିବ । ‘‘ଏକଥା ଶୁଣି ସବୁ ବାଘ ‘ଫାଘ’କୁ ପୂଜା କରି ଗଣ୍ଡା, ମଇଁଷି ପ୍ରଭୃତି ବିଭିନ୍ନ ଦ୍ରବ୍ୟ ଅର୍ପଣ କରିବାକୁ ଲାଗିଲେ । ଲୋକଟି ତା’ଜୀବନରେ କେବେ ହେଲେ ଏତେସଂଖ୍ୟକ ବାଘ ଏକାଠି ହେବାର ଦେଖି ନଥିଲା ।

BSE Odisha 8th Class English Solutions Lesson 2 The Thief and the Tiger

ସେ ଖୁବ୍ ଭୟ ପାଇଗଲା । ସେ ଗଛ ଉପରେ ବସି ଥରିବାକୁ ଲାଗିଲା । ଉଭୟ ଲୋକ ଏବଂ ଗଛ ଦୋହଲୁଥିଲେ । ବାଘମାନେ ମଧ୍ୟ ଭୟ ପାଇ ଯାଇଥିଲେ । ସେମାନେ ଉପରକୁ ଚାହିଁଲେ; ମାତ୍ର ପତ୍ରଗହଳରେ ଲୋକଟିକୁ ଦେଖୁରିଲେ ନାହିଁ । ଲୋକର ପିନ୍ଧାଲୁଗାର ଶେଷ ଅଂଶଟି ଗୋଟିଏ ଡାଳରୁ ଓହଳିଥିଲା । ସେମାନେ ଏହା ପତ୍ରଗହଳ ମଧ୍ୟରେ ଜାଣିପାରିଲେ ନାହିଁ । ସେମାନେ ଏହାକୁ ଏକ ଲାଞ୍ଜ ବୋଲି ଭାବିଲେ । ଏହା ଦେଖ୍ ଗୋଟିଏ ବୁଢ଼ା ବାଘ କହିଲା, ‘ଏହା ଏକ ବିପଜ୍ଜନକ ଜୀବଭଳି ଲାଗୁଛି । ଏହା ନିଶ୍ଚୟ ଲାଇଗର ।’’ ଏହା ଶୁଣି ସମସ୍ତ ବାଘ ଚିତ୍କାର କରି ଉଠିଲେ, ‘‘ସେ ଆମକୁ ଧରିନେବ, ଜୀବନ ବଞ୍ଚାଇ ପଳାଇଯାଅ ।’’ ଏବଂ ସମସ୍ତେ ଯେତେ ଜୋର୍‌ରେ ପାରିଲେ ଦୌଡ଼ବାକୁ ଲାଗିଲେ ।

Word Meaning

dangle: to hang or swing loosely (ଉପରୁ ଓହଳିବା / ଝୁଲିରହିବା)
escape: to get free from something (ଖସି ପଳେଇବା)
flutter: to move by waving quickly and lightly
huge: very big (ବିରାଟ, ଖୁବ୍ ବଡ଼ ଆକାରର, ବିଶାଳ)
pant: to breathe quickly (ଅଣନିଶ୍ୱାସୀ ହୋଇପଡ଼ିବା)
rush out: to go or move suddenly with great speed (ହଠାତ୍ ଧାଇଁବା)
Search: to look for (ଖୋଜିବା)
shelter: a house or a place to stay (ଆଶ୍ରୟସ୍ଥଳୀ)

Class 8 Questions and Answers Part – I

BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 ତ୍ରିକୋଣମିତି Ex 4(b)

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 ତ୍ରିକୋଣମିତି Ex 4(b) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 4 ତ୍ରିକୋଣମିତି Ex 4(b)

Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(i) sin (A – B) = sin A – cos A |
(ii) cos (θ + α) + cos (α – θ) = ……….. |
(iii) cos (60° – A) + ………. = cos A |
(iv) sin (30° + A) + sin (30° – A) = ……..|
(v) 2 sin A. sin B = ………… cos (A + B) |
(vi) tan (45° + θ). tan (45° – θ) = …………. |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 Img 1

BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 ତ୍ରିକୋଣମିତି Ex 4(b)

Question 2.
BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 Img 2
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 Img 3

Question 3.
ତ୍ପମାଣ କର:
(i) cos (A + 45°) = \frac{1}{\sqrt{2}} (cos A – sin A)
(ii) sin (45° – θ) = \frac{1}{\sqrt{2}} (sin θ – cos θ)
(iii) tan (45° + θ) = \frac{1+\tan \theta}{1-\tan \theta}
(iv) cot (45° + θ) = \frac{1+\tan \theta}{1-\tan \theta}
Solution:
(i) L.H.S. = cos (A + 45°) = cos A. cos 45° – sin A. sin 45°
= cos A . \frac{1}{\sqrt{2}} – sin A .\frac{1}{\sqrt{2}}

(ii) L.H.S. = sin (45° + θ) = sin 45° × cos θ – cos 45° 45° × sin θ
= \frac{1}{\sqrt{2}} cos θ – \frac{1}{\sqrt{2}} sin θ = – \frac{1}{\sqrt{2}} (sin θ – cos θ) = R.H.S.

(iii) L.H.S. = tan (45° + θ) = \frac{\tan 45^{\circ}+\tan \theta}{1-\tan 45^{\circ} \cdot \tan \theta} = \frac{1+\tan \theta}{1-1 \times \tan \theta}
= \frac{1+\tan \theta}{1-\tan \theta} = R.H.S.

(iv) L.H.S. = cot (45° – θ) = \frac{\cot 45^{\circ} \cdot \cot \theta+1}{\cot \theta-\cot 45^{\circ}} = \frac{1 \times \cot \theta+1}{\cot \theta-1} = \frac{\cot \theta+1}{\cot \theta-1}

Question 4.
ତ୍ପମାଣ କର:
(i) cos (45° – A). cos (45° – B) – sin (45° – A). sin (45° – B) = sin (A + B)
(ii) sin (45° + A). cos (20° – A) + cos (45° + A). sin (20° – A) = \frac{\sqrt{3}}{2}
(iii) cos (65° + θ). cos (35° + θ) + sin (65° + θ). sin (35° + θ) = \frac{\sqrt{3}}{2}
(iv) cos nθ . cos θ + sin nθ. sin θ = sin (n – 1) θ
(v) tan (60° – A) = \frac{\sqrt{3} \cos A-\sin A}{\cos A+\sqrt{3} \sin A}
Solution:
(i) L.H.S. = cos (45° – A). cos (45° – B) – sin (45° – A). sin (45° – B)
= cos (45° – A + 45° – B) = cos {90° – (A + B)}
= sin (A + B) = R.H.S.

(ii) L.H.S. = sin (40° + A). cos (20° – A) + cos (40° + A). sin (20° – A)
= sin (40° – A + 20° – A) = sin 60° = \frac{\sqrt{3}}{2} = R.H.S.

(iii) L.H.S. = cos (65° + θ). cos (35° + θ) + sin (65° + θ). sin (35° + θ)
= cos {(65° + θ) – (35° + θ)} = cos (65° + θ – 35° – θ)
= cos 30° = \frac{\sqrt{3}}{2} = R.H.S.

(iv) L.H.S. = cos nθ. cos θ + sin nθ. sin θ = cos (nθ – θ)
= sin (n – 1)θ = R.H.S.

BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 Img 4

Question 5.
ତ୍ପମାଣ କର:
(i) tan 62° = \frac{\cos 17^{\circ}+\sin 17^{\circ}}{\cos 17^{\circ}-\sin 17^{\circ}}
(ii) tan 70° = \frac{\cos 25^{\circ}+\sin 25^{\circ}}{\cos 25^{\circ}-\sin 25^{\circ}}
(iii) tan 7A. tan 4A. tan 3A = tan 7A – tan 4A – tan 3A
(iv) tan (x + y) – tan x – tan y = tan (x + y) . tan x . tan y
(v) (1 + tan 15°) (1 + tan 30°) = 2
(vi) (cot 10° – 1) (cot 35° – 1) = 2
(vii) \frac{1}{\cot A+\tan B}\frac{1}{\tan A+\cot B} = tan (A – B)
(viii) √3 + cot 50° + tan 80° = √3 cot 50° . tan 80°
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 Img 5
BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 Img 6

BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 ତ୍ରିକୋଣମିତି Ex 4(b)

Question 6.
cos 75° ଓ sin 15° ର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
Solution:
cos 75° = cos (45° + 30°) = cos 45° . cos 30° – sin 45° . sin 30°
= \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}} × \frac { 1 }{ 2 } = \frac{\sqrt{3}-1}{2 \sqrt{2}}
sin 15° = sin (60° – 45°) = sin 60° . cos 45° – cos 60° . sin 45°
= \frac{\sqrt{3}}{2} . \frac{1}{\sqrt{2}}\frac { 1 }{ 2 } × \frac{1}{\sqrt{2}} = \frac{\sqrt{3}-1}{2 \sqrt{2}}

Question 7.
(i) cos α = \frac { 8 }{ 17 } ଓ sin β = \frac { 5 }{ 13 } ହେଲେ sin (α – β) ର ମାନ ନିଣ୍ଡଯ କର |
(ii) tan A = \frac { 1 }{ 2 }, cot B = 3 ହେଲେ A + B ର ମାନ ନିଣ୍ଡଯ କର |
(iii) tan β = \frac{1-\tan \alpha}{1+\tan \alpha} ହେଲେ (α + β) ର ମାନ ନିଣ୍ଡଯ କର |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 Img 7

Question 8.
A + B + C = 90° ହେଲେ ତ୍ପମାଣ କର ଯେ
(i) cot A + cot B + cot C = cot A . cot B . cot C
(ii) tan A . tan B + tan B . tan C + tan C . tan A = 1
Solution:
(i) A + B + C = 90° ⇒ A + B = 90° – C
⇒ cot (A + B) = cot (90° – C)
\frac{\cot A \cdot \cot B-1}{\cot B+\cot A} = tan C = \frac{1}{\cot \mathrm{C}}
⇒ cot C (cot A. cot B – 1) = 1 (cot B + cot A)
⇒ cot A. cot B. cot C – cot C = cot B + cot A
⇒ cot A. cot B. cot C = cot A + cot B + cot C

(ii) A + B + C = 90° ⇒ B + C = (90° – A)
⇒ tan (B + C) = tan (90° – A)
\frac{\tan B+\tan C}{1-\tan B \cdot \tan C} = cot A = \frac{1}{\tan \mathrm{A}}
⇒ tan A (tan B + tan C) = 1 – tan B. tan C
⇒ tan A. tan B + tan C. tan A = 1 – tan B. tan C
⇒ tan A. tan B + tan B. tan C + tan C. tan A = 1

Question 9.
(i) A + B + C = 180° ଏବଂ sin C = 1 ହେଲେ ପ୍ରମାଣ କର ଯେ tan A . tan B = 1
(ii) A + B + C = 180° ହେଲେ ପ୍ରମାଣ କର ଯେ cot A . cot B + cot B . cot C + cot C . cot A = 1
(iii) A + B + C = 180° ଏବଂ cos A = cot B . cos C ହେଲେ ପ୍ରମାଣ କର ଯେ
(a) tan A – tan B + tan C
(b) tan B . tan C = 2
Solution:
(i) sin C = 1 ⇒ sin C = sin 90° ⇒ C = 90°
A + B + C = 180° ⇒ A + B = 180° – C = 180° = 90° = 90°
⇒ A = 90° – B
∴ tan A . tan B = tan (90° – B) . tan B = cot B × tan B = 1

(ii) A + B + C = 180° ⇒ A + B = 180° – C
⇒ cot(A + B) = cot (180° – C)
\frac{\cot A \cdot \cot B-1}{\cot B+\cot A} = – cot C
⇒ cot A. cot B – 1=-cot C (cot B + cot A)
⇒cot A. cot B – 1=-cot B. cot C – cot C. cot A
⇒ cot A. cot B + cot B. cot C + cot C. cot A = 1

(iii) (a) A + B + C = 180° ⇒ (A + C) = 180° – A
⇒ sin (B + C) = sin (180° – A)
⇒ sin B . cos C + cos B . sin C = sin A
\frac { sin B. cos C }{ cos A } + \frac { cos B. sin C }{ cos A } = \frac { sin A }{ cos A } (ଭଉଯ ପାଣରେ cos A କାମାଗଣ)
\frac { sin B. cos C }{ cos B. cos C } + \frac { cos B. sin C }{ cos B. cos C } = tan A
⇒ tan B + tan C = tan A

(b) A + B + C = 180°
⇒ B + C = 180° – A
⇒ cos (B + C) = cos (180° – A) = – cos A
⇒ cos B. cos C – sin B. sin C = – cos B. cos C
⇒ 2cos B. cos C = sin B. sin C
⇒ 2 = \frac{\sin B \cdot \cos C}{\cos B \cdot \cos C} = tan B. tan C

BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 ତ୍ରିକୋଣମିତି Ex 4(b)

Question 10.
ଜଣାଥ ଯେ : (i) sin (A + B) . sin (A – B) = sin2 A – sin2 B
(ii) cos (A + B) . cos (A – B) = cos2 A – sin2 B
Solution:
(i) L.H.S.= sin (A + B) . sin (A – B)
= (sin A. cos B + cos A· sin B) (sin A · cos B – cos A. sin B)
= (sin A. cos B)2 – (cos A. sin B)2
= sin2A. cos2B – cos2A. sin2B
= sin2A (1 – sin2B) − (1 − sin2A) sin2B
= sin2A – sin2A. sin2B – sin2B + sin2A. sin2B
= sin2A – sin2B = R.H.S.

(ii) L.H.S.= cos (A + B) . cos (A – B)
= (cos A. cos B – sin A. sin B) (cos A. cos B + sin A. sin B)
= (cos A. cosB)2 – (sin A. sin B)2
= cos2A cos2B – sin2A. sin2B
= cos2A (1 − sin2B) − (1 − cos2A) sin2B
= cos2A – cos2A. sin2B – sin2B + cos2A. sin2B
= cos2A – sin2B = R.H.S.

Question 11.
ପ୍ରମାଣ କର :
(i) sin 50° + sin 40° = √2 sin 85°
(ii) cos 50° + cos 40° = √2 cos 5°
(iii) sin 50° – sin 70° + sin 10° = 0
Solution:
(i) ଦାନପାଣ = sin 50° + sin 40°
= sin (45° +5°) + sin (45° – 5°)
= sin 45°. cos 5° + cos 45°. sin 5° + sin 45°. cos 5° – cos 45°. sin 5°
= 2 sin 45° × cos 5° = 2 × \frac{1}{\sqrt{2}} cos (90° – 85°)
= √2 sin 85° = ଦର୍ପଣପାଣ

(ii) ଦାନପାଣ = cos 50° + cos 40°
= cos (45° +5°) + cos (45° – 5°)
= cos 45°. cos 5° – sin 45°. sin 5° + cos 45°. cos 5° + sin 45°. sin 5°
= 2 cos 45°. cos 5°
= 2 × \frac{1}{\sqrt{2}} cos 5° = √2 cos 5° = ଦର୍ପଣପାଣ

(iii) ଦାନପାଣ = sin 50° – sin 70° + sin 10°
= sin (60° – 10°) – sin (60° – 10°) + sin 10°
= (sin 60° . cos 10° – cos 60° . sin 10°)
– (sin 60° . cos 10° + cos 60° . sin 10°) + sin 10°
BSE Odisha 10th Class Maths Solutions Geometry Chapter 4 Img 8

Question 12.
ପ୍ରମାଣ କର :
(i) sin (A + B) = \frac{1}{\sqrt{2}} , cos (A – B) = \frac{1}{\sqrt{2}}
(ii) cos (A + B) = – \frac { 1 }{ 2 } , sin (A – B) = \frac { 1 }{ 2 }
(iii) tan (A – B) = \frac{1}{\sqrt{2}} = cot (A + B)
(iv) tan (A + B) = -1, cosec (A – B) = √2
Solution:
(i) sin (A + B) = \frac{1}{\sqrt{2}} = sin 45°ଦା, sin 135°
⇒ A + B = 45° ଦା, 135°
cos (A – B) = \frac{1}{\sqrt{2}} = cos 45° ⇒ A – B = 45°
A + B = 45° , A – B = 45°
ହେଲେ, (i) ଓ (ii) ରୁ A + B +A – B = 45° + 45° ⇒ 2A = 90° ⇒ A = 45°
B = 45° – 45° = 0°
ଯଦି A + B = 135°, A – B = 45°
∴ (i) ଓ (ii) ରୁ A + B + A – B = 135° + 45°
⇒ 2A = 180° ⇒ A = 90°
B = 135° – 90° = 45°
(∴ A = 45°, B = 0°) ଦା, (A = 90° , B = 45°)

(ii) cos (A + B) = – \frac { 1 }{ 2 } = cos 120°
⇒ A + B = 120°
sin (A – B) = \frac { 1 }{ 2 } = sin 30°
⇒ A – B = 30°
∴ (i) ଓ (ii) ରୁ A + B + A – B = 120° + 30°
⇒ 2A = 150° ⇒ A = 75°
B = 120° – 75° = 45° (∵ A = 75°, B = 45°)

(iii) tan (A – B) = \frac{1}{\sqrt{3}} = tan 30° ⇒ A – B = 30°
cot (A + B) = \frac{1}{\sqrt{3}} = cot 60° ⇒ A – B = 30°
∴ (i) ଓ (ii) ରୁ A + B + A – B = 60° + 30°
⇒ 2A = 90° ⇒ A = 45° , B = 60° – 45° = 15°
∴ A = 45° , B = 15°

(iv) tan (A + B) = -1 = tan 135° ⇒ A + B = 135°
cosec (A – B) = √2 = cosec 45° ⇒ A – B = 45°
(i) ଓ (ii) ରୁ 2A = 180° ⇒ A = 90°
∴ B = 135° – 90° = 45°
∴ A = 90° , B = 45°

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

\left(\frac{{at}_1^2+{at} {t}_2^2}{2}, \frac{2 {at}_1+2 {at}_2}{2}\right)

Question 1.
ନିମ୍ନଲିଖତ ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ବିନ୍ଦୁଦ୍ଵୟ ମଧ୍ଯରେ ଦୂରତା ନିର୍ଣ୍ଣୟ କର ।
(i) (0, 0) ଓ (4, 3)
(ii) (0, 2) ଓ (-6, 2)
(iii) (-3, 0) ଓ (5, 6)
(iv) (2, 4) ଓ (1, 3)
(v) (-2, -2) ଓ (-3, -5)
(vi) (a, b) ଓ (- a, b)
ସମାଧାନ :
ମେନକର O(0, 0) ଓ P (4,3)
ମୂଳବିନ୍ଦୁଠାରୁ p(x, y)ର ଦୂରତା = \sqrt{x^2 + y^2}
OP = \sqrt{4^2 + 3^2}=\sqrt{16 + 9}=\sqrt{25}=5

(ii) ମେନକର A (0, 2), ଓ B (-6, 2)
ଏଠାରେ x1 =0, y1 = 2, x2 = -6, y2 = 2
AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = √(-6-0)² +(2-2)² = √(36+0) = 6

(iii) P(-3, 0) ଓ Q (5, 6)
ଏଠାରେ x1 = 3, y1 = 0, x2 = 5, y2 = 6
PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
= \sqrt{{5- (-3)}^2 +(6-0)^2} = \sqrt{8^2 + 6^2} = √64+36 = √100 = 10

(iv) P (2, 4) ଓ Q (1, 3)
PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(1-2)^2+(3-4)^2} = \sqrt{(-1)^2 +(-1)^2} =√2

(v) A (-2,-2) ଓ B (-3,-5).
AB = \sqrt{{-3-(-2)}^2 + {-5-(-2)}^2} = √1² +3² =√10

(vi) P (a, b) ଓ Q (a, b)
PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(-a-a)^2+{b-(-b)}^2}
\sqrt{(-2a)^2+(2b)^2} = \sqrt{4a^2+4b^2} = 2√(a² + b²)

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 2.
ନିମ୍ନଲିଖ କେଉଁ ବିନ୍ଦୁଦ୍ଵୟ ମୂଳବିନ୍ଦୁ ଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ସ୍ଥିର କର ।
(i) (0, 1) ଓ (- 1, 0)
(ii) (2,3) ଓ (4, \frac{3}{2})
(iii) (√7, √19) ଓ (-√7, – √19)
(iv) (4, – 2) ଓ (2, 4)
(v) (0, 4) ଓ (2, 2)
ସମାଧାନ :
(i) ଏଠାରେ ମୂଳବିନ୍ଦୁ O(0, 0) । A (0, 1) ଓ B(- 1, 0) ।
OA = \sqrt{1^2+0^2} = 1, OB = \sqrt{(-1)^2 +0^2} = 1 ∴ OA = OB
∴ (0, 1) ଓ ( 1, 0) ମୂଳବିନ୍ଦୁ ଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ।

(ii) ଏଠାରେ A(2,3) ଓ B(4, \frac{3}{2})
OA = √2^2+3^2 = √4+9 = √13
OB = \sqrt{4² + (\frac{3}{2})²} = \sqrt{16 + \frac{9}{4}} = \sqrt{\frac{64+9}{4}} = \sqrt{\frac{1}{2}}
OA ≠ OB ∴ A(2, 3) ଓ B(4, \frac{3}{2}) ବିଦୁ୍ୟଦ୍ଵୟ ମୂଳବିନ୍ଦୁଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ନୁହେଁ ।

(iii) ଏଠାରେ A (√7, √19) ଓ B (-√7,-√19) ।
OA = \sqrt{(√7)^2+(√19)^2} = \sqrt{7+19} = √26
OB = \sqrt{(√7)^2+(-√19)^2} = \sqrt{7+19} = √26
∴ OA = OB ଅର୍ଥାତ୍‌ A(√7, √19) ଓ B(-√7, -√19) ବିନ୍ଦୁଦ୍ଵୟ ମୂଳ ବିନ୍ଦୁଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ।

(iv) ଏଠାରେ A (4, – 2) ଓ B (2, 4) ।
OA = \sqrt{4^2 +(-2)^2} = \sqrt{16+4} = √20 = 2√5
OB = \sqrt{2^2 +4^2} = \sqrt{4+16} = √20 = 2√5
∴ OA = OB ଅର୍ଥାତ୍‌ A(4, – 2) ଓ B(2, 4) ବିନ୍ଦୁଦ୍ଵୟ ମୂଳ ବିନ୍ଦୁଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ।

(v) ଏଠାରେ A(0, 4) ଓ B(2, 2)
OA = \sqrt{2^2+3^2} = √16 = 4
OB = \sqrt{2² + 2²} = \sqrt{4+4} = √8 = 2√2.
∴ OA ≠ OB ଅର୍ଥାତ୍ A(0, 4) ଓ B(2, 2) ବିଦୁ୍ୟଦ୍ଵୟ ମୂଳ ବିନ୍ଦୁଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ନୁହେଁ ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 3.
ପ୍ରମାଣ କର ଯେ, ନିମ୍ନୋକ୍ତ ଶୀର୍ଷବିନ୍ଦୁ ବିଶିଷ୍ଟ ABC ତ୍ରିଭୁଜମାନ ସମକୋଣୀ । ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ କେଉଁ କୋଣଟି ସମକୋଣ ଦର୍ଶାଅ I
(i) A (3, 3), B (9, 0) ଓ C (12, 21)
(ii) A (1, 1), B (3, 4) ଓ C(0, 6)
(iii) A (-1, -2), B (5, -2) ଓ C (5, 6)
(iv) A (12, 8), B (- 2, 6) ଓ C (6, 0)
(v) A (1, 6), B (5, – 1) ଓ C (7, 2)
ସମାଧାନ :
P1 = (x1, y1) ଓ P2 (x2, y2) ହେଲେ, P1P2 = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

(i) A (3, 3), B (9, 0) ଓ C (12, 21)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -1
∴ ABC ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜ। ⇒ m∠BAC = 90°

(ii) A (1, 1), B (3, 4), C(0, 6)
AB = \sqrt{(3-1)^2+(4-1)^2} = \sqrt{2^2 +3^2} = √13
BC= \sqrt{(0-3)^2+(6-4)^2} = \sqrt{3^2 +2^2} = √13
AC = \sqrt{(0-1)^2+(6-1)^2} = \sqrt{1^2 +5^2} = √26
AB² + BC² = (√13)² +(√13)² = 13 + 13 = 26 = (√26)² =
∴ ABC ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜ। ⇒ m∠BAC = 90°

(iii) A (-1,- 2), B (5, -2) ଓ C (5, 6)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -2
∴ ABC ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜ। ⇒ m∠BAC = 90°

(iv) A (12, 8), B (- 2, 6) ଓ C (6, 0)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -3
∴ ABC ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜ। ⇒ m∠BAC = 90°

(v) A (1, 6), B (5, – 1) ଓ C (7, 2)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -4
∴ ABC ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜ। ⇒ m∠BAC = 90°

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 4.
ଦର୍ଶାଅ ଯେ ନିମ୍ନୋକ୍ତ ଶୀର୍ଷବିନ୍ଦୁ ବିଶିଷ୍ଟ ABC ତ୍ରିଭୁଜମାନ ସମଦ୍ବିବାହୁ ।
(i) A (8, 2), B(5,- 3) ଓ C (0, 0)
(ii) A (0, 6), B (- 5, 3) ଓ C (3, 1)
(iii) A (8, 9), B(- 6, 1) ଓ C (0,-5)
(iv) A (7, 1), B (11, 4) ଓ C (4, – 3)
(v) A (0, 0), B (4, 0) ଓ C (0, -4)
(vi) A (2, 2) B (- 2, 4) ଓ C (2,6)

(i) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟର ସ୍ଥାନାଙ୍କ A (8, 2), B(5, – 3), C (0, 0) ।
AB = \sqrt{(5-8)^2+(-3-2)^2} = \sqrt{(-3)^2 +(-5)^2} = \sqrt{9+25}=√34
BC = \sqrt{(0-5)^2+{0-(-3)}^2} = \sqrt{25+9} = √34
CA = \sqrt{(8-0)^2+(2-0)^2} = \sqrt{64+4} = √68 = 2√17
∴ ∆ ABCର AB = BC ତେଣୁ ଏହା ଏକ ସମଦ୍ବିବାହୁ ତ୍ରିଭୁଜ ।

(ii) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟ A(0, 6), B (- 5, 3) ଓ C (3, 1) ।
AB = \sqrt{(-5-0)^2+(3-6)^2} = \sqrt{(-5)^2 +(-3)^2} = \sqrt{25+9} = √34
BC= \sqrt{{3-(-5)}^2 +(1–3)^2} = \sqrt{8^2 +(-2)^2} = \sqrt{64+4}= √68
AC = \sqrt{(3-0)^2+(1-6)^2} = \sqrt{3^2 +(-5)^2} = \sqrt{9+25} = √34
∴ ∆ ABCର AB = AC ତେଣୁ ଏହା ଏକ ସମଦ୍ବିବାହୁ ତ୍ରିଭୁଜ ।

(iii) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟ A(8, 9), B(- 6, 1) ଓ C (0, – 5) ।
AB = \sqrt{(-6-8)^2 +(1-9)^2} = \sqrt{(14)^2 +(8)^2} = \sqrt{196+64} = √260 = 2√65
BC = \sqrt{{0-(-6)}2 +(-5-1)^2} = \sqrt{62+(-6)^2} = \sqrt{36+36} = √72 = 6√2
AC = \sqrt{(0-8)^2 + (-5-9)^2} = \sqrt{(-8)^2 +(-14)^2} = \sqrt{64+196} = √260 = 2√65
∴ ∆ ABCର AB = AC ତେଣୁ ଏହା ଏକ ସମଦ୍ବିବାହୁ ତ୍ରିଭୁଜ ।

(iv) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟ A (7, 1), B (11, 4) ଓ C (4, – 3) ।
AB = \sqrt{(11-7)^2+(4-1)^2} = \sqrt{42+32} = \sqrt{16+9}= √25=5
BC = \sqrt{(4-11)^2+(-3-4)^2} = \sqrt{(-7)^2+(-7)^2} = \sqrt{49 +49} = √98 = 7√2
AC = \sqrt{(4-7)^2 +(-3-1)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 +16} = √25 = 5
∴ ∆ ABCର AB = AC ତେଣୁ ଏହା ଏକ ସମଦ୍ବିବାହୁ ତ୍ରିଭୁଜ ।

(v) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟ A (0, 0), B (4, 0), C (0, – 4) ।
AB = \sqrt{4^2 +0^2} = \sqrt{16} = 4,
AC = \sqrt{0^2 +(-4)^2} = √16 = 4
BC= \sqrt{(0-4)^2+(-4-0)^2} = \sqrt{(-4)^2 +(-4)^2} = \sqrt{16+16} = √32 = 4√2
∴ ∆ ABCର AB = AC ତେଣୁ ଏହା ଏକ ସମଦ୍ବିବାହୁ ତ୍ରିଭୁଜ ।

(vi) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟ A (2, 2), B ( 2, 4) ଓ C (2, 6) ।
AB = \sqrt{(-2-2)^2 + (4 -2)^2} = \sqrt{4^2 +2^2} = \sqrt{16+4} = √20 = 2√5
BC = \sqrt{{2-(-2)}^2 +(6-4)^2} = \sqrt{4^2 +2^2} = \sqrt{16+4} = √20 = 2√5
∴ ∆ ABCର AB = BC ତେଣୁ ଏହା ଏକ ସମଦ୍ବିବାହୁ ତ୍ରିଭୁଜ ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 5.
ଦର୍ଶାଅ ଯେ ନିମ୍ନଲିଖୂତ ବିନ୍ଦୁଗୁଡ଼ିକ ପାର୍ଶ୍ଵରେ ସୂଚିତ ଚିତ୍ରକୁ ଗଠନ କରିବ ।
(i) (1, 1), (- 1, – 1), (- √3, √3) (ସମବାହୁ ତ୍ରିଭୁଜ)
(ii) (3, – 3), (- 3, 3), (3√3, 3√3) (ସମବାହୁ ତ୍ରିଭୁଜ)
(iii) (1, 2), (3, 4) ଓ (5, 8) (ସମବାହୁ ତ୍ରିଭୁଜ)
(iv) (1, 2), (2, 4) ଓ (3, 5) (ବିଷମବାହୁ ତ୍ରିଭୁଜ)
(v) (-2, 3), (8, 3) ଓ (6, 7) (ସମକୋଣୀ ତ୍ରିଭୁଜ)
(vi) (-6, -8), (-16, 12) ଓ (- 26,- 18) (ସମକୋଣୀ ସମବାହୁ ତ୍ରିଭୁଜ)
ସମାଧାନ :
(i) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟ A(1, 1), B(- 1, – 1), ଓ C (-√3, √3) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -5
∴ ∆ ABCର AB = BC = AC । ତେଣୁ ଏହା ଏକ ସମବାହୁ ତ୍ରିଭୁଜ ।

(ii) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟ A(3, – 3), B(- 3, 3), C(3√3, 3√3) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -6
∴ ∆ ABCର AB = BC = AC । ତେଣୁ ଏହା ଏକ ସମବାହୁ ତ୍ରିଭୁଜ ।

(iii) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟ A(1, 2), B(3, 4) ଓ C(5, 8) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -7
∴ ∆ ABCର AB ≠ BC ≠ AC । ତେଣୁ ଏହା ଏକ ବିଷମବାହୁ ତ୍ରିଭୁଜ ।

(iv) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟ A(1, 2), B(2, 4) ଓ C(3, 5) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -8
∴ ∆ ABCର AB ≠ BC ≠ AC । ତେଣୁ ABC ଏକ ବିଷମବାହୁ ତ୍ରିଭୁଜ ।

(v) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟ A(-2, 3), B(8, 3) ଓ C(6, 7) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -9
∴ ABC ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜ ।

(vi) ∆ ABCର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟ A(-2, 3), B(8, 3) ଓ C(6, 7) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -10
∴ ABC ଏକ ସମକୋଣୀ ସମବାହୁ ତ୍ରିଭୁଜ ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 6.
ଦର୍ଶାଅ ଯେ ନିମ୍ନଲିଖ୍ ବିନ୍ଦୁଗୁଡ଼ିକ ପାର୍ଶ୍ଵରେ ସୂଚିତ ଚିତ୍ରକୁ ଗଠନ କରିବ ।
(i) (-8, 3), (-2, -1), (6, -2) ଓ (0, 2) (ସାମାନ୍ତରିକ ଚିତ୍ର)
(ii) (-2, -1), (1, 0), (4, 3) ଓ (1, 2) (ସାମାନ୍ତରିକ ଚିତ୍ର)
(iii) (0, -1), (2, 1), (0, 3) ଓ (-2, 1) (ବର୍ଗ ଚିତ୍ର)
(iv) (0,5), (-1, 2), (-4, 3) ଓ (-3, 6) (ବର୍ଗ ଚିତ୍ର)
(v) (-2, 3), (-4, -1), (-6, 0) ଓ (-4, 4) (ଆୟତ ଚିତ୍ର)
ସମାଧାନ :
(i) ଦଉ ବିନ୍ଦୁ ଚତୁର୍ଭୁଜଟି A(-8, 3), B (-2, -1), C (6,-2) ଓ D(0, 2)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -11
∴ ABCD ଚତୁର୍ଭୁଜର AB = CD ଏବଂ AD = BC ।
⇒ ଚତୁର୍ଭୁଜଟି ସାମାନ୍ତରିକ ଚିତ୍ର ଅଟେ । (ପ୍ରମାଣିତ)
(ABCD ଚତୁର୍ଭୁଜର ବିପରୀତ ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ସମାନ)

(ii) ଦଉ ବିନ୍ଦୁ ଚତୁର୍ଭୁଜଟି A(-2, -1), B(1, 0), C(4, 3) ଓ D(1, 2)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -12
∴ ABCD ଚତୁର୍ଭୁଜର AB = CD ଏବଂ AD = BC ।
⇒ ଚତୁର୍ଭୁଜଟି ସାମାନ୍ତରିକ ଚିତ୍ର ଅଟେ । (ପ୍ରମାଣିତ)
(ABCD ଚତୁର୍ଭୁଜର ବିପରୀତ ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ସମାନ)

(iii) ଦଉ ବିନ୍ଦୁ ଚତୁର୍ଭୁଜଟି A(-2, -1), B(1, 0), C(4, 3) ଓ D(1, 2)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -13

(iv) ଦଉ ବିନ୍ଦୁ ଚତୁର୍ଭୁଜଟି A(0,5), B(-1, 2), C(-4, 3) ଓ D(-3, 6)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -14

(v) ଦଉ ବିନ୍ଦୁ ଚତୁର୍ଭୁଜଟି A(-2, 3), B(-4, -1), C(-6, 0) ଓ D(-4, 4)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a) -15

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 7.
ଦର୍ଶାଅ ଯେ P (1, 1) ବିନ୍ଦୁ A (0, 2), B (2, 0) ଓ C (0, 0) ବିନ୍ଦୁମାନଙ୍କଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ।
ସମାଧାନ :
AP = \sqrt{(0-1)^2 +(2-1)^2} = \sqrt{1^2 + 1^2} = √2
BP = \sqrt{(2-1)^2+(0-1)^2} = \sqrt{1^2+(-1)^2} = √2
CP = \sqrt{(0-1)^2+(0-1)^2} = \sqrt{(-1)^2 +(-1)^2} = √2
∴ AP = BP = CP ତେଣୁ ‘P’ ବିନ୍ଦୁ A, B ଓ C ବିନ୍ଦୁମାନଙ୍କଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ।

Question 8.
xର କେଉଁ ମାନ ପାଇଁ C (x, 3) ବିନ୍ଦୁ, A (2, 4) ଓ B (3, 5) ବିନ୍ଦୁଦ୍ଵୟ ଠାରୁ ସମାନ ଦୂରରେ ରହିବ ?
ସମାଧାନ :
C (x, 3), ବିନ୍ଦୁ A (2, 4), B (3, 5) ବିନ୍ଦୁଦ୍ଵୟଠାରୁ ସମଦୂରରେ ହେବେ।
AC² = (2 – x)² + (4 – 3)² = (2 – x)² + 1
BC² = (3 – x)² + (5 – 3)² = (3 – x)² + 4
କିନ୍ତୁ ଦଉ ଅନ୍ଥି AC = BC ⇒ AC2 = BC2
⇒ (2 – x)² + 1 = (3 – x)² + 4
⇒ 4 – 4x + x² + 1 = 9 – 6x + x² + 4
⇒ x² – x² + 6x – 4x = 9+ 4 – 5
⇒ 2x = 8 ⇒ x = \frac{8}{2} = 4
∴ xର ମାନ 4 ପାଇଁ C ବିନ୍ଦୁ A ଓ B ଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ହେବେ ।

Question 9.
P(2, y) ବିନ୍ଦୁ Q (-1, 2) ବିନ୍ଦୁଠାରୁ 5 ଏକକ ଦୂରରେ ରହିଲେ, ଦୁର ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
P ଓ Q ବିନ୍ଦୁ ପରସ୍ପରଠାରୁ 5 ଏକକ ଦୂରରେ ଅବସ୍ଥିତ ।
P (2, y) Q) (-1, 2) ମଧ୍ୟରେ ଦୂରତା = \sqrt{(1-2)^2+(2-y)^2}
ପ୍ରଶ୍ମାନୁସାରେ \sqrt{(1-2)^2+(2-y)^2} = 5
ଉଭୟ ପାର୍ଶ୍ଵର ବର୍ଗ ନେଲେ, (- 1 – 2)² + (2 – y)² = 5²
⇒ (-3)² + 4 – 4y + y² = 25 ⇒ 9 + 4 + y² – 4y – 25 = 0
⇒ y² – 4y + 12 = 0 ⇒ y² – 6y + 2y + 12 = 0
⇒ y (y – 6) + 2 (y – 6) = 0 ⇒ (y – 6) (y + 2) = 0
⇒ y – 6 = 0 ବା y + 2 = 0 ⇒ y = 6 = 0 ବା y = -2
∴ yର ମାନ 6 କିମ୍ବା – 2 ହେଲେ, PQ = 5 ଏକକ ହେବ ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 10.
ଦର୍ଶାଅ A (1, 1),B (2, 2) & C (3, 3) ଯେ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ରହିବେ ।
ସମାଧାନ :
AB = \sqrt{(2-1)^2 +(2-1)^2} = \sqrt{1^2 +1^2} = \sqrt{1+1} = √2
BC = \sqrt{(3-2)^2 +(3-2)^2} = \sqrt{1^2 +1^2} = \sqrt{1+1} = √2
AC = \sqrt{(3-1)^2+(3-1)^2} = \sqrt{2^2 +2^2} = \sqrt{4+4} = √8 = 2√2
AB + BC = √2 + √2 = 2√2 = AC
⇒ A, B, C ବିଦୁ୍ୟତ୍ରୟ ଏକ ସରଳରେଖାରେ ରହିବେ ।

Question 11.
ଦର୍ଶାଅ A (1, 4), B (-1, 6), C (2, 3) ଯେ ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖ୍ୟ ।
ସମାଧାନ :
AB = \sqrt{(-1-1)^2 +(6-4)^2} = \sqrt{(-2)^2 +(2)^2} = \sqrt{4+4} = √8 = 2√2
AC = \sqrt{(2-1)^2+(3-4)^2} = \sqrt{1^2 +1^2} = √2
BC = \sqrt{{2-(-1)}^2 +(3-6)^2} = \sqrt{(2+1)^2 +(-3)^2} = \sqrt{9+9} = √18 = 3√2
AB + AC = 2√2 + √2 = 3√2 = BC
⇒ B, A, C ଏକ ରେଖ୍ୟ ।

Question 12.
ପ୍ରମାଣ କର ଯେ, (1, 0), (2, -3) ଏବଂ (- 1, 6) ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖ୍ୟ ଓ (1, 0) ବିନ୍ଦୁଟି ଅନ୍ୟ ଦୁଇ ବିନ୍ଦୁର ମଧ୍ୟବର୍ତ୍ତୀ ଅଟେ ।
ସମାଧାନ :
ମନେକର ବିଦୁ୍ୟତ୍ରୟ A (1, 0), B(2, -3) 3 C(-1, 6) ।
AB = \sqrt{(2-1)^2+(-3-0)^2} = \sqrt{1^2 +(-3)^2} = \sqrt{1+9} = √10
BC = \sqrt{(-1-2)^2+(6+3)^2} = \sqrt{(-3)^2+9^2} = \sqrt{9+81} = √90 = 3√10
AC = \sqrt{(-1-1)^2 + (6-0)^2} = \sqrt{(-2)^2 +6^2} = \sqrt{4+36} =√40 = 2√10
AB + AC = √10 + 2√10 = 3√10 = BC
∴ AB + AC = BC ⇒ B-A-C
⇒ A (1, 0), B (2, -3) 3 C (1, -6) ବିଦୁ୍ୟତ୍ରୟଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ହେବେ ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 13.
x ଅକ୍ଷ ଉପରେ ଏକ ବିନ୍ଦୁର ସ୍ଥାନଙ୍କ ସ୍ଥିର କର ଯାହା (5, 4) ଓ (-2, 3) ସ୍ଥାନାଙ୍କ ବିଶିଷ୍ଟ ବିନ୍ଦୁଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ହେବ ।
ସମାଧାନ :
ମନେକର x ଅକ୍ଷ ଉପରେ ଅବସ୍ଥିତ ବିନ୍ଦୁଟି A (x,0) ।
ଦିଉ ବିଦୁ୍ଦ୍ଵୟ B (5, 4) ଓ C (- 2, 3) ।
AB² = (5 – x)² + (4 – 0)² = (5 – x)² + 16
AC² = (-2 – x)² + (3 – 0)² = (2 + x)² + 9
ପ୍ରଶ୍ମାବସାକ (2 + x)² + 9 = (5 – x)² + 16 ⇒ 4 + 4x + x² + 9 = 25 – 10x + x² + 16
= 4x + x² + 13 = 41 – 10x + x²
⇒ 14x = 28 ⇒ x =2
= 4x + 10x + x² – x = 41 – 13
∴ x ଅକ୍ଷ ଉପରେ ଅବସ୍ଥିତ ବିନ୍ଦୁଟିର ସ୍ଥାନାଙ୍କ (2, 0) ।

Question 14.
ଯଦି O (0, 0), A (1, 2), B (3, 8) ଏବଂ C (3, – 1) ହୁଏ, ତେବେ ଦର୍ଶାଅ ଯେ, AB = 2CO ।
ସମାଧାନ :
ଦର ବିନ୍ଦୁ ଚାରୋଟି O(0, 0), A(1, 2), B (3, 8) ଏବଂ C (3, – 1) ।
AB = \sqrt{(3-1)^2+(8-2)^2} = \sqrt{2^2+6^2} = \sqrt{4+36} = √40 = 2√10
CO = \sqrt{3^2 +(-1)^2} = \sqrt{9+1} = √10
∴ AB = 2√10 = 2CO ⇒ AB = 2CO ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 15.
ଗୋଟିଏ ସମବାହୁ ତ୍ରିଭୁଜର ଦୁଇ ଶୀର୍ଷବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (0, 3) ବିନ୍ଦୁ (4, 3) ହେଲେ, ତୃତୀୟ ଶୀର୍ଷବିନ୍ଦୁର ସ୍ଥାନଙ୍କ ସ୍ଥିର କର ।
ସମାଧାନ :
ନିମ୍ନସ୍ଥ ଚିତ୍ରରେ ABC ଏକ ସମବାହୁ ତ୍ରିଭୁଜ ।
ଏହାର ଶୀର୍ଷବିନ୍ଦୁ B (0, 3) ଏବଂ C (4, 3) । BC ସମବାହୁ ତ୍ରିଭୁଜର ବାହୁ ।
BC = \sqrt{(4-0)^2 +(3-3)^2} = \sqrt{16+0} = √16 = 4 ଏକକ।
ଶୀର୍ଷବିନ୍ଦୁ Aରୁ BC ପ୍ରତି ଅଙ୍କିତ ଲମ୍ବ AD x ଅକ୍ଷକୁ P ବିନ୍ଦୁରେ ଛେଦ କରୁ ।
∴ D ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{4+0}{2}, \frac{3+3}{2}) = (2,3)

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(b)

Question 1.

Using the ε – δ definition prove that
(i) \lim _{x \rightarrow 0} (2x + 3) = 3
Solution:
Let f(x) = 2x + 3
Here a = 0 and = 3
Let ε be any positive real number however small it may be.
Now |f(x) – ℓ| =| 2x + 3 – 3| =|2x|
Thus |f(x) – | < ε whenever
|2x| < ε i.e |x| < \frac{\varepsilon}{2}
Then |f(x) – ℓ| < ε
whenever |x – 0| < δ
under the condition δ = \frac{\varepsilon}{2}
\lim _{x \rightarrow 0} (2x + 3) = 3

(ii) \lim _{x \rightarrow 1} (2x – 1) = 1
Solution:
Here f(x) = 2x – 1, = l and a = 1
Now |f(x)| = | 2x – 1 – 1|
= |2x – 2| = 2|x – 1|
Thus |f(x) –  ℓ| < ε
whenever 2|x – 1| < ε
i,e. |x – 1| < \frac{\varepsilon}{2} put δ = \frac{\varepsilon}{2}
Then |f(x) – ℓ| < ε
whenever|x – 1| < δ
Hence \lim _{x \rightarrow 1} (2x – 1) = 1

(iii) \lim _{x \rightarrow -2} (3x + 8) = 2
Solution:
|(3x + 8) – 2|
= |3x + 6| = 3|x + 2|
So |3x + 8 – 2| < ε
whenever 3|x + 2| < ε
i.e. |x + 2| < \frac{\varepsilon}{3}
Hence |(3x + 8) – 2| < ε
whenever | x + 2 | < δ
\lim _{x \rightarrow -2} (3x + 8) = 2

(iv) \lim _{x \rightarrow 3} (x2 + 2x – 8) = 7
Solution:
|(x2 + 2x – 8) – 7|
= |(x2 + 2x – 15|
= |(x + 5) (x – 3)|
=| x + 5| | x – 3|
If |x – 3| < 1 then| x + 5| =| x – 3 + 8| < |x – 3| + 8 < 9
Thus |(x2 + 2x – 8) – 7| < 9 |x – 3|
So |(x2 + 2x – 8) – 7| < ε
whenever 9|x – 3| < ε
i.e.| x – 3| < \frac{\varepsilon}{9}
Choose δ = minimum of 1 and \frac{\varepsilon}{9}
Then |(x2 + 2x – 8) – 7| < ε
whenever |x – 3| < δ
\lim _{x \rightarrow 3} (x2 + 2x – 8) = 7   (proved)

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b)

(v) \lim _{x \rightarrow 9} √x = 3
Solution:
|√x – 3| = |\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}+3}|
= \frac{|x-9|}{|\sqrt{x}+3|}
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b)

(v) \lim _{x \rightarrow a} √x = √a, a > 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 1

(vii) \lim _{x \rightarrow 1} |3x + 2| = 5
Solution:
When x → 1, 3x + 2 is always positive.
So |3x + 2| = 3x + 2
Thus ||3x + 2| -5| = |3x + 2 – 5|
= 3|x – 1|
∴ ||3x + 2| – 5 | < ε
whenever 3|x – 1| < ε
i.e. |x – 1| < \frac{\varepsilon}{3}
put δ = \frac{\varepsilon}{3}
Hence ||3x + 2| – 5| < ε
whenever |x – 1| < δ
\lim _{x \rightarrow 1} |3x + 2| = 5

(viii) \lim _{x \rightarrow 2} |5x – 7| = 3
Solution:
Let any arbitrary ε > 0
then |5x – 7 – 3| < ε
If |5(x – 2)| < ε
i.e. if lx – 2| < \frac{\varepsilon}{5}
Choosing δ = \frac{\varepsilon}{5} we have
for any arbitrary ε > 0 there exists a δ > 0 depending on ε
Such that
|x – 2| < δ ⇒ |(5x – 7) – 3| < ε
\lim _{x \rightarrow 2} |5x – 7| = 3

Question 2.
If \lim _{x \rightarrow a} f(x) = ℓ then prove that \lim _{x \rightarrow a} |f(x)| = | ℓ | Is the converse true ? Justify your answer with reasons.
Solution:
Let \lim _{x \rightarrow a} f(x) = ℓ
Then |f(x) – ℓ| < ε whenever |x – a| < δ
Now |f(x)| – ℓ| < |f(x) – ℓ| < ε
whenever |x – a| < δ
So \lim _{x \rightarrow a} |f(x)| = | ℓ |
The converse is not always true because | ℓ | = | -ℓ |
So \lim _{x \rightarrow a} f(x) = ℓ or -ℓ

Question 3.
(i) Prove that \lim _{x \rightarrow a} x = a
Solution:
Let ε is any positive number
Let f(x) = x
Now |f(x) – a| < ε
if |x – a| < ε
Choosing δ = ε we see that for each ε > 0 we find a δ > 0 depending on ε such that
|x – al < d ⇒ |f(x) – a| < ε
\lim _{x \rightarrow a} f(x) = a i,e. \lim _{x \rightarrow a} x = a

(ii) Using (i) and the laws of limits prove that \lim _{x \rightarrow a} x^n=a^n, when n is an integer.
Solution:
Case-1: Let n > 0 and n ε z
Now \lim _{x \rightarrow a} x^n=\lim _{x \rightarrow a} (x. x. x…….. n factors)
= a. a …… n factors = an
Case-2: Let n = 0
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 2

(iii) Using (ii) and the laws of limits prove that \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1} where n is an integer.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 3
Case-3: n = 0  Hence the case is obvious

(iv) Using (iii), the laws of limits and assuming that \lim _{x \rightarrow a} \frac{1}{x^m}=a^{\frac{1}{m}} where m is a non-zero integer prove that for any rational number n, \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 4
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 5

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b)

Question 4.
Evaluate the following :
(i) \lim _{x \rightarrow 1} (1 + 2x – 3x2 + 4x3 – 5x4)
Solution:
\lim _{x \rightarrow 1} (1 + 2x – 3x2 + 4x3 – 5x4)
= 1 + 2 – 3 + 4 – 5 = 7 – 8 = -1

(ii) \lim _{x \rightarrow 0} (3x2 + 4x – 1)(x4 + 2x3 – 3x2 + 5x + 2)
Solution:
\lim _{x \rightarrow 0} (3x2 + 4x – 1)(x4 + 2x3 – 3x2 + 5x + 2)
=(-1). 2 = -2

(iii) \lim _{x \rightarrow 2} \frac{x^2+3 x-9}{x+1}
Solution:
\lim _{x \rightarrow 2} \frac{x^2+3 x-9}{x+1}
\frac{2^2+3 \cdot 2-9}{2+1}=\frac{1}{3}

(iv) \lim _{x \rightarrow 3} \frac{x^2-9}{x-3}
Solution:
\lim _{x \rightarrow 3} \frac{x^2-9}{x-3}
= \lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{x-3}
= \lim _{x \rightarrow 3} (x + 3) = 3 + 3 = 6

(v) \lim _{x \rightarrow 1} \frac{x^3-1}{x-1}
Solution:
\lim _{x \rightarrow 1} \frac{x^3-1}{x-1}
= \lim _{x \rightarrow 1} \frac{(x-1)\left(x^2+x+1\right)}{x-1}
= \lim _{x \rightarrow 3} (x2 + x + 1)
= 1 + 1 +1 = 3

(vi) \lim _{x \rightarrow 2} \frac{x-2}{x^4-16}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 6

(vii) \lim _{x \rightarrow 2} \frac{x^3-8}{x^5-32}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 7

(viii) \lim _{x \rightarrow 3} \frac{x^2+2 x-15}{x^2-x-6}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 8

(ix) \lim _{x \rightarrow 0} \frac{(3+x)^3-27}{x}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 9

(x) \lim _{x \rightarrow 2} \frac{\frac{1}{x^2}-\frac{1}{4}}{x-2}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 10

(xi) \lim _{x \rightarrow 1} \frac{1}{(x-1)}\left\{\frac{1}{x+3}-\frac{2}{3 x+5}\right\}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 11

(xii) \lim _{h \rightarrow 0} \frac{(x+h)^3-x^3}{h}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 12

(xiii) \lim _{h \rightarrow 0} \frac{(x+h)^4-x^4}{h}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 13

(xiv) \lim _{x \rightarrow 1} \frac{x^m-1}{x^n-1}, where m, n are integers.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 14

(xv) \lim _{x \rightarrow 1} \frac{x^2-2 x+1}{x^2-x}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 15

(xvi) \lim _{x \rightarrow 1} \frac{x^2+x-2}{x^3-x^2-x+1}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 16

Question 5.
Evaluate the following :
(i) \lim _{x \rightarrow \infty} \frac{2 x+1}{3 x-2}
Solution:
\lim _{x \rightarrow \infty} \frac{2 x+1}{3 x-2}
= \lim _{x \rightarrow \infty} \frac{2+\frac{1}{x}}{3-\frac{2}{x}}=\frac{2}{3}
[ ∵ As x → ∞, \frac{1}{x} → 0]

(ii) \lim _{x \rightarrow \infty} \frac{3 x^2+x-1}{2 x^2-7 x+5}
Solution:
\lim _{x \rightarrow \infty} \frac{3 x^2+x-1}{2 x^2-7 x+5}
=\lim _{x \rightarrow \infty} \frac{3+\frac{1}{x}-\frac{1}{x^2}}{2-\frac{7}{x}+\frac{5}{x^2}}=\frac{3}{2}

(iii) \lim _{x \rightarrow \infty} \frac{x^3+2 x^2+3}{x^4-3 x^2+1}
Solution:
\lim _{x \rightarrow \infty} \frac{x^3+2 x^2+3}{x^4-3 x^2+1}
\lim _{x \rightarrow\infty}\frac{\frac{1}{x}+\frac{2}{x^2}+\frac{3}{x^4}}{1-\frac{3}{x^2}+\frac{1}{x^4}}=\frac{0}{1} =0
[ ∵ As x → ∞, \frac{1}{x} → 0]

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b)

(iv) \lim _{x \rightarrow \infty} \frac{x^4-5 x+2}{x^3-3 x+1}
Solution:
\lim _{x \rightarrow \infty} \frac{x^4-5 x+2}{x^3-3 x+1}
\lim _{x \rightarrow \infty} \frac{x-\frac{5}{x^2}+\frac{2}{x^3}}{1-\frac{3}{x^2}+\frac{1}{x^3}} = ∞

(v) \lim _{x \rightarrow \infty}\left(\frac{x^3}{2 x^2-1}-\frac{x^2}{2 x+1}\right)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 17

(vi) \lim _{n \rightarrow \infty} \frac{n}{n+1}
Solution:
\lim _{n \rightarrow \infty} \frac{n}{n+1}
= \lim _{n \rightarrow \infty} \frac{n}{1+\frac{1}{n}} = 1

(vii) \lim _{n \rightarrow \infty}\left(\frac{n^2+n+1}{5 n^2+2 n+1}\right)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 18

(viii) \lim _{n \rightarrow \infty}\left(\frac{\sqrt{n}-1}{\sqrt{n}+1}\right)
Solution:
\lim _{n \rightarrow \infty}\left(\frac{\sqrt{n}-1}{\sqrt{n}+1}\right)
= \lim _{n \rightarrow \infty} \frac{1-\frac{1}{\sqrt{n}}}{1+\frac{1}{\sqrt{n}}} = 1

(ix) \lim _{n \rightarrow \infty}\left(\frac{6 n^5+2 n+1}{n^5+n^4+3 n^3+2 n^2+n+1}\right)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 19

(x) \lim _{n \rightarrow \infty} \frac{1+2+3+\cdots+n}{n^2}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 20

(xi) \lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 21

(xii) \lim _{n \rightarrow \infty} \frac{1^3+2^3+3^3+\ldots+n^3}{n^4}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 22

(xiii)  \lim _{n \rightarrow \infty} \frac{1+\frac{1}{2}+\frac{1}{2^2}+\ldots+\frac{1}{2^n}}{1+\frac{1}{3}+\frac{1}{3^2}+\ldots \frac{1}{3^n}}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 23

(xiv) \lim _{n \rightarrow \infty} \frac{\lfloor n}{\mid n+1-\lfloor n}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 24

Question 6.
Examine the existence of the following limits :
(i) \lim _{x \rightarrow \sqrt{3}} [x]
Solution:
L.H.L. = \lim _{x \rightarrow \sqrt{3}-} [x] = \lim _{h \rightarrow 0} [√3 – h] = 1
R.H.L. = \lim _{x \rightarrow \sqrt{3}+} [x] = \lim _{h \rightarrow 0} [√3 + h] = 1
Thus L.H.L., R.H.L both
exist and L.H.L. = R.H.L.
So the limit exists and its value is 1.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b)

(ii) \lim _{x \rightarrow 0}[x]
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 25

(iii) \lim _{x \rightarrow-2}[x]
Solution:
L.H.L. = \lim _{x \rightarrow-2-} \frac{x-2}{|x-2|}
= \lim _{h \rightarrow 0}[-2 – h] = -3
R.H.L. \lim _{x \rightarrow-2+} [x] = \lim _{h \rightarrow 0}[-2 + h] = -2
Thus L.H.L. ≠ R.H.L.
So the limit does not exist.

(iv) \lim _{x \rightarrow 0} \frac{|x|}{x}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 26

(v) \lim _{x \rightarrow 2} \frac{x-2}{|x-2|}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 27

(vi) \lim _{x \rightarrow \frac{1}{2}} \frac{|2 x-1|}{2 x-1}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 28

(vii) \lim _{x \rightarrow 1}[2 x+3]
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 29

(viii) \lim _{x \rightarrow \infty} \frac{x}{[x]}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 30

(ix) \lim _{x \rightarrow \infty} \frac{x^2-x}{\left[x^2-x\right]}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 31

(x) \lim _{x \rightarrow 1} \frac{\left|x^2-3 x+2\right|}{x^2-3 x+2}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 32

(xi) \lim _{x \rightarrow \infty}(-1)^{[x]}
Solution:
\lim _{x \rightarrow \infty}(-1)^{[x]}
[Put n ≤ n + 1,As n→ ∞, x → ∞
= \lim _{x \rightarrow \infty}(-1)^n [ [x] = n
= ± 1 [If n is odd, (-1)n = – 1 and if n is even (-1)n = 1 ]
We know that whenever the limit exists it must be unique.
So \lim _{x \rightarrow \infty}(-1)^{[x]} does not exist.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b)

(xii) \lim _{x \rightarrow \infty} \sin x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 33

(xiii) \lim _{x \rightarrow \infty} \cos x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 34

(xiv) \lim _{x \rightarrow 0} \cos \frac{1}{x}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 35

(xv) \lim _{x \rightarrow \infty} \sin \frac{1}{x}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 36

(xvi) \lim _{x \rightarrow 1} f(x) \text { if } f(x)= \begin{cases}2 x-1, & x \leq 1 \\ 2 x+1, & x>1\end{cases}
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 37

(xvii) \lim _{x \rightarrow 0} f(x) \text { and } \lim _{x \rightarrow 1} f(x)
if f(x)=\left\{\begin{array}{l} 0 . x \leq 0 \\ 1-2 x, 0<x \leq 1 \\ 3-4 x, x>1 \end{array}\right.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(b) 38

Question 7.
Let f(x) = {1 if x is rational, 0 if x is irrational then show that \lim _{x \rightarrow a} f(x) does not exist for any a ∈ R.
Solution:
Let x → a through rational numbers.
Then \lim _{x \rightarrow a} f(x) = 1
If x → a through rational numbers.
Then \lim _{x \rightarrow a} f(x) = 0
Thus \lim _{x \rightarrow a} f(x) does not exist.

BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a)

Odisha State Board BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 9 Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a)

Question 1.
ଭୁଲ୍ ଥିଲେ (F) ଓ ଠିକ୍ ଥିଲେ (T) ଲେଖ ।

(i) -1 ଦ୍ଵାରା -201 ବିଭାଜ୍ୟ ।
ସମାଧାନ:
T

(ii) 1 ଦ୍ଵାରା 0 ବିଭାଜ୍ୟ ।
ସମାଧାନ:
T

(iii) 0 ଦ୍ଵାରା 5 ବିଭାଜ୍ୟ ।
ସମାଧାନ:
F

(iv) ପୂର୍ଣ୍ଣସଂଖ୍ୟା ପରିମେୟ ନୁହେଁ ।
ସମାଧାନ:
F

(v) -5 < -3
ସମାଧାନ:
T

(vi) 0.9 ଏକ ପରିମେୟ ସଂଖ୍ୟା ।
ସମାଧାନ:
T

(vii) 0 ଏକ ଗଣନ ସଂଖ୍ୟା ।
ସମାଧାନ:
F

(viii) -1/2 ଏକ ପରିମେୟ ସଂଖ୍ୟା ।
ସମାଧାନ:
T

(ix) a, b ∈ N ହେଲେ ab ∈ N ।
ସମାଧାନ:
T

(x) a, b ∈ N ହେଲେ a – b ∈ N ।
ସମାଧାନ:
F

(xi) a, b ∈ N ହେଲେ a – b ∈ Z ।
ସମାଧାନ:
T

(xii) a, b ∈ Z ହେଲେ a/b ∈ Z Q ।
ସମାଧାନ:
F

BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a)

Question 2.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।

(i) \frac{1}{2} ଯୋଗାତ୍ମକ ବିଲୋମୀ । _____ ।
ସମାଧାନ:
-\frac{1}{2}

(ii) -7ରଗୁଣନାତ୍ମକ ବିଲୋମୀ ______ ।
ସମାଧାନ:
-\frac{1}{7}

(iii) _____ ତା’ ନିଜର ଯୋଗାତ୍ମକ ବିଲୋମୀ ।
ସମାଧାନ:
0

(iv) _____ ତା’ ନିଜର ଯୋଗାତ୍ମକ ବିଲୋମୀ ।
ସମାଧାନ:
1, -1

(v) ପୂର୍ବ ସଂଖ୍ୟା ସେଟ୍‌ରେ ଯୋଗାତ୍ମକ ଅଭେଦ ______
ସମାଧାନ:
0

(vi) ଯୁଗ୍ମ ଓ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଯୋଗଫଳ _____
ସମାଧାନ:
ଅଯୁଗ୍ମ

(vii) _____ ଏକମାତ୍ର ଯୁଗ୍ମ ମୌଳିକ ସଂଖ୍ୟା ଅଟେ ।
ସମାଧାନ:
2

(viii) ସର୍ବନିମ୍ନ ଅଯୁଗ୍ମ ମୌଳିକ ସଂଖ୍ୟାଟି _____ ଅଟେ ।
ସମାଧାନ:
3

(ix) ଗୁଣନ ପ୍ରକ୍ରିୟା _____ ପ୍ରକ୍ରିୟାକୁ ବଣ୍ଟନ କରେ ।
ସମାଧାନ:
ପୋଗ

(x) ପ୍ରକ୍ରିୟାକୁ ବଣ୍ଟନ କରେ । ଯୋଗାତ୍ମକ ବିଲୋମୀ ଉପାଦାନକୁ ମିଶାଇଲେ _____ ପକ ଅଟେ ।
ସମାଧାନ:
0

(xi) N ∩ N * = _____
ସମାଧାନ:
N

(xii) ସେଟ୍‌ରେ – 1ର ଗୁଣନାତ୍ମକ ବିଲୋମୀ _____ ।
ସମାଧାନ:
-1

Question 3.
ନିମ୍ନଲିଖତ ପ୍ରତ୍ୟେକ ପ୍ରଶ୍ନପାଇଁ ପ୍ରଦତ୍ତ ସମ୍ଭାବ୍ୟ ଉତ୍ତରରୁ ଠିକ୍ ଉତ୍ତରଟିକୁ ବାଛ ।

(i) n, m ∈ Z ହେଲେ ନିମ୍ନଲିଖ ମଧ୍ୟରୁ କେଉଁଟି ଅସତ୍ୟ ?
(a) m + n ∈ Z
(b) m – n ∈ Z
(c) m × n ∈ Z
(d) n ÷ m ∈ Z
ସମାଧାନ:
n + m ∈ Z

(ii) Z ସେଟ୍‌ରେ କେଉଁଟି ସତ୍ୟ ?
(a) ଯୋଗାତ୍ମକ ଅଭେଦ 0
(b) ଯୋଗାତ୍ମକ ଅଭେଦ 1
(c) ଗୁଣନାତ୍ମକ ଅଭେଦ 0
(d) ଯୋଗାତ୍ମକ ବିଲୋମୀ (−1)
ସମାଧାନ:
ଯୋଗାତ୍ମକ ଅଭେଦ 0

(iii) ନିମ୍ନଲିଖ ମଧ୍ୟରୁ କେଉଁଟି ସତ୍ୟ ?
(a) ସବୁଠାରୁ କ୍ଷୁଦ୍ରତମ ମୌଳିକ ସଂଖ୍ୟାଟି 3 (b)
(b) ଦୁଇଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଗୁଣଫଳ ଅଯୁଗ୍ମ
(c) ଦୁଇଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଗୁଣଫଳ ଅଯୁଗ୍ମ (d)
(d) ଦୁଇଟି ମୌଳିକ ସଂଖ୍ୟାର ଗୁଣଫଳ ମୌଳିକ
ସମାଧାନ:
ଦୁଇଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଗୁଣଫଳ ଅଯୁଗ୍ମ

(iv) ନିମ୍ନଲିଖ ମଧ୍ୟରୁ କେଉଁଟି ସତ୍ୟ ?
(a) x < y ଓ y < z ହେଲେ x < z
(b) x< y 3 z ∈ Q ହେଲେ xz < yz
(c) x < y ଓ z ∈ Q ହେଲେ x + z < y + z ନ ହୋଇପାରେ ।
(d) ଦୁଇଟି ପରିମେୟ ସଂଖ୍ୟା ମଧ୍ୟରେ ସସୀମ ସଂଖ୍ୟକ ପରିମେୟ ବିଦ୍ୟାମନ ।
ସମାଧାନ:
x < y ଓ y < z ହେଲେ x < z ଓ x < z

(v) ନିମ୍ନଲିଖ ମଧ୍ୟରୁ କେଉଁଟି ଠିକ୍ ?
(a) 0.9999 ……. < 1.0
(b) 1/5 ର ଦଶମିକ ପରିପ୍ରକାଶଟି 0.19999 …….
(c) 1/3 ର ଦଶମିକ ପରିପ୍ରକାଶ
(d) n ଏକ ମୌଳିକ ସଂଖ୍ୟା ହେଲେ 1/n ର ଦଶମିକ ପରିପ୍ରକାଶ ସର୍ବଦା ପୌନଃପୁନିକ ।
ସମାଧାନ:
1/5 ର ଦଶମିକ ପରିପ୍ରକାଶଟି 0.19999 …..

(vi) \frac{1}{2}, \frac{2}{3}, \frac{3}{5}, \frac{4}{7} ମଧ୍ୟରେ ବୃହତ୍ତମ ପରିମେୟ ସଂଖ୍ୟାଟି କେଉଁଟି ?
(a) \frac{1}{2}
(b) \frac{2}{3}
(c) \frac{3}{5}
(d) \frac{4}{7}
ସମାଧାନ:
\frac{2}{3}

(vii) \frac{1}{2}, \frac{2}{3}, \frac{3}{5}, \frac{4}{7} ମଧ୍ୟରେ କ୍ଷୁଦ୍ରତମ ସଂଖ୍ୟା କେଉଁଟି ?
(a) \frac{1}{2}
(b) \frac{2}{3}
(c) \frac{3}{5}
(d) \frac{4}{7}
ସମାଧାନ:
\frac{1}{2}

(viii) 1ର ଯୋଗାତ୍ମକ ବିଲୋମୀ କେଉଁଟି ? 
(a) 1
(b) 0
(c) -1
(d) ଏଥରୁ କୌଣସିଟି ନୁହେଁ
ସମାଧାନ:
-1

BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a)

(ix) ନିମ୍ନଲିଖ ମଧ୍ୟରୁ କେଉଁ ଉଦ୍ଭଟି ଅସତ୍ୟ ?
(a) p ଓ q ମୌଳିକ ହେଲେ ସେମାନଙ୍କର ଗ.ସା.ଗୁ. = 1 ।
(b) p ଓ ୟୁ ଗଣନ ସଂଖ୍ୟା ହେଲେ p + g + pg ଏକ ଗଣନ ସଂଖ୍ୟା ।
(c) p ଓ ୟୁ ମୌଳିକ ସଂଖ୍ୟା ହେଲେ p + q ମଧ୍ୟ ଏକ ମୌଳିକ ସଂଖ୍ୟା ।
(d) p ଏକ ପୂର୍ବ ସଂଖ୍ୟା ଓ ଠୁ ଏକ ଗଣନ ସଂଖ୍ୟା ହେଲେ pg ଏକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟା ।
ସମାଧାନ:
p ଓ ୟୁ ମୌଳିକ ସଂଖ୍ୟା ହେଲେ p + g ମଧ୍ଯ ଏକ ମୌଳିକ ସଂଖ୍ୟା ହେବ ।

Question 4.
ପ୍ରତି ଯୁଗ୍ମ ସଂଖ୍ୟା ଯୌଗିକ ଅଟେ କି ? କାରଣ ସହ ଉତ୍ତର ଦିଅ ।
ସମାଧାନ:
2ର ଯୁଗ୍ମସଂଖ୍ୟା ଯାହା ଏକ ମୌଳିକ ସଂଖ୍ୟା ।
ପ୍ରତି ଯୁଗ୍ମସଂଖ୍ୟା ଯୌଗିକ ନୁହେଁ । କାରଣ 2 ବ୍ୟତୀତ ପ୍ରତି ଯୁଗ୍ମ ସଂଖ୍ୟା ଯୌଗିକ ।

Question 5.
କେଉଁ କେଉଁ ବୀଜଗାଣିତିକ ଧର୍ମଗୁଡ଼ିକ ପୂର୍ବ ସଂଖ୍ୟା ସେଟ୍ Zରେ ସତ୍ୟ, ମାତ୍ର ଗଣନ ସଂଖ୍ୟା ସେଟ୍‌ରେ ସତ୍ୟ ନୁହେଁ ସେଗୁଡ଼ିକ ଲେଖ ।
ସମାଧାନ:
ଯୋଗାତ୍ମକ ଅଭେଦ ଧର୍ମ ଏବଂ ଯୋଗାତ୍ମକ ବିଲୋମୀ ଧର୍ମ ଦ୍ଵୟ ପୂର୍ବସଂଖ୍ୟା ସେଟ୍ Zରେ ସତ୍ୟ ହେଲେ ହେଁ ତାହା ଗଣନ ସଂଖ୍ୟା ସେଟ୍ (N) ରେ ସତ୍ୟ ନୁହେଁ ।

Question 6.
କେଉଁ କେଉଁ ବୀଜଗାଣିତିକ ଧର୍ମଗୁଡ଼ିକ ପରିମେୟ ସଂଖ୍ୟା ସେଟ୍ ( ରେ ସତ୍ୟ, ମାତ୍ର ପୂର୍ବ ସଂଖ୍ୟା ସେଟ୍‌ରେ ଅସତ୍ୟ ସେଗୁଡ଼ିକ ଲେଖ ।
ସମାଧାନ:
ଗୁଣନାତ୍ମକ ବିଲୋମୀ ଧର୍ମ ପରିମେୟ ସଂଖ୍ୟା ସେଟ୍ (Q) ରେ ସତ୍ୟ ହେଲେ ହେଁ ପୂର୍ବସଂଖ୍ୟା ସେଟ୍ (Z) ରେ ଅସତ୍ୟ ।

Question 7.
x ଓ y ଅଯୁଗ୍ମ ହେଲେ ପ୍ରମାଣ କର ଯେ, xy ଅଯୁଗ୍ମ ମାତ୍ର x + y ଯୁଗ୍ମ ।
ସମାଧାନ:
ପ୍ରମାଣ : ମନେକର x ଓ y ଅଯୁଗ୍ମ ସଂଖ୍ୟା ଏବଂ x = 2n + 1 ଓ y = 2m + 1 (m, n ∈ Z) ।
xy = (2n + 1) (2m + 1 ) = 4mn + 2m + 2n + 1
= 2 (2 mn + m + n) + 1 (∴ m, n ∈ Z = 2mn + m + n ∈ Z)
x ଓ y ଅଯୁଗ୍ମ ⇒ xy ଅଯୁଗ୍ମ  … (i)  (ପ୍ରମାଣିତ)
x + y = (2n + 1) + (2m + 1 ) = 2n + 2m + 2
= 2(n + m + 1 ) (∴ m, n ∈ Z = m + n + 1 ∈ Z)
∴ x + y ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟା  … (ii)
x ଓ y ଅଯୁଗ୍ମ = x + y ଯୁଗ୍ମ ।  (ପ୍ରମାଣିତ)
(i) ଓ (ii)ରୁ x ଓ y ଅଯୁଗ୍ମ ହେଲେ xy ଅଯୁଗ୍ମ ଏବଂ x + y ଯୁଗ୍ମ ହେବ ।

Question 8.
ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନେ ଯୋଗ ଜନିତ ସଂବୃତ୍ତି ନିୟମ ପାଳନ କରନ୍ତି କି ? କାରଣ ସହ ଉତ୍ତର ଦିଅ ।
ସମାଧାନ:
ମନେକର ଦୁଇଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟା, x = 2n + 1 ଏବଂ
y = 2n+ 3 (n ∈ Z)
x + y ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା
ଦର୍ଶାଇବାକୁ ପଡ଼ିବ ।
2n + 1 + 2n + 3 = 4n + 4 = 2 (2n + 2)
ଅର୍ଥାତ୍ x + y ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟା ।
ତେଣୁ ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନେ ଯୋଗ ଜନିତ ସଂବୃତ୍ତି ନିୟମ ପାଳନ କରନ୍ତି ନାହିଁ ।

Question 9.
15 ଅପେକ୍ଷା ବୃହତ୍ତର ଓ 100 ଠାରୁ କ୍ଷୁଦ୍ରତର ଯେଉଁ ପୂର୍ଣ ସଂଖ୍ୟାଗୁଡ଼ିକର ସାଧାରଣ ରୂପ 3x2 +  2, n ∈ Z ସେଗୁଡ଼ିକ ଲେଖ ।
ସମାଧାନ:
ଦତ୍ତ ପୂର୍ଣ୍ଣସଂଖ୍ୟାର ସାଧାରଣ ରୂପ 3x? + 2 (n ∈ Z)
ଏଠାରେ ଲର ମାନ 1, 2, 3, ….. ହେଲେ, ଦତ୍ତ ପୂର୍ଣ୍ଣସଂଖ୍ୟାଗୁଡ଼ିକ ଯଥାକ୍ରମେ….
3(1)2 + 2 = 5, 3(2)2 + 2 = 12, 3(3)2 + 2 = 29, 3(4)2 + 2 = 50, 3(5)2 + 2 = 77, 3(6)2 + 2 = 110,
ଏଠାରେ nର ମାନ 3, 4 ଓ 5 ପାଇଁ ପୂର୍ବସଂଖ୍ୟାର ମାନ ଯଥାକ୍ରମେ 20, 50 ଓ 77 ହେବ ଯହା 15 ଓ 100 ମଧ୍ୟସ୍ଥ ପୂର୍ଣ୍ଣସଂଖ୍ୟା । ପ୍ରକାଶ ଥାଉକି nର ମାନ –3, –4 ଓ –5 ମଧ୍ଯ ପୂର୍ବସଂଖ୍ୟାର ମାନଗୁଡ଼ିକ ଯଥାକ୍ରମେ 29, 50, 77 ପ୍ରତ୍ୟେକ 15 ରୁ ବଡ଼ ଏବଂ 100 ରୁ ସାନ ।

BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a)

Question 10.
0.123 123 123 …. ସଂଖ୍ୟାଟି ପରିମେୟ ସଂଖ୍ୟା ହେବ କି ? କାରଣ ସହ ଉତ୍ତର ଦିଅ ।
ସମାଧାନ:
0.123 123 123 …..= 0.123
ପନେକବ x = 0. 123 ⇒ 1000 x = 123, \overline{123}
∴ 1000x -x= 123.0 – 0.0 = 123
⇒ 999x = 123 ⇒ x = \frac{123}{999}=\frac{41}{333}
ବି.ଦ୍ର. : ଏଥରୁ ସ୍ପଷ୍ଟ ଯେ, 0.123 123 123… ବ ପରିମେୟ ବୁପ \frac{41}{333} ହେତୁ ଦତ୍ତ ରାଶିଟି ଏକ ପରିମେୟ ସଂଖ୍ୟା ।

Question 11.
0.131 ସଂଖ୍ୟାକୁ, \frac{p}{q} ପରିମେୟ ରୂପରେ ପ୍ରକାଶ କର ।
ସମାଧାନ:
0.131 = \frac{131}{1000}
0.131କୁ \frac{p}{q} ପରିମେୟ ରୂପରେ ପ୍ରକାଶ କଲେ ହେବ \frac{131}{1000}

Question 12.
\frac{1}{3} ପରିମେୟ ସଂଖ୍ୟାଟିକୁ ଅସରନ୍ତି ପୌନଃପୁନିକ ଦଶମିକ ରୂପେ ଲେଖ ।
ସମାଧାନ:
\frac{1}{3} = 1 ÷ 3 = 0. 333333…….= 0.3
\frac{1}{3} ର ଅସରନ୍ତି ପୌନଃପୁନିକ ଦଶମିକ ରୂପ 0.333333….. ।

Question 13.
\frac{1}{3} ପରିମେୟ ସଂଖ୍ୟାଟିକୁ ଲଘିଷ୍ଠାକୃତି ନ ହୋଇଥବା \frac{100}{q_1}, \frac{p_1}{-102}, \frac{6 \times p_3}{q_3} ରୂପରେ ପ୍ରକାଶ କର ।
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a)

Question 14.
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a) 1
ମେହିପରି \frac{-15}{15}=\frac{-15}{15} = 1 (n = 15 ପାଇଁ)
ପରିମେୟ ସଂଖ୍ୟାମାନ = -15, -7.5, – ….. -1
ଏଠାରେ –15 ଓ –1 ଯଥାକ୍ରମେ କ୍ଷୁଦ୍ରତମ ଏବଂ ବୃହତ୍ତମ ପରିମେୟ ସଂଖା ।

Question 15
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a) 2

Question 16.
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a) 3

Question 17.
ସମାଧାନ:
BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a) 4

Question 18.
(i) 0 . \overline{9} = 1
ସମାଧାନ:
ମନେଳର x = 0 . \overline{9} = 0.999…
⇒ 10 x = 9.999…
∴ 10x – x = 9.9999 …. – 0.999 … = 9
⇒ 9x = 9 ⇒ x = \frac{9}{9} = 1
0 . \overline{9} = 1 (ପ୍ରମାଶିତ)

(ii) 1.29 = 1.3
ସମାଧାନ:
ମନେଳର x = 1.29
⇒ 10x = 12.9 = 12.999 ….
⇒ 100x = 129.9
⇒ 100x – 10x = 129.9 – 12.9
⇒ 90x = 117
⇒ x = \frac{117}{90}=\frac{13}{10} = 1.3
∴ 1.29 = 1.3 (ପ୍ରମାଶିତ)

(iii) 2.349 = 2.35
ସମାଧାନ:
ମନେଳର x = 2.349 ⇒ 100x = 234.9
⇒ 1000 x = 2349.9
∴ 1000 x – 100 x = 2349.9 – 234.9
⇒ 900 x = 2115 ⇒ x = \frac{2115}{900}=\frac{235}{100} = 2.35
∴ 234.9 = 2.35 (ପ୍ରମାଶିତ)

BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a)

Question 19.
(i) 0 . \overline{1}
ସମାଧାନ:
ମନେଳର x = 0 . \overline{1} = 0.11111 ….
⇒ 10x = 1.11…. = 1 . \overline{1}
∴ 10x – x = 1 . \overline{1}0 . \overline{1}
⇒ 9x = 1 ⇒ x = \frac{1}{9}
0 . \overline{1} = \frac{1}{9}

(ii) 0 . \overline{11}
ସମାଧାନ:
ମନେଳର x = 0 . \overline{11} = 0.11111111….
⇒ 100x = 11.11
∴ 100x – x = 11.11 – 0.11
⇒ 99x = 11 ⇒ x = \frac{11}{99}=\frac{1}{9}
0 . \overline{11} \frac{1}{9}

(iii) 0 . \overline{89}
ସମାଧାନ:
ମନେଳର 0 . \overline{89} = 0.898989 …..
⇒ 10x = 8 . 9\overline{89} = 8.9898989….
⇒ 100 x = 89.898989 …..
∴ 100x – x = 89.898989 ….. – 0.898989 ….
⇒ 99 x = 89
⇒ x = \frac{89}{99}

(iv) 0 . \overline{37}
ସମାଧାନ:
ମନେଳର x – 0 . \overline{37} = 0.373737….
⇒ 100x = 37 . \overline{37}
∴ 100x – x = 37.37 – 0.37
⇒ 99x = 37 ⇒ x = \frac{37}{99}
∴ 0.37 = \frac{37}{99}

(v) 0 . \overline{123}
ସମାଧାନ:
ମନେଳର x = 0 . \overline{123} = 0.123123123….
⇒ 10x = 1.23123 …..
⇒ 100x = 12.3123 …
⇒ 1000x = 123 .123
∴ 1000x – x = 123.123 – 0.123
⇒ 999x = 123 ⇒ x = \frac{123}{999}=\frac{41}{333}
0 . \overline{123} = \frac{123}{999}=\frac{41}{333}

(vi) 0 . 32\overline{1}
ସମାଧାନ:
ମନେଳର 0 . 32\overline{1} = 0.3211111….
⇒ 100x = 32\overline{1} ⇒ 1000x = 321\overline{1}
∴ 1000x – 100x = 321\overline{1}32\overline{1}
⇒ 900x = 289 ⇒ x = \frac{289}{900}
0 . 32\overline{1} = \frac{289}{900}

(vii) –0. 5\overline{4}
ସମାଧାନ:
ମନେଳର x = –0. 5\overline{4} = -0.5444….
⇒ 10x = –5\overline{4}
⇒ 100x = –54.\overline{4}
100x – 10x = –54.\overline{4} + 5\overline{4}
⇒ 90x = -49 ⇒ x = \frac{-49}{90}
∴ –0. 5\overline{4} = \frac{-49}{90}

(viii) 6. \overline{89}
ସମାଧାନ:
ମନେଳର x = 6. \overline{89} = 6.8999 ….
⇒ 10x = 68. \overline{9}
⇒ 100x = 689. \overline{9}
∴ 100 x – 10 x = 689. \overline{9}68. \overline{9}
⇒ 90 x = 621
⇒ x = \frac{621}{90}=\frac{69}{10}

(ix) –0 . \overline{12}
ସମାଧାନ:
ମନେଳର –0 . \overline{12} = – 0.12121212….
⇒ 100x = –12 . \overline{12}
∴ 100x – x = –12 . \overline{12} + 0 . \overline{12}
⇒ 99x = -12 ⇒ x = \frac{-12}{99}=\frac{-4}{33}
∴ –0 . \overline{12} = \frac{-4}{33}

(x) 0 . 013\overline{05}
ସମାଧାନ:
ମନେଳର x = 0 . 013\overline{05}
⇒ 1000x = 13. \overline{05}
⇒ 100000 x = 1305.05
⇒ 100000 x – 1000 x = = 1305.05 – 13.05
⇒ 99000 x = 1292
⇒ x = \frac{1292}{99000}=\frac{323}{24750}

Question 20.
ମୂଲ୍ୟ ନିରୂପଣ କର (ପୂର୍ଣ ସଂଖ୍ୟା କିମ୍ବା ଭଗ୍ନ ସଂଖ୍ୟା ରୂପରେ)
(i) 0 . \overline{6} + 0 . \overline{3}
ସମାଧାନ: 
0 . \overline{6} + 0 . \overline{3} = \frac{6}{9}=\frac{3}{9} = \frac{6+3}{9} = \frac{9}{9} = 1
(∵ 0 . \overline{6} = \frac{6-0}{10-1} = \frac{6}{9})

(ii) 0 . \overline{6} + (0 . \overline{3}) × 2
ସମାଧାନ: 
0 . \overline{6} + (0 . \overline{3}) × 2 = \frac{6}{9}-\frac{3}{9} × 2 = \frac{6}{9}-\frac{6}{9} = 0

BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a)

(iii) (0 \cdot \overline{6})^2+(0 \cdot \overline{3})^2+2 \times(0 \cdot \overline{6}) \times(0 \cdot \overline{3})
ସମାଧାନ: 
BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a) 5

(iv) (0 . \overline{6})^2+(0 \cdot \overline{3})^2-2 \times(0 . \overline{6}) \times(0 \cdot \overline{3})+0 . \overline{6}
ସମାଧାନ: 
BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a) 6

(v) (0 \cdot \overline{6})^2-(0 \cdot \overline{3})^2
ସମାଧାନ: 
(0 \cdot 6)^2-(0 \cdot 3)^2=\left(\frac{6}{9}\right)^2-\left(\frac{3}{9}\right)^2=\frac{36}{81}-\frac{9}{81}=\frac{4}{9}-\frac{1}{9}=\frac{4-1}{9}=\frac{3}{9}=\frac{1}{3}

(vi) (0 \cdot \overline{6})^3+(0 \cdot \overline{3})^3+3 \times(0 \cdot \overline{6}) \times(0 \cdot \overline{3})(0 \cdot \overline{6}+0 \cdot \overline{3})
ସମାଧାନ: 
BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a) 7

(vii) (0 . \overline{6})^3-(0 \cdot \overline{3})^3-3 \times(0 \cdot \overline{3}) \times(0 . \overline{6}) \times(0 \cdot \overline{6}-0 . \overline{3})
ସମାଧାନ: 
BSE Odisha 9th Class Maths Solutions Algebra Chapter 2 ବାସ୍ତବ ସଂଖ୍ୟା Ex 2(a) 8

BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3 Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3

Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(i) ଏକ ବୃତ୍ତର କେନ୍ଦ୍ର O, ବୃତ୍ତ ବହିଃସ୍ଥ P କୌଣସି ଏକ ବିନ୍ଦୁ ଏବଂ \overline{\mathrm{PT}} ଉକ୍ତ ବୃତ୍ତର ଏକ ସ୍ପର୍ଶକଖଣ୍ଡ ହେଲେ, m∠OTP = _____ |
(ii) ଏକ ବୃତ୍ତର କେନ୍ଦ୍ର ଠ । ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P ଏବଂ \overline{\mathrm{PX}}\overline{\mathrm{PY}} ଉକ୍ତ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ଦୁଇଟି ସ୍ପର୍ଶକଖଣ୍ଡ । ∠XPY ଏକ ସୂକ୍ଷ୍ମକୋଣ ହେଲେ, ∠XOY ଏକ
କୋଣ ।
(iii) ଏକ ବୃତ୍ତର କେନ୍ଦ୍ର ଠ, ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P ଏବଂ \overline{\mathrm{PT}} ଉକ୍ତ ବୃତ୍ତପ୍ରତି ଏକ ସ୍ପର୍ଶକଖଣ୍ଡ ହେଲେ, m∠TOP + m∠TPO = ____ |
(iv) ଏକ ବୃତ୍ତର କେନ୍ଦ୍ର ଠ, ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P ଏବଂ \overline{\mathrm{PX}}\overline{\mathrm{PY}} ଉକ୍ତ ବୃତ୍ତ ପ୍ରତି ଦୁଇଟି ସ୍ପର୍ଶକ ଖଣ୍ଡ ହେଲେ,
(a) XOP କୋଣ ଓ ………………… କୋଣ ସମପରିମାଣ ବିଶିଷ୍ଟ;
(b) YPO କୋଣ ଓ ………………… କୋଣ ସମପରିମାଣ ବିଶିଷ୍ଟ ।
(v) ଏକ ବୃତ୍ତର କେନ୍ଦ୍ର ୦ ଏବଂ ବ୍ୟାସାର୍ଦ୍ଧ r ଏକକ । ବୃତ୍ତର ସମତଳରେ P ଏକ ବିନ୍ଦୁ ଏବଂ OP ଓ 1 ମଧ୍ଯରେ – ବୃହତ୍ତର ହେଲେ, P ବିନ୍ଦୁରୁ ବୃତ୍ତ ପ୍ରତି ଏକ ସ୍ପର୍ଶକ ଖଣ୍ଡ ଅଙ୍କନ ସମ୍ଭବ ।
(vi) 5 ସେ.ମି. ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଏକ ବୃତ୍ତର କେନ୍ଦ୍ରଠାରୁ 13 ସେ.ମି. ଦୂରରେ ଓ ବୃତ୍ତର ସମତଳରେ ଅବସ୍ଥିତ ଏକ ବିନ୍ଦୁ P ହେଲେ, PT ସ୍ପର୍ଶକଖଣ୍ଡର ଦୈର୍ଘ୍ୟ ………………. ସେ.ମି.|
(vii) କେନ୍ଦ୍ର ୦ ଏବଂ 1 ସେ.ମି. ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଏକ ବୃତ୍ତର ସମତଳରେ ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P ଠାରୁ ଉକ୍ତ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ଏକ ସ୍ପର୍ଶକ ଖଣ୍ଡର ଦୈର୍ଘ୍ୟ t ସେ.ମି. ହେଲେ OP = ………………. ସେ.ମି.|
(viii) ଦୁଇଟି ବହିଃସ୍ପର୍ଶୀ ବୃତ୍ତର (a) ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ ସଂଖ୍ୟା = ………………… ଏକ
(b) ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ ସଂଖ୍ୟା = …………………
(ix) ଦୁଇଟି ଅନ୍ତସ୍ପର୍ଶୀ ବୃତ୍ତର (a) ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ ସଂଖ୍ୟା = ………………..
(b) ତୀର୍ଯ୍ୟକ୍ ସାଧାରଣ ସ୍ପର୍ଶକ ସଂଖ୍ୟା = ……………..
(x) ପରସ୍ପର ବହିଃସ୍ଥ ହୋଇଥ‌ିବା ଦୁଇଟି ଅଣଛେଦୀ ବୃତ୍ତର
(a) ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ ସଂଖ୍ୟା = ……………………
(b) ତୀର୍ଯ୍ୟକ୍ ସାଧାରଣ ସ୍ପର୍ଶକ ସଂଖ୍ୟା = ………………..
(xi) ପରସ୍ପର ବହିଃସ୍ଥ ହୋଇ ନ ଥ‌ିବା ଦୁଇଟି ଅଣଛେଦୀ ବୃତ୍ତର
(a) ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ ସଂଖ୍ୟା = …………….
(b) ତୀର୍ଯ୍ୟକ୍ ସାଧାରଣ ସ୍ପର୍ଶକ ସଂଖ୍ୟା = ……………
(xii) △ABC ର AB =AC । △ABC ର ପରିବୃତ୍ତ ଉପରିସ୍ଥ A ବିନ୍ଦୁରେ ଅଙ୍କିତ ସ୍ପର୍ଶକ ଉପରେ P ଏକ ବିନ୍ଦୁ, ଯେପରି P ଓ B ବିନ୍ଦୁଦ୍ଵୟ \overline{\mathrm{AC}} ର ବିପରୀତ ପାର୍ଶ୍ଵରେ ଅବସ୍ଥିତ ।
m∠PAC = 70° ଦେଲେ, m∠ABC =
(xiii) ଗୋଟିଏ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ୫ ସେ.ମି. ହେଲେ ଏହାର ଦୁଇଟି ସମାନ୍ତର ସ୍ପର୍ଶକ ମଧ୍ୟବର୍ତ୍ତୀ ଦୂରତା ………………. ସେ.ମି.|
(xiv) ଦୁଇଟି ବର୍ହିସ୍ପର୍ଶୀ ବୃତ୍ତର କେନ୍ଦ୍ରଦ୍ଵୟ ମଧ୍ୟବର୍ତ୍ତୀ ଦୂରତା ହେଉଛି ବୃତ୍ତଦ୍ୱୟର ବ୍ୟାସାର୍ଦ୍ଧମାନଙ୍କର ………………. ସଦ୍ ପମାନ |
(xv) ଦୁଇଟି ଅନ୍ତଃସ୍ପର୍ଶୀ ବୃତ୍ତର କେନ୍ଦ୍ରଦ୍ଵୟ ମଧ୍ୟବର୍ତ୍ତୀ ଦୂରତା ହେଉଛି ବୃତ୍ତଦ୍ଵୟର ବ୍ୟାସାର୍ଦ୍ଧମାନଙ୍କର ……………… ସମାନ |
(xvi) ଏକ ସରଳରେଖା ଉପରିସ୍ଥ ଗୋଟିଏ ବିନ୍ଦୁ P ଠାରେ ସରଳରେଖାଟି ସର୍ବାଧିକ ……………… ହୋଇପାରିବ ।
Solution:
(i) 90°
(ii) ସ୍ଥୂଳକୋଣ
(iii) 90°
(iv) (a) YOP (b) XPO
(v) OP
(vi) 12
(vii) \sqrt{\mathrm{r}^2+\mathrm{t}^2}
(viii) (a) 2 (b) 1
(ix) (a)1 (b) 0
(x) (a) 2 (b) 0
(xi) (a) 0 (b) 0
(xii) 70°
(xiii) 16 ସେ.ମି.|
(xiv) ସମପୁ
(xv) ଥନ୍ତ୍ରର
(xvi) ଅସଂଖ୍ୟ

BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3

Question 2.
ଦତ୍ତ ଥ‌ିବା ଉକ୍ତି ଭୁଲ୍‌ଲେ (ଏହାକୁ ଦତ୍ତ ଉକ୍ତିର ନାସ୍ତିବାଚକ ଉକ୍ତି (Negative Statement) ବ୍ୟବହାର ନ କରି) ସଂଶୋଧନ କର ।
(i) r ଏକକ ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଏକ ବୃତ୍ତର L ରେଖା ଏକ ଛେଦକ ହେଲେ, ବୃତ୍ତର କେନ୍ଦ୍ରଠାରୁ L ର ଦୂରତା = r ଏକକ ।
(ii) ଗୋଟିଏ ବୃତ୍ତର କେନ୍ଦ୍ର O ଏବଂ ବୃତ୍ତର ସମତଳରେ ବୃତ୍ତ ବହିଃସ୍ଥ କୌଣସି ଏକ ବିନ୍ଦୁ P | P ବିନ୍ଦୁରୁ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ଏକ ସ୍ପର୍ଶକ ଖଣ୍ଡ \overline{\mathrm{PT}} ହେଲେ △OPT ରେ ∠POT ଏକ ସମକୋଣ ।
(iii) ଗୋଟିଏ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ r ଏକକ । ବୃତ୍ତର ସମତଳରେ ବୃତ୍ତ ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P ଠାରୁ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ ଖଣ୍ଡ \overline{\mathrm{PT}} ର ଦୈର୍ଘ୍ୟ t ଏକକ ଏବଂ ବୃତ୍ତର କେନ୍ଦ୍ର ୦ ଠାରୁ P ର ଦୂରତା d ଏକକ ହେଲେ, d2 + r2 = t2|
(iv) ଏକ ବୃତ୍ତର ସମତଳରେ ବୃତ୍ତ ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ Pରୁ ଉକ୍ତ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ \overline{\mathrm{PT}}; P ବିନ୍ଦୁଗାମୀ ଏକ ଛେଦକ, ବୃତ୍ତଟିକୁ A ଓ B ବିନ୍ଦୁରେ ଛେଦ କରେ, ଯେପରି P – A – B । ତେବେ PT2 = PA × AB |
(v) ଏକ ବୃତ୍ତର ଅନ୍ତଃସ୍ଥ କୌଣସି ଏକ ବିନ୍ଦୁ Q ଠାରୁ ଉକ୍ତ ବୃତ୍ତ ପ୍ରତି ଦୁଇଟି ସ୍ପର୍ଶକଖଣ୍ଡ ଅଙ୍କନ କରାଯାଇ ପାରିବ ।
(vi) ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ବ୍ୟାସାର୍ଦ୍ଧ ଗୋଟିଏ ବୃତ୍ତର ବହିଃସ୍ଥ କେବଳ ଗୋଟିଏ ବିନ୍ଦୁ P ଅଛି, ଯେଉଁଠାରୁ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ଦୈର୍ଘ୍ୟ ବିଶିଷ୍ଟ ହେବ ।
(vii) ଦୁଇଟି ସ୍ପର୍ଶକ ବୃତ୍ତର କେନ୍ଦ୍ରବିନ୍ଦୁ ଦ୍ଵୟ ମଧ୍ୟବର୍ତ୍ତୀ ଦୂରତା ସହ ଉକ୍ତ ବୃତ୍ତଦ୍ଵୟର ବ୍ୟାସାର୍ଦ୍ଧର ସମଷ୍ଟି ସମାନ ହେଲେ, ବୃତ୍ତ ଦ୍ଵୟ ଅନ୍ତଃସ୍ପର୍ଶୀ ହେବେ ।
(viii) ଦୁଇଟି ଅନ୍ତଃସ୍ପର୍ଶୀ ବୃତ୍ତର କେନ୍ଦ୍ରନ୍ବୟର ମଧ୍ୟବର୍ତ୍ତୀ ଦୂରତା, ବୃତ୍ତଦ୍ଵୟର ବ୍ୟାସାର୍ଦ୍ଧ ଦ୍ବୟର ପାର୍ଥକ୍ୟ ସହ ସମାନ ।
(ix) ଦୁଇଟି ବୃତ୍ତ ମଧ୍ୟରୁ ଗୋଟିଏ ଅନ୍ୟଟିର ଅନ୍ତର୍ଦେଶରେ ଅବସ୍ଥିତ ହେଲେ, ସେ ଦୁଇଟି ବୃତ୍ତର ଗୋଟିଏ ମାତ୍ର ସାଧାରଣ ସ୍ପର୍ଶକ ରହିବ ।
(x) ଦୁଇଟି ବିନ୍ଦୁରେ ପରସ୍ପରକୁ ଛେଦ କରୁଥିବା ଦୁଇଟି ବୃତ୍ତର କେବଳ ଗୋଟିଏ ତୀର୍ଯ୍ୟକ ସାଧାରଣ ସ୍ପର୍ଶକ ଥାଏ ।
(xi) ଦୁଇଟି ଅନ୍ତଃସ୍ପର୍ଶୀ ସ୍ପର୍ଶକ ବୃତ୍ତର ସ୍ପର୍ଶବିନ୍ଦୁ, ଅନ୍ତଃସ୍ଥ ବୃତ୍ତର ଏକ ବହିଃସ୍ଥ ବିନ୍ଦୁ ନୁହେଁ ।
(xii) ଦୁଇଟି ବହିଃସ୍ପର୍ଶୀ ସ୍ପର୍ଶକ ବୃତ୍ତର ସ୍ପର୍ଶ ବିନ୍ଦୁ, ଉଭୟ ବୃତ୍ତ ମଧ୍ୟରୁ କୌଣସିଟିର ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ ନୁହେଁ ।
Solution:
(i) <r
(ii) ∠OTP ଟି ∠PTO
(iii) d2 = t2 + r2
(iv) PT2 = PA × PB
(v) ଅନ୍ତର୍ ବଦଲରେ ଦହିମ
(vi) ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଗୋଟିଏ ବୃତ୍ତର ବହିଃସ୍ଥ ଅସଂଖ୍ୟ ବିନ୍ଦୁ P ଅଛି ଯେଉଁଠାରୁ ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ଦୈର୍ଘ୍ୟବିଶିଷ୍ଟ ହେବ ।
(vii) ସମଷ୍ଟି ବଦଳରେ ଅନ୍ତର ହେବ ।
(viii) ପାର୍ଥକ୍ୟ ବଦଳରେ ଅନ୍ତର ହେବ ।
(ix) ଗୋଟିଏ ମାତ୍ର ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ ହେବ ।
(x) ଗୋଟିଏ ତୀର୍ଯ୍ୟକ ବଦଳରେ ଦୁଇଟି ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ ହେବ ।
(xi) ଉଭୟର ଏକ ସାଧାରଣ ବିନ୍ଦୁ ।
(xii) ଉଭୟ ବୃତ୍ତର ଏକ ସାଧାରଣ ବିନ୍ଦୁ ।

Question 3.
ଗୋଟିଏ ବୃତ୍ତର କେନ୍ଦ୍ରବିନ୍ଦୁ O ଏବଂ ବ୍ୟାସାର୍ଦ୍ଧ ୫ ସେ.ମି. । ଉକ୍ତ ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P ଏବଂ PO = 17 ସେ.ମି. ହେଲେ, P ବିନ୍ଦୁରୁ ଉକ୍ତ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡର ଦୈର୍ଘ୍ୟ କେତେ ?
Solution:
ମନେକର S ବୃତ୍ତର କେନ୍ଦ୍ର 80 1
\overline{\mathrm{PT}} ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ।
ଓ ବ୍ୟାସାର୍ଦ୍ଧ (OT) = 8 ସେ.ମି., P ବହିଃସ୍ଥ ବିନ୍ଦୁ S ।
PO = 17 ସେ.ମି.|
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 1
\overline{\mathrm{PT}}\overline{\mathrm{OT}}
POT ସମକୋଣା ତ୍ରିଭୁଜରେ PT = \sqrt{\mathrm{OP}^2-\mathrm{OT}^2}
= \sqrt{17^2-8^2} ସେ.ମି. = \sqrt{289-64} = \sqrt{225} = 15 ସେ.ମି.|
∴ P ବିନ୍ଦୁରୁ ଉକ୍ତ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ pତୁର ଦୈର୍ଘ୍ୟ 15 ସେ.ମି. |

Question 4.
ଦୁଇଟି ବହିଃସ୍ପର୍ଶୀ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ 4.5 ସେ.ମି. ଓ 12.5 ସେ.ମି. । ବୃତ୍ତ ଦ୍ଵୟର ଏକ ସାଧାରଣ ସ୍ପର୍ଶକ ବୃତ୍ତ ଦ୍ଵୟକୁ P ଓ Q ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କଲେ, \overline{\mathrm{PQ}}ର ଦୈର୍ଘ୍ୟ କେତେ ?
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 2

Question 5.
ଦୁଇଟି ଅଣଛେଦୀ ବୃତ୍ତର ଏକ ତୀର୍ଯ୍ୟକ ସାଧାରଣ ସ୍ପର୍ଶକ ବୃତ୍ତ ଦ୍ଵୟକୁ P ଓ ଠୁ ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରନ୍ତି । କେନ୍ଦ୍ରଦ୍ୱୟ ମଧ୍ୟବର୍ତ୍ତୀ ଦୂରତା 20 ସେ.ମି. ଏବଂ ବ୍ୟାସାର୍ଦ୍ଧ ଦ୍ୱୟ 7 ସେ.ମି. ଓ 5 ସେ.ମି. ହେଲେ, PQ କେତେ ସେ.ମି. ?
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 3

BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3

Question 6.
ଚିତ୍ରରେ ବୃତ୍ତର ବହିଃସ୍ଥ ବିନ୍ଦୁ P | P ବିନ୍ଦୁଗାମୀ ଗୋଟିଏ ଛେଦକ ଦର ବୃତ୍ତକୁ A ଓ B ବିନ୍ଦୁରେ ଛେଦ କରେ, ଯେପରିକି P – A – B | P ବିଦୁଗାମୀ ଅନ୍ୟ ଏକ ଛେଦକ ଉକ୍ତ ବୃତ୍ତକୁ C ଓ D ବିନ୍ଦୁରେ ଛେଦ କରେ ଯେପରିକି P – C – D|
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 4
(i) ସ୍ପର୍ଶକ-ସଂପୃକ୍ତ ଉପପାଦ୍ୟ ପ୍ରୟୋଗ କରି ପ୍ରମାଣ କର ।
PA × PB = PC × PD
(ii) PA = 10 ସେ.ମି., PB = 16 ସେ.ମି. ଓ PD = 20 ସେ.ମି. 6ହଲେ, CD କଣ୍ଡଯ କର |
(iii) PA = 8 ସେ.ମି. ଓ AB =10 ସେ.ମି. ହେଲେ, P ବିନ୍ଦୁଗାମୀ ସ୍ପର୍ଶକଖଣ୍ଡର ଦୈର୍ଘ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
Solution:
ବଭ : ABCD ଦୂଇରେ P ଦହିମ ଏକ ଦିନ୍ଦୁ |
P – C – D ଓ P – A – B ଦୁଇଟି ଛେଦନ |
ତ୍ପ।ମାଣ୍ୟ : (i) PA × PB = PC × PD
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 5
ଅକନ : \overline{\text { PT }} ମ୍ଟଣକଖଣ୍ଡ ଅକନ କର |
ପ୍ତମାଣ: (i) PT2 = PA × PB
PT2 = PC × PD
∴ PA × PB = PC × PD

(ii) PC = \frac{PA \times PB}{PD} = \frac{10 \times 16}{20} ସେ.ମି = 8 ସେ.ମି
∴ CD = PD – PC = 20 ସେ.ମି – 8 ସେ.ମି = 12 ସେ.ମି |

(iii) PT2 = PA × PB = PA (PA + AB)
= 8 × (8 + 10) = 8 × 18 ଦଗ ସେ.ମି |

Question 7.
ଚିତ୍ରରେ ଥ‌ିବା ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P | P ବିନ୍ଦୁଗାମୀ ଏକ ଛେଦକ ପୂର୍ବୋକ୍ତ ବୃତ୍ତକୁ A ଓ B ବିନ୍ଦୁରେ ଛେଦ କରେ । ଯେପରି P – A – b | P ବିନ୍ଦୁଗାମୀ ସ୍ପର୍ଶକରଶ୍ମିର ସ୍ପର୍ଶବିନ୍ଦୁ T |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 7
(i) m\overparen{\text { AXT }} = 60° m\overparen{\text { BYT }} = 130° ଛେଦକ, m∠ATP, m∠APT, m∠ATB ଓ m∠BTQ ନିଣ୍ଡୟ କର |
(ii) m∠BTQ = 2m∠ATP ହେଲେ ,ପ୍ରମାଣ କର : (a) BT = TP (b) TA = AP
(iii) PA = 8 ସେ.ମି. ଓ PT = 12 ସେ.ମି. ହେଲେ , AB ନିଣ୍ଟୟ କର |
(iv) PT = 2AP ଏବଂ AB = 18 ସେ.ମି. ହେଲେ , PT ନିଣ୍ଟୟ କର |
(v) PT = 2AP ଏବଂ PB = 24 ସେ.ମି. ହେଲେ , PT ନିଣ୍ଟୟ କର |
Solution:
\overline{\mathrm{B} T}\overline{\mathrm{A} T} ଅଙ୍କନ କର ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 8
(i) m∠ABT = \frac { 1 }{ 2 }m\overparen{\text { AXT }} = \frac { 1 }{ 2 } × 60° = 30°
⇒ m∠ATP = m∠ABT = 30°
m∠BAT = \frac { 1 }{ 2 }m\overparen{\text { BYT }} = \frac { 1 }{ 2 } × 130° = 65°
△ATP ରେ ଦଦ୍ୱିମ m∠BAT = m∠ATP + m∠APT
⇒ m∠APT = m∠BAT – m∠ATP = 65° – 30° = 35°
m∠ATB = 180° – 30° – 65° = 85° (: ATB ଏକ ତ୍ରିଭୁବ)
m∠BTQ = m∠BAT = 65°

(ii) ମନେକର m∠ATP = θ ⇒ m∠ABT = θ
m∠BTQ = 2m∠ATP = 2θ
m∠BAT = m∠BTQ = 2θ
⇒ m∠APT = 2θ – θ = θ
△BTP ରେ m∠TBP = m∠TPB
⇒ BT = TP
△TAP ରେ m∠ATP = m∠APT ଦ୍ରେତୁ AT = AP

(iii) PT2 = PA × PB
⇒ PB = \frac{\mathrm{PT}^2}{\mathrm{PA}} = \frac{12 \times 12}{8} = 18 ସେ.ମି
∴ AB = PB – PA = 18 ସେ.ମି – 8 ସେ.ମି = 10 ସେ.ମି |

(iv) PT2 = PA × PB = PA × (PA + AB)
⇒ (2AP)2 = PA (PA + 18)
\frac{4 \mathrm{AP}^2}{\mathrm{PA}} = PA + 18 ⇒ 4AP – AP = 18 ସେ.ମି.
⇒ 3AP = 18 ସେ.ମି. ⇒ PA = \frac { 18 }{ 3 } ସେ.ମି. = 6 ସେ.ମି.

(v) PT2 = PA × PB
⇒ (2AP)2 × (PA + AB)
⇒ PB = \frac{4 \mathrm{AP}^2}{\mathrm{PA}} = 4AP = 24 ସେ.ମି.
∴ PT = 2AP = \frac{4 \mathrm{AP}}{2} = \frac { 24 }{ 2 } ସେ.ମି. = 12 ସେ.ମି. |

Question 8.
(a) ଦୁଇଟି ବୃତ୍ତ ବହିଃସ୍ପର୍ଶୀ ହେଲେ, ପ୍ରମାଣ କର ଯେ, ଏହାର ତୀର୍ଯ୍ୟକ୍ ସାଧାରଣ ସ୍ପର୍ଶକ ଉପରିସ୍ଥ ଯେକୌଣସି ବିନ୍ଦୁରୁ ବୃତ୍ତଦ୍ଵୟ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ଦ୍ବୟ ସର୍ବସମ ।
(b) ଦୁଇଟି ବୃତ୍ତ ଅନ୍ତସ୍ପର୍ଶୀ ହେଲେ, ପ୍ରମାଣ କର ଯେ, ସେମାନଙ୍କର ସାଧାରଣ ସ୍ପର୍ଶକ ଉପରିସ୍ଥ ଯେକୌଣସି ବିନ୍ଦୁରୁ ଉକ୍ତ ବୃତ୍ତ ଦ୍ଵୟ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ଦ୍ଵୟ ସର୍ବସମ ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 9
Solution:
(a) ଦତ୍ତ : S ଓ S ବୃତ୍ତଦ୍ୱୟ ବହିଃସ୍ପର୍ଶୀ । ସେମାନଙ୍କର ସାଧାରଣ ସ୍ପର୍ଶକ L | L ଉପରିସ୍ଥ M ଏକ ବିନ୍ଦୁ | M ବିନ୍ଦୁରୁ ବୃତ୍ତଦ୍ଵୟ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ ଖଣ୍ଡଦ୍ଵୟ \overline{\mathrm{MN}}\overline{\mathrm{MR}} |
ପ୍ରାମାଣ୍ୟ : MN = MR
ପ୍ରମାଣ : M ବିନ୍ଦୁରୁ S1 ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ସମାନ ।
⇒ MN = MP …(i)
ପୁନଶ୍ଚ M ବିନ୍ଦୁରୁ S2 ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ସମାନ ।
⇒ MR = MP …(ii)
(i) ଓ (ii)ରୁ MN = MR (ପ୍ରମାଣିତ)

(b) ଦତ୍ତ : S1 ଓ S2 ବୃତ୍ତଦ୍ୱୟ ଅନ୍ତଃସ୍ପର୍ଶୀ । ସେମାନଙ୍କର ସାଧାରଣ ସ୍ପର୍ଶକ L । L ଉପରିସ୍ଥ M ଏକ ବିନ୍ଦୁ । M ବିନ୍ଦୁରୁ ବୃତ୍ତଦ୍ଵୟ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ ଖଣ୍ଡଦ୍ଵୟ \overline{\mathrm{MN}}\overline{\mathrm{MR}} |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 10
ପ୍ରାମାଣ୍ୟ : MN = MR
ପ୍ରମାଣ : M ବିନ୍ଦୁରୁ ଅନ୍ତବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ ଖଣ୍ଡଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ସମାନ ।
⇒ MR = MP
ସେହିପରି M ବିନ୍ଦୁରୁ ବହିଃସ୍ଥ ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ
ସ୍ପର୍ଶକଖଣ୍ଡ ଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ସମାନ ।
⇒ MN = MP
(i) ଓ (ii) ରୁ MN = MR (ପ୍ରମାଣିତ)

Question 9.
ପରସ୍ପରଛେଦୀ ଦୁଇଟି ବୃତ୍ତର ଛେଦବିନ୍ଦୁ A ଓ B । \overleftrightarrow{\mathbf{A B}} ଉପରିସ୍ଥ P ଏକ ବିନ୍ଦୁ ଯେପରି A – B – P। ପ୍ରମାଣ କର ଯେ, ବୃତ୍ତଦ୍ଵୟ ପ୍ରତି P ବିନ୍ଦୁରୁ ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ଦ୍ଵୟ ସର୍ବସମ ।
ସମାଧାନ :
ଦତ୍ତ : ଦୁଇଟି ବୃତ୍ତ S1 ଓ S2, ପରସ୍ପରକୁ A ଓ B ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । \overleftrightarrow{\mathbf{A B}} ଉପରିସ୍ଥ ଏକ ବିନ୍ଦୁ P ଠାରୁ ବୃତ୍ତଦ୍ଵୟ ପ୍ରତି \overrightarrow{\mathrm{PQ}}\overrightarrow{\mathrm{PR}} ସ୍ପର୍ଶକ ଅଙ୍କିତ ହୋଇଛି ।
ପ୍ରାମାଣ୍ୟ : PQ = PR
ପ୍ରମାଣ : S1 ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ \overrightarrow{\mathrm{PQ}} |
S1 ବୃତ୍ତର ଏକ ଛେଦକ \overleftrightarrow{\mathbf{P A B}} |
∴ PQ2 = PA.PB …(i)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 11
ସେହିପରି S2 ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ \overrightarrow{\mathrm{PR}}
S2 ବୃତ୍ତର ଏକ ଛେଦକ \overleftrightarrow{\mathbf{P A B}} |
∴ PR2 = PA·PB …(ii)
(i) ଓ (ii) ତି PQ2 = PR2 ⇒ PQ = PR (ପ୍ରମାଣିତ)

Question 10.
ଚିତ୍ରରେ r1 ଓ x2 ଏକକ ବ୍ୟାସାର୍କ ବିଶିଷ୍ଟ ବୃତ୍ତ S1 ଓ S2 ର କେନ୍ଦ୍ର ଯଥାକ୍ରମେ A ଓ B । ଚିତ୍ର (a)ରେ ବୃତ୍ତଦ୍ଵୟର ଗୋଟିଏ ତୀର୍ଯ୍ୟକ୍ ସାଧାରଣ ସ୍ପର୍ଶକ \overline{\mathbf{A B}} କୁ M ବିନ୍ଦୁରେ ଛେଦ କରେ । ପ୍ରମାଣ କର ଯେ,
AM : MB = r1 : r2 |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 12
ଚିତ୍ର (b)ରେ ବୃତ୍ତ ଦ୍ଵୟର ଗୋଟିଏ ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ \overrightarrow{\mathbf{A B}} କୁ M ବିନ୍ଦୁରେ ଛେଦ କରେ, ଯେପରିକି A-B-M | ତ୍ପମାଣ କର ସେ AM : BM = r1 : r2 |
ସମାଧାନ :
ଦତ୍ତ : ଦୁଇଟି ଅଣଛେଦୀ ବୃତ୍ତର କେନ୍ଦ୍ର ଯଥାକ୍ରମେ O1 ଓ O2 |
\overline{\mathrm{PQ}} ଏକ ସାଧାରଣ ତୀର୍ଯ୍ୟକ ସ୍ପର୍ଶକ \overleftrightarrow{\mathrm{PQ}} ଓ O1O2 ର ଛେଦବିନ୍ଦୁ R |
ପ୍ରାମାଣ୍ୟ : \frac{\mathrm{O}_1 \mathrm{R}}{\mathrm{O}_2 \mathrm{R}} = \frac{r_1}{r_2}
ଅଙ୍କନ : \overline{\mathrm{O}_1 \mathrm{P}}\overline{\mathrm{O}_1 \mathrm{Q}} ଅକନ କର |
ପ୍ରମାଣ : O1PR ଓ A O2OR ମଧ୍ୟରେ
m∠O1PR= m∠02QR (ପ୍ତତ୍ୟେକ ସମୟରେ)
m∠O1RP = m∠O∠02RQ (ପ୍ତତ୍ୟେକ ସମୟରେ)
⇒ △O1PR ~ △O2QR (କୋ-କୋ ଗାଦଶ୍ୟ)
\frac{\mathrm{O}_1 \mathrm{R}}{\mathrm{O}_2 \mathrm{R}} = \frac{\mathrm{O}_1 \mathrm{P}}{\mathrm{O}_2 \mathrm{Q}}\frac{r_1}{r_2} (ପ୍ରମାଣିତ)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 13
(b) ଦତ୍ତ : S1 ଓ S2 ବୃତ୍ତଦ୍ୱୟ ଅଣଛେଦୀ । ସେମାନଙ୍କ କେନ୍ଦ୍ର ଯଥାକ୍ରମେ A ଓ B |
\overline{\mathrm{PQ}} ସେମାନଙ୍କର ଏକ ସାଧାରଣ ସ୍ପର୍ଶକ ।
\overrightarrow{\mathrm{PQ}}\overrightarrow{\mathrm{AB}} ର ଛେଦ ବିନ୍ଦୁ M, A – B – M |
ପ୍ରାମାଣ୍ୟ : \frac { AM }{ BM } = \frac{r_1}{r_2}
ଅଙ୍କନ : \overline{\mathrm{AP}}\overline{\mathrm{BQ}} ଅଙ୍କନ କର ।
ସ୍ତମାଣ: m∠APM = 90°
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 14
(\overline{\mathrm{PM}}, S1 ବୃତ୍ତର ସ୍ପର୍ଶକ ହେତୁ)
ସେହିପରି m∠BQM = 90°
⇒ m∠APM = = m∠BQM
m∠PMA = m∠QMB (ପାଧାରଣ କୋଣ)
⇒ △APM ~ △BQM (କୋ-କୋ ଗାଦଶ୍ୟ)
\frac { AM }{ BM } = \frac{r_1}{r_2}

BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3

Question 11.
ଗୋଟିଏ ବୃତ୍ତରେ \overline{\mathrm{PQ}}\overline{\mathrm{PR}} ଜ୍ୟା ଦ୍ଵୟ ସର୍ବସମ । ପ୍ରମାଣ କର ଯେ, ବୃତ୍ତ ପ୍ରତି P ବିନ୍ଦୁରେ ଅଙ୍କିତ ସ୍ପର୍ଶକ, \overline{\mathrm{QR}} ସହ ସମାନ୍ତର ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 15
Solution:
ଦତ୍ତ : ବୃତ୍ତର \overline{\mathrm{PQ}}\overline{\mathrm{PR}} ଦୁଇଟି ସର୍ବସମ ଜ୍ୟା ।
P ବିନ୍ଦୁରେ ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ \overleftrightarrow{M N} |
ପ୍ରାମାଣ୍ୟ : \overleftrightarrow{M N} || \overline{\mathrm{QR}}
ପ୍ରମାଣ : m∠MPQ = m∠PRQ (ଏକାନ୍ତର ଚାପାନ୍ତର୍ଲିଖ କୋଣ)
PQ = PR ⇒ m∠PQR = m∠PRQ
∴ m∠MPQ = m∠PQR
କିନ୍ତୁ ଏମାନେ ଏକାନ୍ତର ହେତୁ \overleftrightarrow{M N} || \overline{\mathrm{QR}} | (ପ୍ରମାଣିତ)

Question 12.
ଦୁଇଟି ଏକକେନ୍ଦ୍ରିକ ବୃତ୍ତ ମଧ୍ୟରୁ ଗୋଟିକର ଏକ ଜ୍ୟା \overline{\mathrm{AB}} ଅନ୍ୟ ବୃତ୍ତକୁ P ବିନ୍ଦୁରେ ସ୍ପର୍ଶକଲେ, ପ୍ରମାଣ କର ଯେ P ବିନ୍ଦୁରେ \overline{\mathrm{AB}} ସମଦ୍ଵିଖଣ୍ଡିତ ହୁଏ ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 16
ଦତ୍ତ : S1 ଓ S2 ଦୁଇଟି ଏକକେନ୍ଦ୍ରିକ ବୃତ୍ତ । S1 ବୃତ୍ତର ଜ୍ୟା \overline{\mathrm{AB}},
ଯାହା S2 ବୃତ୍ତପ୍ରତି M ବିନ୍ଦୁରେ ସ୍ପର୍ଶକ ଅଟେ ।
ପ୍ରାମାଣ୍ୟ : \overline{\mathrm{AB}} ଜ୍ୟା M ବିନ୍ଦୁରେ ସମଦ୍ବିଖଣ୍ଡିତ ହେବ । ଅର୍ଥାତ୍ AM = MB |
ଅଙ୍କନ : \overline{\mathrm{OM}} ଅଙ୍କନ କର ।
ପ୍ରମାଣ : S2 ବୃତ୍ତର ‘M’ ସ୍ପର୍ଶବିନ୍ଦୁ । \overline{\mathrm{OM}} ସ୍ପର୍ଶବିନ୍ଦୁଗାମୀ ବ୍ୟାସାର୍ଦ୍ଧ । ⇒ \overline{\mathrm{OM}}\overline{\mathrm{AB}}
ପୁନଶ୍ଚ S2 ବୃତ୍ତର \overline{\mathrm{AB}} ଜ୍ୟା ପ୍ରତି \overline{\mathrm{OM}} ଲମ୍ବ ହେତୁ M, \overline{\mathrm{AB}} ର ମଧ୍ୟବିନ୍ଦୁ ।
ଅର୍ଥାତ୍ \overline{\mathrm{AB}} କ୍ୟାଟି M ବିନ୍ଦୁରେ ସମଦ୍ଵିଖଣ୍ଡିତ ହେବ । (ପ୍ରମାଣିତ)

Question 13.
ପ୍ରମାଣ କର ଯେ, ବୃତ୍ତର ଦୁଇ ସମାନ୍ତର ସ୍ପର୍ଶକର ସ୍ପର୍ଶବିନ୍ଦୁ ଦ୍ଵୟର ସଂଯୋଜକ ରେଖାଖଣ୍ଡ ଉକ୍ତ ବୃତ୍ତର ଏକ ବ୍ୟାସ ।
Solution:
ଦତ୍ତ : S ବୃତ୍ତର ଠ କେନ୍ଦ୍ର ।
\overline{\mathrm{AB}} || \overline{\mathrm{CD}} ଏବଂ \overline{\mathrm{AB}}\overline{\mathrm{CD}} ବୃତ୍ତ Sର ଦୁଇଟି ସ୍ପର୍ଶକ ।
ସେମାନଙ୍କ ସ୍ପର୍ଶବିନ୍ଦୁ ଯଥାକ୍ରମେ E ଓ F |
ପ୍ରାମାଣ୍ୟ : E-O-F ଏକ ରେଖ୍ୟ ଅର୍ଥାତ୍‌ \overline{\mathrm{EF}} ବୃତ୍ତର ବ୍ୟାସ ।
ଅଙ୍କନ : \overline{\mathrm{EO}}\overline{\mathrm{OF}} ଅଙ୍କନ କର । O ବିନ୍ଦୁରେ \overline{\mathrm{AB}}\overline{\mathrm{CD}} ସହ ସମାନ୍ତର କରି \overleftrightarrow{\mathrm{MON}} ଅଙ୍କନ କର ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 17
ତ୍ପମାଣ : m∠OEA = 90° ଓ \stackrel{\leftrightarrow}{\mathrm{AB}} || \stackrel{\leftrightarrow}{\mathrm{MN}}
⇒ m∠EOM = 90°
ସେଦ୍ୱିପରି m∠FOM = 90°
∴ m∠EOM + m∠FOM = 90° + 90° = 180°
⇒ E-O-F ଏକରେଖ୍ୟ ଅର୍ଥାତ୍ \overline{\mathrm{EF}} ବୃତ୍ତର ବ୍ୟାସ ।

Question 14.
△ABC ସମ୍ପୃକ୍ତ \overline{\mathrm{BC}} ବାହୁ, \overrightarrow{\mathbf{A B}} ରଶ୍ମି ଓ \overrightarrow{\mathbf{A C}} ରଶ୍ମିକୁ POR ବୃତ୍ତ ଯଥାକ୍ରମେ P, Q ଓ R ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରେ | ତ୍ପମାଣ କର ଯେ, AQ = \frac { 1 }{ 2 } (AB + BC + AC)|
Solution:
ଦତ୍ତ : △ABC ର ବହିଃସ୍ଥ ଏକ ବୃତ୍ତ POR, \overline{\mathrm{BC}} ବାହୁକୁ P ବିନ୍ଦୁରେ ସ୍ପର୍ଶକରେ ଏବଂ \overrightarrow{\mathrm{AB}}\overrightarrow{\mathrm{AC}} ରଶ୍ମି ଦ୍ଵୟକୁ ଯଥାକ୍ରମେ Q ଓ R ବିନ୍ଦୁରେ ସ୍ପର୍ଶକରେ ।
ପ୍ରାମାଣ୍ୟ : AQ = \frac { 1 }{ 2 } (AB + BC + CA)
ପ୍ରମାଣ : ବହିଃସ୍ଥ ବିନ୍ଦୁ A ରୁ ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ସମାନ ।
∴ AQ = AR
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 18
ପୁନଶ୍ଚ BQ = BP ଏବଂ CP = CR
2AQ = AQ + AQ = AQ + AR = AB + BQ + AC + CR
= AB + BP + AC + CP = AB + (BP + CP) + AC = AB + BC + AC
∴ AQ = \frac { 1 }{ 2 } (AB + BC + AC) (ପ୍ରମାଣିତ)

BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3

Question 15.
ଏକ ସାମାନ୍ତରିକ ଚିତ୍ରର ସମସ୍ତ ବାହୁକୁ ଗୋଟିଏ ବୃତ୍ତ ସ୍ପର୍ଶ କଲେ, ପ୍ରମାଣ କର ଯେ ସାମାନ୍ତରିକ ଚିତ୍ରଟି ଏକ ରମ୍ବସ୍ ।
Solution:
ଦତ୍ତ :
ABCD ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର । ଏହାର \overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}\overline{\mathrm{DA}} ଯଥାକ୍ରମେ ଏକ ବୃତ୍ତକୁ P, Q, R, S ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରେ ।
ପ୍ରାମାଣ୍ୟ : ABCD ଏକ ରମ୍ବସ୍ ।
ପ୍ରମାଣ : ବହିଃସ୍ଥ ବିନ୍ଦୁ A ରୁ AP = AS
ସେହିପରି BP = BQ, DR = DS ଓ CR = CQ
∴ AP + BP + DR + CR = AS + BQ + DS + CQ = AS + DS + BQ + CQ
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 19
⇒ AB + CD = AD + BC
⇒ AB + AB = AD + AD (‘.’ AB = CD ଓ AD = BC)
⇒ 2AB = 2AD
⇒ AB = AD
ଅର୍ଥାତ୍ ABCD ଏକ ରମ୍ବସ୍ । (ପ୍ରମାଣିତ)

Question 16.
ଗୋଟିଏ ବୃତ୍ତର କେନ୍ଦ୍ର O ଏବଂ ଏହି ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P | P ଠାରୁ ପୂର୍ବୋକ୍ତ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ ଖଣ୍ଡ ଦ୍ଵୟ ହେଉଛନ୍ତି \overline{\mathrm{PA}}\overline{\mathrm{PB}} | \overline{\mathrm{OP}} ର ଦୈର୍ଘ୍ୟ ବୃତ୍ତଟିର ବ୍ୟାସ ସହ ସମାନ ହେଲେ, ପ୍ରମାଣ କର ଯେ, △ABP ଏକ ସମବାହୁ ତ୍ରିଭୁଜ ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 20
ଦତ୍ତ : ABC ବୃତ୍ତର କେନ୍ଦ୍ର ଠ ଓ P ଏକ ବହିଃସ୍ଥ ବିନ୍ଦୁ । \overline{\mathrm{PA}}\overline{\mathrm{PB}} ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ।
OP = ବୃତ୍ତର ବ୍ୟାସ ।
ପ୍ରାମାଣ୍ୟ : △ABP ଏକ ସମବାହୁ ଅର୍ଥାତ୍ AP = BP = AB |
ଅଙ୍କନ : \overline{\mathrm{OA}}, \overline{\mathrm{OB}}\overline{\mathrm{AR}} ଅଙ୍କନ କର ।
∴ OP = 2r = 2OR ⇒ R, \overline{\mathrm{OP}} ର ମଧ୍ୟବିନ୍ଦୁ ।
∴ △OAP ସମକୋଣୀ (∵ m∠OAP = 90°)
ସମକୋଣରୁ କର୍ଣ୍ଣର ମଧ୍ୟବିନ୍ଦୁ Rକୁ ସଂଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡ \overline{\mathrm{AR}}, \overline{\mathrm{OP}} ର ଅର୍ଦ୍ଧେକ ଦୈର୍ଘ୍ୟ ବିଶିଘୁ |
∴ AR = OA = OR
∴ AROA = OR
m∠AOR = 60°
ସେହିପରି m∠BOR = 60°
∴ m∠AOB 120° ⇒ m∠APB = 60°
ବର୍ତ୍ତମାନ △APBରେ PA = PB ଏବଂ m∠APB = 60°
∴ △APB ସମବାହୁ ।

Question 17.
ଗୋଟିଏ ବୃତ୍ତର କେନ୍ଦ୍ର O ଏବଂ P ବୃତ୍ତର ବହିଃସ୍ଥ ଏକବିନ୍ଦୁ । \overrightarrow{\mathbf{P T}} ସ୍ପର୍ଶକରଶ୍ମିର ସ୍ପର୍ଶବିନ୍ଦୁ T, \overline{\mathrm{OP}}ର ମଧ୍ୟବିନ୍ଦୁ Q ହେଲେ ପ୍ରମାଣ କର ଯେ, QT = QP |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 21
ଦତ୍ତ : S ବୃତ୍ତର ‘O’ କେନ୍ଦ୍ର । P ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ ।
\overline{\mathrm{OP}} ର ମଧ୍ୟବିନ୍ଦୁ Q । \overrightarrow{\mathbf{P T}} ବୃତ୍ତପ୍ରତି P ବିନ୍ଦୁରୁ ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ।
ପ୍ରାମାଣ୍ୟ : QT = OP
ଅଙ୍କନ : \overline{\mathrm{OT}} ଅଙ୍କନ କର ।
ପ୍ରମାଣ : m∠OTP = 90° (ସ୍ପର୍ଶକ, ସ୍ପର୍ଶବିଦୁଗାମୀ ବ୍ୟାସାର୍ଦ୍ଧ ପ୍ରତି ଲମ୍ବ)
∴ △OTP ସମକୋଣୀ । ଉକ୍ତ ତ୍ରିଭୁଜର କର୍ଣ୍ଣର ମଧ୍ୟବିନ୍ଦୁ Q |
∴ ସମକୋଣା ତ୍ରିଭୁକର ସମକୋଣରୁ କଣ୍ଡର ମଧ୍ୟ ବିନ୍ଦୁକୁ ଯୋଗ କରୁଥିବା ସରକରେଖା କଣ୍ଡର ବୈଶ୍ୟର ଅର୍ଦ୍ଧେକ ।
∴ QT = \frac { 1 }{ 2 } OP = QP (ପ୍ରମାଣିତ)

Question 18.
ଗୋଟିଏ ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P ଏବଂ ସ୍ପର୍ଶକ ରଶ୍ମି \overrightarrow{\mathbf{P T}} ର ସ୍ପର୍ଶବିନ୍ଦୁ T | P ବିନ୍ଦୁଗାମୀ ଏକ ରେଖା ଉକ୍ତ ବୃତ୍ତକୁ A ଓ B ବିନ୍ଦୁରେ ଛେଦ କରେ, ଯେପରିକି P-A-B | \overline{\mathrm{AB}} ଉପରେ A ଓ Bର ମଧ୍ୟବର୍ତ୍ତୀ C ଏକ ବିନ୍ଦୁ ।
ପ୍ରମାଣ କର :
(a) \overrightarrow{\mathbf{T C}}, ∠ATB , ସମଦ୍ବିଖଣ୍ଡିତ ତ୍ରିଭୁଜର PC = PT
(b) PC = PT ହେଲେ \overrightarrow{\mathbf{T C}} ଦ୍ଵାରା ∠ATB ସମଦ୍ବିଖଣ୍ଡିତ ହୁଏ ।
Solution:
(a)
ଦତ୍ତ : ଏକ ବୃତ୍ତର ବହିଃସ୍ଥ ବିନ୍ଦୁ P । PT ଏକ ସ୍ପର୍ଶକଖଣ୍ଡ ଏବଂ P – A – B ଏକ ଛେଦକରେଖା ।
\overrightarrow{\mathbf{T C}}, ∠ATBର ସମଦ୍ବିଖଣ୍ଡକ ।
ପ୍ରାମାଣ୍ୟ : PC = PT
ପ୍ରମାଣ : m∠PTA = m∠ABT (ଏକାନ୍ତର ବୃତ୍ତଖଣ୍ଡସ୍ଥ )
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 22
m∠ATC = m∠BTC (ଦଇ)
⇒ m∠PTA + m∠ATC = m∠ABT + m∠BTC
⇒ m∠ABT + m∠BTC = m∠CBT + m∠BTC = m∠TCA
(△BTCର ବହିଃସ୍ଥ କୋଣ)
ପୁନଶ୍ଚ m∠PTA + m∠ATC = m∠PTC
∴ m∠PTC = m∠TCA ⇒ m∠PTC = m∠TCP ⇒ PT = PC

(b) ଦତ୍ତ : ଏକ ବୃତ୍ତର P ବହିଃସ୍ଥ ବିନ୍ଦୁ । \overline{\mathrm{PT}} ଏକ ସ୍ପର୍ଶକଖଣ୍ଡ ଏବଂ P-A–B ଏକ ଛେଦକ ।
\overline{\mathrm{AB}} ର ମଧ୍ୟସ୍ଥ C ଏକ ବିନ୍ଦୁ ଓ PT = PC |
ପ୍ରାମାଣ୍ୟ : \overrightarrow{\mathbf{T C}}, ∠ATBର ସମଦ୍ଵିଖଣ୍ଡକ ।
ପ୍ରମାଣ : △PTCରେ PT = PC (ଦତ୍ତ)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 23
⇒ m∠PTC= m∠PCT
⇒ m∠PTA + m∠ATC = m∠CBT + m∠BTC
(∵ △TBC ରେ ଦହିମ କୋଣ ∠PCT)
m∠PTA = m∠CBT (ଏକାନ୍ତ୍ରର ଦ୍ବରଖଣ୍ଡମ)
⇒ m∠ATC = m∠BTC ⇒ \overrightarrow{\mathbf{T C}}, ∠ATBର ସମଦ୍ଵିଖଣ୍ଡକ (ପ୍ରମାଣିତ)

BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3

Question 19.
△ABCର ବାହୁ AB ଓ AC ଉପରେ ଯଥାକ୍ରମେ X ଓ Y ବିନ୍ଦୁ ଅବସ୍ଥିତ, ଯେପରିକି △ABCର ଅନ୍ତଃବୃତ୍ତକୁ \overline{\mathbf{XY}} ସ୍ପର୍ଶ କରିବ । ପ୍ରମାଣ କର ଯେ AX + XY+YA = AB + AC – BC |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 24
Solution:
ଦତ୍ତ : △ABC ର ଅନ୍ତଃ ବୃ ତ୍ତ ବାହୁ ମାନ ଙ୍କୁ AB, BC, AC କୁ ଯଥାକ୍ରମେ P, Q, R ବିନ୍ଦୁ ରେ B ସ୍ପର୍ଶକରେ । △ABC ର AB ଓ AC ବାହୁ ଉପରେ X ଓ Y ଏପରି ଦୁଇରି ବିନ୍ଦୁ ଯେ XY, AABC Q ଅନ୍ତଃବୃତ୍ତକୁ M ବିନ୍ଦୁରେ ସ୍ପର୍ଶକରେ ।
ହ୍ମାଣ୍ୟ: AX + XY + YA = AB + AC – BC
ପ୍ରମାଣ: AB+ AC – BC
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 25
= AP + BP + AR + RC − (BQ + QC)
= (AX + XP) + BP + (AY + YR) + RC – BQ – QC
[ଦତ୍ତ XM = XP, YM = YR, BP = BQ 19° RC = QC]
= AX + XM + BQ + AY + MY + QC − BQ – QC
= AX + AY + (XM + MY) = AX + AY + XY
∴ AX + AY + XY = AB + AC – BC

Question 20.
ବହିଃସ୍ପର୍ଶୀ ଦୁଇଟି ବୃତ୍ତ S1 ଓ S2 ପରସ୍ପରକୁ P ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରନ୍ତି ।
ବୃତ୍ତଦ୍ଵୟର ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ S1 ଓ S2 ବୃତ୍ତ ଦ୍ଵୟକୁ ଯଥାକ୍ରମେ A ଓ B ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରନ୍ତି । P ବିନ୍ଦୁ ଦେଇ ଅଙ୍କିତ ସାଧାରଣ ସ୍ପର୍ଶକ \stackrel{\leftrightarrow}{\mathbf{A B}} କୁ C ବିନ୍ଦୁରେ ଛେଦ କଲେ, ପ୍ରମାଣ କର :
(a) AC = BC ଏର୍ତ (b) m∠APB = 90° |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 26
Solution:
ଦତ୍ତ : ଦୁଇଟି ବୃତ୍ତ ପରସ୍ପରକୁ P ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରନ୍ତି । ସାଧାରଣ ସ୍ପର୍ଶକ \stackrel{\leftrightarrow}{\mathbf{A B}}, ବୃତ୍ତଦ୍ଵୟକୁ A ଓ B ବିନ୍ଦୁରେ ସ୍ପର୍ଶକରେ । P ବିନ୍ଦୁଗାମୀ ସ୍ପର୍ଶକ \stackrel{\leftrightarrow}{\mathbf{A B}} କୁ ‘C’ ବିନ୍ଦୁରେ ଛେଦ କରୁଛି ।
ପ୍ରାମଣ୍ୟ: (i) AC = CB (ii) m∠APB = 90°
ପ୍ରମାଣ : C ବହିଃସ୍ଥ ବିନ୍ଦୁରୁ ଅଙ୍କିତ ସ୍ପର୍ଶକ ଖଣ୍ଡ ଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ସମାନ । ଅର୍ଥାତ୍ CA = CP |
ସେହିପରି ଅନ୍ୟ ବୃତ୍ତରେ CP = CB ∴CA = CB .. (i) (ପ୍ରମାଣିତ)
∵ CA = CP → m∠CAP = m∠CPA
∵ CP = CB → m∠CBP = m∠CPB
∴ m∠CAP + m∠CBP = m∠CPA + m∠CPB
⇒ m∠CAP +m∠CBP = m∠APB
⇒ m∠CAP+m∠CBP + m∠APB = 2m∠APB
⇒ 180° = 2m∠APB ⇒ m∠APB = 90° …(ii) (ପ୍ରମାଣିତ)

Question 21.
S1 ଓ S2 ବୃତ୍ତଦ୍ଵୟ ପରସ୍ପରକୁ A ଓ B ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । S1 ଉପରିସ୍ଥ ଏକ ବିନ୍ଦୁ P ଦେଇ ଅଙ୍କିତ \overrightarrow{\mathbf{P A}}\overrightarrow{\mathbf{P B}} S, ବୃତ୍ତକୁ ଯଥାକ୍ରମେ C ଓ D ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । ପ୍ରମାଣ କର ଯେ P ବିନ୍ଦୁରେ S1 ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ, \overline{\mathbf{CD}} ସହ ସମାନ୍ତର ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 27
ଦତ୍ତ : S1 ଓ S2 ବୃତ୍ତ ଦ୍ବୟ ପରସ୍ପରକୁ A ଓ B ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । P, S1 ଉପରିସ୍ଥ ଏକ ବିନ୍ଦୁ । \overrightarrow{\mathbf{P A}}\overrightarrow{\mathbf{P B}}, S2 କୁ ଯଥାକ୍ରମେ C ଓ D ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି ।
ପ୍ରାମାଣ୍ୟ : \overleftrightarrow{\mathrm{XY}} || \overline{\mathbf{CD}}
ଅଙ୍କନ : \overline{\mathbf{AB}} ଅଙ୍କନ କର ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 28
ପ୍ରମାଣ : m∠XPA = m∠ABP (ଏକାନ୍ତର ବୃତ୍ତଖଣ୍ଡସ୍ଥ କୋଣ)
ପୁନଶ୍ଚ ABDC ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖତ ଚତୁର୍ଭୁଜ ।
∠ABP ଚତୁର୍ଭୁଜଟିର ବହିଃସ୍ଥ କୋଣ ।
ଏଠାରେ m∠ABP = m∠ACD
∴ m∠XPA = m∠ACD କିନ୍ତୁ ଏମାନେ ଏକାନ୍ତର ।
\overleftrightarrow{\mathrm{XY}} || \overline{\mathbf{CD}} (ପ୍ରମାଣିତ)

Question 22.
ଦୁଇଟି ପରସ୍ପର ଅଣଛେଦୀ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ r1 ଓ r2 ଏକକ ଏବଂ r1 > r2 ବୃତ୍ତଦ୍ଵୟର କେନ୍ଦ୍ରଦ୍ଵୟ ମଧ୍ୟରେ ଦୂରତା d ଏକକ ହେଲେ ଏବଂ
(a) ଉଭୟ ବୃତ୍ତର ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକର ସ୍ପର୍ଶବିନ୍ଦୁ A ଓ B ହେଲେ, ପ୍ରମାଣ କର ଯେ
AB2 = d2(r1 – r2) ଏବଂ
(b) ଉଭୟ ବୃତ୍ତର ତୀର୍ଯ୍ୟକ୍ ସାଧାରଣ ସ୍ପର୍ଶକର ସ୍ପର୍ଶବିନ୍ଦୁ C ଓ D ହେଲେ, ପ୍ରମାଣ କର ଯେ CD2 = d2(r1 + r2)2
Solution:
(a) ଦତ୍ତ : O ଓ P ଯଥାକ୍ରମେ S1 ଓ S2 ଦୁଇଟି ବୃତ୍ତର କେନ୍ଦ୍ରବିନ୍ଦୁ ଓ OP = d |
S1 ଓ S2 ବୃତ୍ତର ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକର ସ୍ପର୍ଶବିନ୍ଦୁ
ଯଥାକ୍ରମେ A ଓ B|
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 29
OA = r, PB = r2 , r1 > r2
ପ୍ରାମାଣ୍ୟ : AB2 = d2 – (r1 – r2)2
ଅଙ୍କନ : \overline{\mathbf{OA}} ଏବଂ \overline{\mathbf{PB}} ଅଙ୍କନ କର । \overline{\mathbf{PD}}\overline{\mathbf{OA}} ଅଙ୍କନ କର ।
\overline{\mathbf{PD}} || \overleftrightarrow{\mathrm{AB}} ଅଙ୍କନ କର ।

ପ୍ରମାଣ : ADPB ଏକ ଆୟତକ୍ଷେତ୍ରର ପ୍ରତ୍ୟେକ କୋଣ ସମକୋଣ
∴ AD = PB = r2
⇒ OD = OA – AD = r1 – r2 ଓ AB = PD
△ODPରେ m∠ODP = 90°
⇒ OP2 = OD2 + PD2 (ପିଥାଗୋରାସ୍ ଉପପାଦ୍ୟ)
⇒ PD2 = OP2 – OD2 ⇒ AB = d2 – (r1 – r2)2

(b) ଦତ୍ତ: O ଓ P ଯଥାକ୍ରମେ S1 ଓ S2 ବୃତ୍ତର କେନ୍ଦ୍ରବିନ୍ଦୁ ଏବଂ OP = d |
S1 ଓ S2 ବୃତ୍ତର ତୀର୍ଯ୍ୟକ୍ ସାଧାରଣ ସ୍ପର୍ଶକ ଯଥାକ୍ରମେ S1 ଓ S2 ବୃତ୍ତକୁ C ଓ D ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରେ ।
OC= r1, PD = r2 ଓ r1 > r2 |
ପ୍ରାମାଣ୍ୟ : CD2 = d2 – (r1 + r2)2
ଅଙ୍କନ : \overline{\mathbf{OC}}, \overline{\mathbf{PD}} ଅଙ୍କନ କର ।
\overline{\mathbf{PE}} || \overline{\mathbf{CD}} ଅକନ କର ଯେବେକି \overline{\mathbf{PE}}
\overrightarrow{\mathrm{OF}} କୁ E ବିନ୍ଦୁରେ ଛେଦ କରିବ ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 30
ପ୍ରମାଣ : CEPD ଏକ ଆୟତଚିତ୍ର (ପ୍ରତ୍ୟେକ କୋଣ ସମକୋଣ)
CE = PD = r2
∴ OE = OC+CE = r1 + r2
OP = d (ଦତ୍ତ) ଓ PE = CD
m∠OCD = 90°, ∠CDP = 90°, m∠CEP = 90°
△OEPରେ OP2 = OE2 + PE2 (ପିଥାଗୋରାସ୍ ଉପପାଦ୍ୟ)
∴PE2 = OP2 – OE2
⇒ CD2 = OP2 – OE2 = d2 – (r1 + r2)2 (ପ୍ରମାଣିତ)

BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3

Question 23.
ଗୋଟିଏ ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P ଏବଂ P ବିନ୍ଦୁଗାମୀ ସ୍ପର୍ଶକ ରଶ୍ମିଦ୍ଵୟର ସ୍ପର୍ଶବିନ୍ଦୁ ଯଥାକ୍ରମେ Q ଏବଂ R | \overline{\mathbf{QR}} ଜ୍ୟାଦ୍ଵାରା ଛେଦିତ କ୍ଷୁଦ୍ରଚାପର ମଧ୍ୟବିନ୍ଦୁ S ହେଲେ, ପ୍ରମାଣ କର ଯେ, \overrightarrow{\mathrm{QS}} ଦ୍ଵାରା ∠PQR ସମଦ୍ବିଖଣ୍ଡିତ ହୁଏ |
Solution:
ଦତ୍ତ : ଗୋଟିଏ ବୃତ୍ତର ବହିଃସ୍ଥ ଏକ ବିନ୍ଦୁ P ଏବଂ P ବିନ୍ଦୁଗାମୀ ସ୍ପର୍ଶକ ରଶ୍ମିଦ୍ବୟର ସ୍ପର୍ଶବିନ୍ଦୁ ଯଥାକ୍ରମେ Q ଏବଂ R । \overline{\mathbf{QR}} ଜ୍ୟାଦ୍ଵାରା ଛେଦିତ କ୍ଷୁଦ୍ରଚାପର ମଧ୍ୟବିନ୍ଦୁ S ।
ପ୍ରାମାଣ୍ୟ : \overline{\mathbf{QS}}, ∠PORର ସମଦ୍ବିଖଣ୍ଡକ ।
ଅଙ୍କନ : \overline{\mathbf{SR}} ଅଙ୍କନ କର ।
ପ୍ରମାଣ : \overparen{\mathrm{QSR}} ଚାପର ମଧ୍ୟବିନ୍ଦୁ S |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 31
\overline{\mathbf{QS}}\overline{\mathbf{SR}} ⇒ m∠SRQ = m∠SQR
\overline{\mathbf{PQ}} ଏକ ସ୍ପର୍ଶକ ଏବଂ \overline{\mathbf{QS}} ସ୍ପର୍ଶକ ବିନ୍ଦୁଗାମୀ ଜ୍ୟା ।
⇒ m∠POS = m∠SRQ (ଏକାନ୍ତର ବୃତ୍ତଖଣ୍ଡସ୍ଥ )
∴ m∠PQS = m∠SQR
ଅର୍ଥାତ୍ \overrightarrow{\mathrm{SQ}}, ∠PORର ସମଦ୍ବିଖଣ୍ଡକ ।

Question 24.
ଚିତ୍ରରେ ଥିବା ବୃତ୍ତର AT ଏକ ବ୍ୟାସ । ବୃତ୍ତ ଉପରିସ୍ଥ ଅନ୍ୟ ଏକ ବିନ୍ଦୁ B । \overrightarrow{\mathrm{AB}} ଏବଂ I ବିନ୍ଦୁରେ ଅଙ୍କିତ ସ୍ପର୍ଶକ ପରସ୍ପରକୁ P ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । B ବିନ୍ଦୁରେ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ \overleftrightarrow{ T P} କୁ ( ବିନ୍ଦୁରେ ଛେଦ କଲେ, ପ୍ରମାଣ କର ଯେ Q ବିନ୍ଦୁ ହେଉଛି PT ର ମଧ୍ୟବିନ୍ଦୁ ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 32
Solution:
ଦତ୍ତ : \overline{\mathbf{AT}} ବୃତ୍ତର ଏକ ବ୍ୟାସ । ABT ବୃତ୍ତ ଉପରିସ୍ଥ B ଏକ ବିନ୍ଦୁ |
\overrightarrow{\mathrm{AB}} ଓ T ବିନ୍ଦୁରେ ଅଙ୍କିତ ସ୍ପର୍ଶ \overleftrightarrow{ T P} ର ଛେଦବିନ୍ଦୁ Q |
ପ୍ରାମାଣ୍ୟ : Q, \overline{\mathbf{TP}}ର ମଧ୍ୟବିନ୍ଦୁ ।
ଅଙ୍କନ : \overline{\mathbf{BT}} ଅଙ୍କନ କର ।
ପ୍ରମାଣ : ବହିଃସ୍ଥ ( ବିନ୍ଦୁରୁ ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକଖଣ୍ଡ ଦ୍ଵୟ \overline{\mathbf{QB}}\overline{\mathbf{QT}} |
⇒ QB = QT
⇒ m∠QBT = m∠QTB
m∠ABT = 90° (∵ \overline{\mathbf{AT}} ବୃତ୍ତ ଏକ ବ୍ୟାସ)
m∠PBT = 90°
⇒ m∠PBQ + m∠QBT = 90°
ପୁନମ୍ନ m∠BTQ + m∠BPQ = 90°
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 33
⇒ m∠PBQ + m∠QBT = m∠BTQ + m∠BPQ
(: m∠QBT = m∠QTB)
⇒ m∠PBQ = m∠BPQ ⇒ BQ = QP
ପୁନମ୍ନ BQ = QT ⇒ QP = QT (ପ୍ରମାଣିତ)

Question 25.
ଗୋଟିଏ ବୃତ୍ତରେ \overline{\mathbf{AB}} ଏକ ବ୍ୟାସ । B ବିନ୍ଦୁରେ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ ଉପରେ C ଏପରି ଏକ ବିନ୍ଦୁ ଯେପରି \overline{\mathbf{CA}}, ବୃତ୍ତକୁ D ବିନ୍ଦୁରେ ଛେଦ କରେ । ପ୍ରମାଣ କର ଯେ AB2 = AC × AD |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 34
Solution:
ଦତ୍ତ : \overline{\mathbf{AB}}, S ବୃତ୍ତର ଏକ ବ୍ୟାସ ।
\overrightarrow{\mathrm{BC}} ବୃତ୍ତପ୍ରତି B ବିନ୍ଦୁରେ ଅଙ୍କିତ ସ୍ପର୍ଶକ C – D – A ଏକ ଛେଦକ ।
ପ୍ରାମାଣ୍ୟ : AB2 = AC × AD
ଅଙ୍କନ : \overline{\mathbf{BD}} ଅଙ୍କନ କର ।
ପ୍ରମାଣ : BC ସ୍ପର୍ଶକ ⇒ ∠ABC = ସମକୋଣ
△ACBରେ AB2 + BC2 = AC2 ⇒ AB2 = AC2 – BC2
ପୁନଶ୍ଚ CB2 = CD · CA
⇒ AB2 = AC2 – BC2 = AC2 – CD · CA
= AC (AC – CD) = AC × AD (ପ୍ରମାଣିତ)

Question 26.
ଗୋଟିଏ ବୃତ୍ତରେ \overline{\mathbf{AB}} ଏକ ବ୍ୟାସ । B ବିନ୍ଦୁରେ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ ଉପରେ C ଓ D ଦୁଇଟି ବିନ୍ଦୁ ଯେପରି C-B-D | ଯଦି CA ଓ \overline{\mathbf{DA}} ଯଥାକ୍ରମେ ବୃତ୍ତକୁ P ଓ Q ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି, ପ୍ରମାଣ କର ଯେ AC × AP = AD × AQ |
Solution:
ଦତ୍ତ : \overline{\mathbf{AB}} ଏକ ବୃତ୍ତର ବ୍ୟାସ । B ବିନ୍ଦୁରେ ବୃତ୍ତପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ \stackrel{\leftrightarrow}{C D} । AC ଓ \overline{\mathbf{AD}} ବୃତ୍ତକୁ P ଓ Q ବିନ୍ଦୁରେ ଛେଦକରନ୍ତି ।
ପ୍ରାମାଣ୍ୟ : AC × AP = AD × AQ
ଅଙ୍କନ : \overline{\mathbf{BP}}\overline{\mathbf{BQ}} ଅଙ୍କନ କର ।
ପ୍ରମାଣ : △APB ଓ △ABC ମଧ୍ୟରେ
m∠APB m∠ABC (ପ୍ରତ୍ୟେକ ସମକୋଣ)
m∠PAB = m∠BAC (ସାଧାରଣ କୋଣ)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 35
⇒ △APB ~ △ABC (କୋ.କୋ, ସାଦୃଶ୍ୟ)
\frac { AP }{ AB } = \frac { AB }{ AC } ⇒ AB2 = AP × AC …(i)
ସେହିପରି △ABQ ~ △ABD
AB2 = AD × AQ …(ii)
∴ (i) ଓ (ii)ରୁ AP × AC = AD × AQ (ପ୍ରମାଣିତ)

Question 27.
ଚିତ୍ରରେ S1 ଓ S2 ବୃତ୍ତ ଦୁଇଟି ବହିଃସ୍ପର୍ଶୀ ଏବଂ G ସେମାନଙ୍କର ସ୍ପର୍ଶବିନ୍ଦୁ । ବୃତ୍ତଦ୍ଵୟର ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ ରଶ୍ମି \overrightarrow{\mathrm{PX}}\overrightarrow{\mathrm{PY}} ଦ୍ଵୟର ସାଧାରଣ ମୂଳ ବିନ୍ଦୁ P | S1 ଓ S, ବୃତ୍ତକୁ \overrightarrow{\mathrm{PX}} ଯଥାକ୍ରମେ C ଓ ଯ ବିନ୍ଦୁରେ ଏବଂ \overrightarrow{\mathrm{PY}} ଯଥାକ୍ରମେ D ଓ F ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରନ୍ତି ।
(a) ପ୍ରମାଣ କର :
(i) P, A, G, B ଏକ ସରଳରେଖାରେ ଅବସ୍ଥିତ ଓ
(ii) CE = DF
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 36
(b) ଉଭୟ ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସାଧାରଣ ସ୍ପର୍ଶକ \overrightarrow{\mathrm{PX}}\overrightarrow{\mathrm{PY}} କୁ ଯଥାକ୍ରମେ M ଓ N ବିନ୍ଦୁରେ ଛେଦକଲେ, ପ୍ରମାଣ କର :
(i) PM = PN, (ii) MG = NG ।
Solution:
ଦତ୍ତ : S1 ଓ S2 ଦୁଇଟି ବହିଃସ୍ପର୍ଶୀ ବୃତ୍ତ ପରସ୍ପରକୁ G ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରନ୍ତି । ବୃତ୍ତଦ୍ୱୟର ସରଳ ସାଧାରଣ ସ୍ପର୍ଶକ ରଶ୍ମି \overrightarrow{\mathrm{PX}}\overrightarrow{\mathrm{PY}}, S1 ଓ S2 ବୃତ୍ତକୁ ଯଥାକ୍ରମେ C ଓ E ଏବଂ D ଓ F ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରନ୍ତି । ଓ ବିନ୍ଦୁରେ ଅଙ୍କିତ ସାଧାରଣ ସ୍ପର୍ଶକ PX ଓ PY କୁ ଯଥାକ୍ରମେ M ଓ N ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । ବୃତ୍ତଦ୍ୱୟର କେନ୍ଦ୍ର ଯଥାକ୍ରମେ A ଓ B |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 37
ପ୍ରାମାଣ୍ୟ : (a) (i) P, A, G ଓ B ସରଳରେଖାରେ ଅବସ୍ଥିତ । (ii) CE = DF
(b) (i) PM = PN, (ii) MG = NG
ଅଙ୍କନ : \overline{\mathbf{AC}}\overline{\mathbf{AD}} ଅଙ୍କନ କର ।
ପ୍ରମାଣ : PE = PD (ବହିଃସ୍ଥ P ବିନ୍ଦୁରୁ S2 ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ ଖଣ୍ଡ)
ପୁନଶ୍ଚ PC = PD (ବହିଃସ୍ଥ P ବିନ୍ଦୁରୁ S1 ବୃତ୍ତ ପ୍ରତି ଅଙ୍କିତ ସ୍ପର୍ଶକ ଖଣ୍ଡ)
∴ PE – PC = PF – PD ⇒ CE = DF (ପ୍ରମାଣିତ)
CM = MG ଏବଂ ME = MG ⇒ CM = ME
ସେହିପରି DN = NG ଏବଂ NG = NF
DN = NF, CE = DF (ପୂର୍ବରୁ ପ୍ରମାଣିତ)
⇒ 2 CM = 2 DN ⇒ CM = DN ….(i)
PC = PD = PC + CM = PD + DN → PM = PN
(i)ରୁ CM= DN →:MG = NG
(∵ MC = MG, DN = NG)
PM = PN (ପୂର୍ବରୁ ପ୍ରମାଣିତ)
⇒ m∠PMG = m∠PNG ⇒ m∠CMG = m∠DNG
ଦଇମାନ CAGM ଚତୁରିକରେ m∠ACM + m∠AGM = 180°
∴ m∠CMG + m∠CAG = 180° … (ii)
ସେହିପରି m∠DNG + m∠DAG = 180° …(iii)
m∠CMG + m∠CAG = m∠DNG + m∠DAG
m∠CAG = m∠DAG (∵ m∠CMG = m∠DNG)
m∠PAC = m∠PAD ….(iv)
(iv)ରୁ 2m∠CAG + 2m∠PAC = 360° ⇒ m∠CAG + m∠PAC = 180°
⇒ P, A, G ଏକରେଖ କିନ୍ତୁ A, G, B ଏକ ରେଞ୍ଜ
∴ P, A, G, B ଏକସରଳରେଖାରେ ଅବସ୍ଥିତ । (ପ୍ରମାଣିତ)

BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 ବୃତ୍ତର ସ୍ପର୍ଶକ Ex 3

Question 28.
ପରସ୍ପର ଅନ୍ତଃସ୍ପର୍ଶୀ ଦୁଇଟି ବୃତ୍ତର ସ୍ପର୍ଶବିନ୍ଦୁ P । ଏକ ସରଳରେଖା ଗୋଟିଏ ବୃତ୍ତକୁ A ଓ B ବିନ୍ଦୁରେ ଓ ଅନ୍ୟ ବୃତ୍ତକୁ C ଓ Ð ବିନ୍ଦୁରେ ଛେଦ କରେ । ପ୍ରମାଣ କର ଯେ ∠APC ଓ ∠BPD ସର୍ବସମ । [A-C-D ଓ A-D-C ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ପ୍ରମାଣ ଯୋଗ୍ୟ ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 38
ଦତ୍ତ : ପରସ୍ପର ଅନ୍ତଃସ୍ପର୍ଶୀ ଦୁଇଟି ବୃତ୍ତର ସ୍ପର୍ଶବିନ୍ଦୁ P । ଏକ ସରଳରେଖା ବୃତ୍ତଦ୍ଵୟକୁ ଯଥାକ୍ରମେ A, B ଓ C, D ବିନ୍ଦୁରେ ଛେଦକରନ୍ତି ।
ପ୍ରାମାଣ୍ୟ : ∠APC ≅ ∠BPD
ଅଙ୍କନ : P ବିନ୍ଦୁରେ ବୃତ୍ତପ୍ରତି ସାଧାରଣ ସ୍ପର୍ଶକ X – P – Y ଅଙ୍କନ କର ।
ମନେକର \overline{\mathrm{PC}}ର ଏକ ପାର୍ଶ୍ବରେ A ଓ X ଅବସ୍ଥାନ କରୁ ।
\overline{\mathrm{PA}}, \overline{\mathrm{PD}}, \overline{\mathrm{PC}}\overline{\mathrm{PB}} ଅଙ୍କନ କର ।
ପ୍ରମାଣ : ମନେକର A – C – D – B|
m∠DPY = m∠DCP (ଏକାନ୍ତର ବୃତ୍ତଖଣ୍ଡସ୍ଥ)
ସେହିପରି m∠BPY = m∠DAP
∴ m∠DPY – m∠BPY = m∠DCP – m∠DAP
⇒ m∠DPB = m(∠CAP + m∠APC) – m∠DAP
(∵ ବହିଃସ୍ଥ m∠DCP = m∠CAP + m∠APC)
= m∠CAP + m∠APC – m∠CAP = m∠APC
ସେହିପରି A – D – C – B ହେଲେ ପ୍ରମାଣ ଅନୁରୂପ ।

Question 29.
△ABC ର ଅନ୍ତଃବୃତ୍ତ, \overline{\mathrm{AB}}, \overline{\mathrm{BC}}\overline{\mathrm{CA}} କୁ ଯଥାକ୍ରମେ P, Q ଓ R ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରେ । ଚିତ୍ରରେ BQ = 8 ସେ.ମି., CQ = 6 ସେ.ମି. ଏବଂ △ABCର ପରିସୀମା 36 ସେ.ମି. ହେଲେ, AB ଓ AC ନିର୍ଣ୍ଣୟ କର ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 39
Solution:
ଦତ୍ତ : △ABCର ଅନ୍ତଃବୃତ୍ତ \overline{\mathrm{AB}}, \overline{\mathrm{BC}}\overline{\mathrm{CA}} କୁ ଯଥାକ୍ରମେ P, Q ଓ R ବିନ୍ଦୁରେ ଛେଦକରେ ।
BQ = 8 ସେ.ମି., CQ = 6 ସେ.ମି.
△ABCର ପରିସୀମା = 36 ସେ.ମି.
ନିର୍ଦେୟ : BQ = 8 ସେ.ମି. ⇒ BP = 8 ସେ.ମି.
CQ = 6 ସେ.ମି. ⇒ CR = 6 ସେ.ମି.
ମନେକର AP = AR = x ସେ.ମି.
△ABCର ପରିସୀମା = 36 ସେ.ମି.
⇒ AP + PB + BQ + QC +CR + RA = 36
⇒ (x + 8 + 8 + 6 + 6 + x) = 36 ⇒ 2x + 28 = 36
⇒ 2x = 36 – 28 = 8 ⇒ x = \frac { 8 }{ 2 } = 4
∴ AB = AP + BP = 4 ସେ.ମି. + 8 ସେ.ମି. = 12 ସେ.ମି.
AC = AR + CR = 4 ସେ.ମି. + 6 ସେ.ମି. = 10 ସେ.ମି. |

Question 30.
ଗୋଟିଏ ବୃତ୍ତର କେନ୍ଦ୍ର O ଏବଂ ପରିଲିଖ୍ ଚତୁର୍ଭୁଜ ABCD ହେଲେ, ପ୍ରମାଣ କର ଯେ ∠AOB ଓ ∠COD ପରସ୍ପର ପରିପୂରକ । ∠B0C ଏବଂ ∠AOD ମଧ୍ୟରେ ଥ‌ିବା ସମ୍ପର୍କ ନିର୍ଣ୍ଣୟ କର ।
Solution:
ଦତ୍ତ : ABCD ଚତୁର୍ଭୁଜ S ବୃତ୍ତର ପରିଲିଖ । ବୃତ୍ତର କେନ୍ଦ୍ର O |
gla: (i) m∠AOB + m∠COD = 180°
(ii) ∠BOC ଏବଂ ∠COD ମଧ୍ୟରେ ସଂପର୍କ ।
ପ୍ରମାଣ : m∠ABO = m∠CBO ⇒ m∠ABO = \frac { 1 }{ 2 } m∠B
m∠BAO = m∠DA0 ⇒ m∠BAO = \frac { 1 }{ 2 } m∠A
m∠OCD = m∠OCB ⇒ m∠OCD = \frac { 1 }{ 2 } m∠C
m∠ODC = m∠ODA ⇒ m∠ODC = \frac { 1 }{ 2 } m∠D
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 40
∴ m∠ABO + m∠BAO + m∠OCD + m∠ODC
= \frac { 1 }{ 2 } (m∠A + m∠B + m∠C + m∠D) = \frac { 1 }{ 2 } × 360° = 180°
ପୁନଶ୍ଚ m∠ABO + m∠BAO + m∠OCD + m∠ODC + m∠AOB + ∠COD = 180° + 180° = 360° …(ii)
∴ (i) ଓ (ii) ରୁ m∠AOB + m∠COD = 180° (ପ୍ରମାଣିତ)
ସେହିପରି m∠AOD + m∠BOC = 180° ଦେବ |

Question 31.
ଗୋଟିଏ ବୃତ୍ତର ଏକ ଜ୍ଯା \overline{\mathrm{AB}}, ଏହି ବୃତ୍ତ ଉପରିସ୍ଥ ଏକ ବିନ୍ଦୁ P ଠାରେ ଅଙ୍କିତ ସ୍ପର୍ଶକ ସହ ସମାନ୍ତର ହେଲେ, ପ୍ରମାଣ କର ଯେ P ବିନ୍ଦୁଠାରେ \overparen{\mathbf{A P B}} ସମଦ୍ବିଖଣ୍ଡିତ ହୁଏ ।
ସମାଧାନ :
ଦତ୍ତ : ABP ବୃତ୍ତର AB ଏକ ଜ୍ୟା । P ଠାରେ \overline{\mathrm{AB}} ସହ ସମାନ୍ତର ଭାବେ ଅଙ୍କିତ ସ୍ପର୍ଶକ \overleftrightarrow{\mathrm{XY}}, \overline{\mathrm{PB}}ର ଏକ ପାର୍ଶ୍ଵରେ X ଓ A ଅବସ୍ଥିତ ।
ପ୍ରାମାଣ୍ୟ : \widehat{\mathrm{AP}}\widehat{\mathrm{BP}}
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 41
ଅଙ୍କନ : \overline{\mathrm{PA}}\overline{\mathrm{PB}} ଅଙ୍କନ କର ।
ପ୍ରମାଣ : m∠XPA = m∠PBA (ଏକାନ୍ତର ବୃତ୍ତଖଣ୍ଡସ୍ଥ)
m∠XPA = m∠PAB (ଏକାନ୍ତର କରେ)
⇒ m∠PBA = m∠PAB
\widehat{\mathrm{PA}}\widehat{\mathrm{PB}}

Question 32.
ଚିତ୍ରରେ ଥ‌ିବା ବୃତ୍ତର କେନ୍ଦ୍ର 0, L1 ଓ L2 ଦୁଇଟି ସ୍ପର୍ଶକ ଏବଂ L1 || L2 | ବୃତ୍ତର K ବିନ୍ଦୁରେ ଅଙ୍କିତ ସ୍ପର୍ଶକ \stackrel{\leftrightarrow}{\mathbf{P Q}}, L1 ଓ L2 କୁ ଯଥାକ୍ରମେ M ଓ N ବିନ୍ଦୁରେ ଛେଦ କରେ । ପ୍ରମାଣ କର ଯେ ∠MON ଏକ ସମକୋଣ ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 42
Solution:
ଦତ୍ତ : S ବୃତ୍ତର କେନ୍ଦ୍ର O,
L1 ଓ L2 ବୃତ୍ତପ୍ରତି ଦୁଇଟି ସ୍ପର୍ଶକ ଏବଂ L1 || L2| ବୃତ୍ତର K ବିନ୍ଦୁରେ ଅଙ୍କିତ ସ୍ପର୍ଶକ PO
L1 ଓ L2 କୁ ଯଥାକ୍ରମେ M ଓ N ବିନ୍ଦୁରେ ଛେଦକରେ ।
ପ୍ଵାମଣ୍ୟ: m∠MON = 90°
ଅଙ୍କନ : \overline{\mathrm{CO}}, \overline{\mathrm{DO}}, \overline{\mathrm{OK}}, \overline{\mathrm{MO}}\overline{\mathrm{NO}} ଅଙ୍କନ କର ।
L1 ଓ L2, S ବୃତ୍ତକୁ ଯଥାକ୍ରମେ C ଓ D ବିନ୍ଦୁରେ ସ୍ପର୍ଶକରୁ ।
ପ୍ରମାଣ : △OCM ଓ △OKMରେ
MC = MK. \overline{\mathrm{MO}} ସାଧାରଣ ବାହୁ ।
ଓ m∠OCM= m∠OKM (ସାଧାରଣ)
∴ △OCM = △OKM = m∠CMO = m∠OMK
BSE Odisha 10th Class Maths Solutions Geometry Chapter 3 Img 43
ସେହିପରି ପ୍ରମାଣ କରାଯାଇପାରେ ଯେ △OKN ≅ △ODN ⇒ m∠ONK = m∠OND L1 || L2, \stackrel{\leftrightarrow}{\mathrm{PQ}} ଛେଦକ ।
ତେଣୁ m∠CMK + m∠DNK= 180° ⇒ \frac { 1 }{ 2 } m∠CMK + \frac { 1 }{ 2 } m∠DNK = 90°
⇒ m∠KMO + m∠KNO = 90° …(i)
କିନ୍ତୁ m∠KMO + m∠KNO + m∠MON = 180°
90° + m∠MON = 180° [(i) ରୁ]
m∠MON = 180° – 90° = 90° (ପ୍ରମାଣିତ)

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b)

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b)

Question 1.
ବନ୍ଧନୀ ମଧ୍ଯରୁ ସଠିକ୍ ଉତ୍ତରଟି ବାଛି ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(i) ଯଦି (1, – 2) ବିନ୍ଦୁଟି (4, 2) ଓ (K, – 6) ବିନ୍ଦୁଦ୍ବୟର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ହୁଏ, ତେବେ k = ………. । [-2, 2, – 4, 4]
(ii) (- 2, 3) ଓ (3, – 2) ବିନ୍ଦୁଦ୍ଵୟର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ହେଉଛି ………. । [(1, 1), (\frac{1}{2}, \frac{1}{2}), (\frac{5}{2}, \frac{5}{2}) (-\frac{1}{2}, -\frac{1}{2})]
(iii) ଏକ ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁ ହେଉଛି ମୂଳବିନ୍ଦୁ; ଯଦି ରେଖାଖଣ୍ଡଟିର ଏକ ପ୍ରାନ୍ତବିନ୍ଦୁ (2, 3) ହୁଏ, ତେବେ ………. । [(-2, 3), (2,-3),(-2,-3) (\frac{1}{2}, \frac{3}{2})]
(iv) (0, 2) ଓ (2, 0) ବିନ୍ଦୁଦ୍ଵୟର ରେଖାଖଣ୍ଡକୁ 1 : 2 ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥ‌ିବା ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ………. । [(\frac{4}{3}, \frac{2}{3}), (\frac{4}{3}, \frac{2}{3}), (-2, 4),(4, -2)]
ଉତ୍ତର:
(i) -2
(ii) (\frac{1}{2}, \frac{1}{2})
(iii) (-2, -3)
(iv) (\frac{2}{3}, \frac{4}{3})

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 2.
ନିମ୍ନଲିଖ ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ଦତ୍ତ ବିନ୍ଦୁମାନଙ୍କୁ ଯୋଗକରୁଥିବା ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ନିର୍ଣ୍ଣୟ କର ।
(i) (3, 4), (1,-2)
(ii) (-1,3), (4, 0)
(iii) (\frac{1}{2}, \frac{1}{3}), (\frac{1}{3}, \frac{1}{2})
(iv) (0,-3), (-4, 0)
(v) (-1,-2), (3, -1)
(vi) (a, b), (c, d)
(vii) (-2, 1), (-3, -4)
(viii) (at1², 2at1), (at2², 2at2)
ସମାଧାନ :
P (x,, y,) ଓ Q(x, y) ହେଲେ PQର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})
(i) P (3, 4) ଓ Q (1, -2) PQର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{3+1}{2}, \frac{4+(-2)}{2}) = (2, 1)
(ii) (-1,3) ଓ (4, 0)ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{-1+4}{2}, \frac{3+0)}{2}) = \frac{3}{2}, \frac{3}{2}
(iii) (\frac{1}{2}, \frac{1}{3}) ଓ (\frac{1}{3}, \frac{1}{2})ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ \left(\frac{\frac{1}{2}+\frac{1}{3}}{2}, \frac{\frac{1}{3}+\frac{1}{2}}{2}\right)=\left(\frac{5}{12}, \frac{5}{12}\right)
(iv) (0,-3) ଓ (-4,0)ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{1-4}{2}, \frac{-3+0}{2}) = (-2, \frac{-3}{2})
(v) (-1, -2) ଓ (3, -1)ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{-1+3}{2}, \frac{-2-1}{2}) = (1, \frac{-3}{2})
(vi) (a, b) ଓ (c, d)ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{a+c}{2}, \frac{b+d}{2})
(vii) (-2, 1) ଓ (-3, -4)ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{-1+3}{2}, \frac{-2-1}{2}) = (1, \frac{-3}{2})
(viii) (at1², 2at1) ଓ (at2², 2at2)ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ \left(\frac{{at}_1^2+{at}_2^2}{2}, \frac{2 {at}_1+2 {at}_2}{2}\right) = (\frac{a({t}_1^2+{t}_2^2)}{2}, a(t1+t2))

Question 3.
ନିମ୍ନଲିଖତ ପ୍ରତ୍ୟେକ ପ୍ରଶ୍ନରେ ଦତ୍ତ ଦୁଇବିନ୍ଦୁକୁ ଯୋଗକରୁଥିବା ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ହେଉଛି (-1, 2) । ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ । ଓ kର ମାନ ନିର୍ଣ୍ଣୟ କର ।
(i) (h, -1), (2, k)
(ii) (5, 3), (h, k)
(iii) (1+h, k), (k, -h – 1)
(iv) (h – k, k – h), (2h, 2k)
ସମାଧାନ :
(i) (h, 1 ) ଓ (2, k) ବିନ୍ଦୁଦ୍ଵୟକୁ ଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (-1, 2) ।
\frac{h+2}{2} = -1 ⇒ h + 2 = -2 ⇒ h = -4
ଏବଂ \frac{-1+k}{2} = 2 ⇒ -1 + k = 4 ⇒ k = 5
∴ h ଓ kର ମାନ ଯଥାକ୍ରମେ – 4 ଓ 5 ।

(ii) 5, 3) ଓ (h, k) ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (-1, 2) ।
\frac{5+h}{2} = -1 ⇒ 5 + h = -2 ⇒ h = -7
ଏବଂ \frac{3+k}{2} = 2 ⇒ 3 + k = 4 ⇒ k = 1
∴ h ଓ kର ମାନ ଯଥାକ୍ରମେ – 7 ଓ 1 ।

(iii) (1+h, k), (k, – h – 1) ବିଦୁ୍ୟଦ୍ୱୟକୁ ଯୋଗକରୁଥିବା ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (-1, 2) ।
\frac{1+h+k}{2} = -1 ⇒ 1 + h + k = -2 ⇒ h + k = -3
ଏବଂ \frac{k-h-1}{2} = 2 ⇒ k – h – 1 = 4 ⇒ k – h = 5
ସମୀକରଣ (i) ଓ (ii) କୁ ଯୋଗକଲେ,
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -1
kର ମାନ ସମୀକରଣ (i)ରେ ପ୍ରୟୋଗ କଲେ, h + k = – 3
⇒ h + 1 = -3 ⇒ h – 1 – 3 = -4
∴ h = -4 ଓ k = 1

(iv) (h – k, k – h), (2h, 2k) ବିନ୍ଦୁଦ୍ଵୟକୁ ଯୋଗକରୁଥିବା ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (-1, 2) ।
\frac{h-k+2h}{2} = -1 ⇒ 3h – k = -2 ……(i)
ଏବଂ \frac{k-h-1}{2} = 2 ⇒ 3k – h = 4 …….(ii)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -2
hର ମାନ ସମୀକରଣ (i)ରେ ପ୍ରୟୋଗ କଲେ, 3h – k = -2
\frac{-3}{4} – k = -2 = -k = -2 + \frac{-3}{4} = \frac{-8+3}{4}=\frac{-5}{4} ⇒ k = \frac{5}{4}
∴ h = \frac{-1}{4} ଓ k = \frac{5}{4}

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 4.
(0, 0) ଏକ ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ । ଯଦି ରେଖାଖଣ୍ଡର ଏକ ପ୍ରାନ୍ତବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (2, 3) ହୁଏ, ତେବେ ଅନ୍ୟ ପ୍ରାନ୍ତବିନ୍ଦୁଟିର ସ୍ଥାନଙ୍କ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଏକ ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (0, 0) ଏବଂ ଏକ ପ୍ରାନ୍ତ ବିଦୁର ସ୍ଥାନାଙ୍କ (2, 3) ।
ମନେକର ଅନ୍ୟ ପ୍ରାନ୍ତବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ।
ପ୍ରଶାନୁସାରେ, \frac{x+2}{2} = 0 ଏବଂ \frac{y+3}{2} = 0 ⇒ x = – 2 ଏବଂ y = -3
∴ ଅନ୍ୟ ପ୍ରାନ୍ତବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) = (-2, -3)

Question 5.
ଏକ ରେଖାଖଣ୍ଡର ଏକ ପ୍ରାନ୍ତବିନ୍ଦୁ ଓ ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ (-2, 4) ଏବଂ (1, 2), ତେବେ ଅନ୍ୟ ପ୍ରାନ୍ତବିନ୍ଦୁଟିର ସ୍ଥାନଙ୍କ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଏକ ରେଖାଖଣ୍ଡର ଏକ ପ୍ରାନ୍ତବିନ୍ଦୁ ଓ ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ (-2, 4) ଓ (1, 2) ।
ମନେକର ଅନ୍ୟ ପ୍ରାନ୍ତ ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ।
ପ୍ରଶାନୁସାରେ, \frac{-2+x}{2} = 1 ⇒ x = 2 + 2 = 4 ଏବଂ \frac{4+ y}{2} = 2 ⇒ y = 4 – 4 = 0
∴ ଅନ୍ୟ ପ୍ରାନ୍ତବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (4, 0) ।

Question 6.
ଗୋଟିଏ ରେଖାଖଣ୍ଡର ଏକ ପ୍ରାନ୍ତବିନ୍ଦୁ ଓ ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ (3, 5) ଏବଂ (2, 1) ହେଲେ, ଅନ୍ୟ ପ୍ରାନ୍ତବିନ୍ଦୁଟିର ସ୍ଥାନଙ୍କ ସ୍ଥିର କର ।
ସମାଧାନ :
ଗୋଟିଏ ରେଖାଖଣ୍ଡର ଏକ ପ୍ରାନ୍ତବିନ୍ଦୁ ଓ ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ (3, 5) ଏବଂ (2, 1) |
ମନେକର ଅନ୍ୟ ପ୍ରାନ୍ତବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ।
ପ୍ରଶାନୁସାରେ, \frac{x+3}{2} = 2 ⇒ x = 4 – 3 = 1 ଏବଂ \frac{y+5}{2} = 1 ⇒ y = 2 – 5 = -3
∴ ଅନ୍ୟ ପ୍ରାନ୍ତବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (1, -3) ।

Question 7.
x ଓ yର କେଉଁ ମୂଲ୍ୟ ପାଇଁ (6, -2) ଓ (2, -4) ବିନ୍ଦୁଦ୍ଵୟକୁ ସଂଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡ ଏବଂ (x, 1) ଓ (-2, y) ବିନ୍ଦୁଦ୍ଵୟକୁ ସଂଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡ ପରସ୍ପରକୁ ସମଦ୍ବିଖଣ୍ଡ କରିବେ ।
ସମାଧାନ :
ଗୋଟିଏ ରେଖାଖଣ୍ଡ PQର ପ୍ରାନ୍ତବିନ୍ଦୁଦ୍ଵୟର ସ୍ଥାନାଙ୍କ P(6, – 2) ଓ Q(2, – 4) ।
ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{6+2}{2}, \frac{-2-4}{2}) = (4, -3)
ଅନ୍ୟ ଏକ ରେଖାଖଣ୍ଡ \bar{RS}ର ପ୍ରାନ୍ତବିନ୍ଦୁ ଦ୍ବୟର ସ୍ଥାନାଙ୍କ R(x, 1) ଓ S(-2, y) ।
RS ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{x+(-2)}{2}, \frac{y+1}{2}) ।
PQ ଓ \bar{RS} ପରସ୍ପରକୁ ସମଦ୍ଵିଖଣ୍ଡ କରିବେ ।
ଅର୍ଥାତ୍ ରେଖାଖଣ୍ଡ ଦ୍ଵୟର ମଧ୍ୟବିନ୍ଦୁ ଅଭିନ୍ନ ଅଟେ ।
ପ୍ରଶାନୁସାରେ, \frac{x+(-2)}{2} = 4 ⇒ x = 8 + 2 = 10 ଏବଂ \frac{1+y}{2} = -3 ⇒ y = – 6 – 1 = -7
∴ x = 10 ଓ y = -7।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 8.
(2, 3) ଓ (1, 4) ବିନ୍ଦୁଦ୍ଵୟକୁ ଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡକୁ 3 : 2 ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥିବା ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
A (x1, y1) ଓ B(x2,y2) ବିନ୍ଦୁଦ୍ଵୟର ସଂଯୋଗକାରୀ ରେଖାଖଣ୍ଡ \bar{AB} କୁ m : n ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥିବା
ବିଦୁର ସ୍ଥାନାଙ୍କ (x, y) = \left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)
(2, 3) ଓ (1, 4) ବିନ୍ଦୁଦ୍ଵୟକୁ ଯୋଗକରୁଥିବା ରେଖାଖଣ୍ଡକୁ 3 : 2 ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥ‌ିବା ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ହେଉ ।
ଏଠାରେ x1 = 2, y1 = 3, x2 = 1, y2 = 4; m = 3 ଓ n = 2
x-ସ୍ଥାନାଙ୍କ = \frac{m x_2+n x_1}{m+n}=\frac{3×1+2×2}{3+2}=\frac{3+4}{5}=\frac{7}{5}
y-ସ୍ଥାନାଙ୍କ = \frac{m y_2+n y_1}{m+n}=\frac{3×4+2×3}{3+2}=\frac{12+6}{5}=\frac{18}{5}
∴ (2, 3) ଓ (1, 4) ରେଖାଖଣ୍ଡକୁ 3 : 2 ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥିବା ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{7}{5}, \frac{18}{5})

Question 9.
(-2, 3) ଓ (5, -7) ବିନ୍ଦୁଦ୍ଵୟକୁ ଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡକୁ 3 : 4 ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥିବା ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର (-2, 3) ଓ (5, -7) ବିନ୍ଦୁଦ୍ୱୟକୁ ଯୋଗକରୁଥିବା ରେଖାଖଣ୍ଡକୁ 3 : 4 ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥିବା ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ।
ଏଠାରେ x1 = -2, y1 = 3, x2 = 5, y2 = -7; m = 3, n = 4 ।
∴ x-ସ୍ଥାନାଙ୍କ = \frac{m x_2+n x_1}{m+n}=\frac{3×5+4×(-2)}{3+4}=\frac{15-8}{7}=\frac{7}{7}=1
y-ସ୍ଥାନାଙ୍କ = \frac{m y_2+n y_1}{m+n}=\frac{3×(-7)+4(3)}{3+4}=\frac{-21+12}{7}=\frac{-9}{7}
∴ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥିବା ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (1, \frac{-9}{7}) ।

Question 10.
ଯଦି (5, 9) ବିନ୍ଦୁଟି, (7, -3) ଓ (4, k)କୁ ସଂଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡକୁ 2 : 1 ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରେ, ତେବେ kର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
(7, -3) ଓ (4, k) କୁ ଯୋଗକରୁଥିବା ରେଖାଖଣ୍ଡକୁ 2 : 1 ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥିବା ବିନ୍ଦୁଟି (5, 9) ।
ଏଠାରେ x1 = 7, y1 = -3, x2 = 4, y2 = k; m = 2, n = 1
y-ସ୍ଥାନାଙ୍କ = \frac{m y_2+n y_1}{m+n}=\frac{2k+1(-3)}{2+1}
ପ୍ରଶାନୁସାରେ, \frac{2k+1(-3)}{2+1} = 9 = 2k – 3 = 9 × 3
⇒ 2k = 27 + 3 = 30 ⇒ k = \frac{30}{2} = 15
∴ kର ମୂଲ୍ୟ 15 ଅଟେ ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 11.
ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି ସାହାଯ୍ୟରେ ପ୍ରମାଣ କର ଯେ, କୌଣସି ତ୍ରିଭୁଜର ମଧ୍ଯମାତ୍ରୟ ଏକ ବିନ୍ଦୁଗାମୀ ।
ସମାଧାନ :
ମନେକର ABC ତ୍ରିଭୁଜର A, B ଓ C ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ A (x1,y1), B (x2, y2) ଓ C (x3, y3) ।
ABC ତ୍ରିଭୁଜର ମଧ୍ଯମାତ୍ରୟ AD, BE, CF । BC ର ମଧ୍ୟବିନ୍ଦୁ D।
∴ D ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}) ।
ସେହିପରି E ଓ F ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ
E(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}), F(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -3
ମଧ୍ୟମାତ୍ରୟର ଛେଦବିନ୍ଦୁ Gକୁ ତ୍ରିଭୁଜର ଭରକେନ୍ଦ୍ର କୁହାଯାଏ । ଭରକେନ୍ଦ୍ରଠାରେ ମଧ୍ୟମାତ୍ରୟ 2 : 1 ଅନୁପାତରେ ପରସ୍ପରକୁ ଅନ୍ତର୍ବିଭକ୍ତ କରନ୍ତି ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -4
∴ ତିନୋଟି ମଧ୍ଯମାଉପରିସ୍ଥ ଓ ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଅଭିନ୍ନ ଅଟେ । ଏଥୁରୁ ସ୍ପଷ୍ଟ ଯେ କୌଣସି ତ୍ରିଭୁଜର ମଧ୍ଯମାତ୍ରୟ ଏକ ବିନ୍ଦୁଗାମୀ ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 12.
(h, 5), (-4, k) ଓ (8, 9) ବିନ୍ଦୁମାନଙ୍କଦ୍ୱାରା ଗଠିତ ତ୍ରିଭୁଜର ଭରକେନ୍ଦ୍ରର ସ୍ଥାନାଙ୍କ (-2, 6) ହେଲେ h ଓ kର ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
(h, 5), (-4, k) 8 (8, 9) ବିନ୍ଦୁମାନଙ୍କଦ୍ୱାରା ଗଠିତ ତ୍ରିଭୁଜର ଭରକେନ୍ଦ୍ରର ସ୍ଥାନାଙ୍କ (-2, 6) ।
ଏଠାରେ x1 = h, x2, = – 4, x3 = 8, y1 = 5, y2 = k, y3 = 9
ଭରକେନ୍ଦ୍ରର ସ୍ଥାନାଙ୍କ (\frac{x_1+x_3+x_3}{3}, \frac{y_1+y_3+y_3}{3}
ପ୍ରଶ୍ନନୁସାରେ, \frac{x_1+x_3+x_3}{3}=-2 ଏବଂ \frac{y_1+y_3+y_3}{3} = 6
\frac{h-4+8}{3}=-2\frac{5+k+9}{3}=6
⇒ h + 4 = -6 ⇒ k + 14 = 18
⇒ h = -6 – 4 =- 10 ⇒ k = 18 – 14 = 4
∴ h = 10 ଓ k = 4 ।

Question 13.
∆ ABCର ଭରକେନ୍ଦ୍ରର ସ୍ଥାନାଙ୍କ (1, 1) । A (3, -4), B (-4, 7) ହେଲେ, C ବିନ୍ଦୁର ସ୍ଥାନଙ୍କ ସ୍ଥିର କର ।
ସମାଧାନ :
∆ ABCର ଭରକେନ୍ଦ୍ରର ସ୍ଥାନାଙ୍କ (1, 1) । A ଓ B ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ A (3, -4) ଓ B(-4, 7) ।
ମନେକର ୯ ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ।
ପ୍ରଶ୍ନନୁସାରେ, \frac{3+(-4)+x}{3}=1 ଏବଂ \frac{-4+7+y}{3}=1
⇒ -1 + x = 3 ⇒ y + 3 = 3
⇒ x = 3 + 1 = 4 ⇒ y = 3 – 3 = 0
∴ C ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ C (4, 0) ।

Question 14.
ଗୋଟିଏ ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ମାନ (-4, 1) ଓ (3,-4) ଏବଂ (1, 3) ହେଲେ, ଏହାର ଭରକେନ୍ଦ୍ର ସ୍ଥାନାଙ୍କ ହେବ ।
ସମାଧାନ :
ଗୋଟିଏ ∆ ର ଶୀର୍ଷବିଦୁ୍ୟତ୍ରୟର ସ୍ଥାନାଙ୍କ ( 4, 1), (3, -4) ଏବଂ (1, 3) ।
ଏହାର ଭରକେନ୍ଦ୍ରର ସ୍ଥାନାଙ୍କ x = \frac{-4+3+1}{3}=\frac{0}{3}=0 ଏବଂ y = \frac{1-4+3}{3}=\frac{4-4}{3}=0
∴ ∆ର ଭରକେନ୍ଦ୍ରର ସ୍ଥାନାଙ୍କ (0, 0) ଅର୍ଥାତ୍ ମୂଳବିନ୍ଦୁ ଅଟେ । (ପ୍ରମାଣିତ)

Question 15.
A ଓ B ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ (1, 2) ଓ (5, -4) । AB ରେଖାଖଣ୍ଡ ଉପରେ ଏକ ବିନ୍ଦୁ ସ୍ଥିର କର, ଯେପରି ବିନ୍ଦୁଟିର A ବିନ୍ଦୁଠାରୁ ଦୂରତା, B ବିନ୍ଦୁଠାରୁ ଦୂରତାର 3 ଗୁଣ ହେବ ।
ସମାଧାନ :
AB ର A ଓ B ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ (1, 2) ଓ (5,-4) ।
AB ଉପରେ ଏକ ବିନ୍ଦୁ P ଯେପରି AP = 3 BP ⇒ AP : BP=3 : 1
ଏଠାରେ x1 = 1, y1 = 2, x2 = 5, y2 = -4; m = 3, n = 1
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -5

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 16.
(1, 5) ଓ (7, 2) ସ୍ଥାନାଙ୍କ ବିଶିଷ୍ଟ ବିଦୁ୍ୟଦ୍ୱୟକୁ ଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡକୁ ସମତ୍ରିଖଣ୍ଡ କରୁଥିବା ବିନ୍ଦୁଦ୍ବୟର ସ୍ଥାନାଙ୍କ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
AB ର A ଓ B ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ (1, 5) ଓ (7, 2) ।
ମନେକର AB କୁ ସମର୍ଦ୍ଦିଖଣ୍ଡ କରୁଥିବା ବିଦୁ୍ୟଦ୍ଵୟ P ଏବଂ Q ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -6
∴ (1, 5) ଓ (7, 2) ସ୍ଥାନାଙ୍କ ବିଶିଷ୍ଟ ବିଦୁ୍ୟତ୍ବକୁ ଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡକୁ ସମତ୍ରିଖଣ୍ଡ କରୁଥିବା ବିଦୁ୍ୟଦ୍ୱୟର ସ୍ଥାନାଙ୍କ (3, 4) ଏବଂ (5, 3) ।

Question 17.
O(0, 0), A (2a, 0) ଓ B (0, 2b) ହେଲେ ଦର୍ଶାଅ ଯେ OAB ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜ ଏବଂ ଏହାର କର୍ଣ୍ଣର ମଧ୍ୟବିନ୍ଦୁ ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁମାନଙ୍କଠାରୁ ସମାନ ଦୂରରେ ଅବସ୍ଥିତ ।
ସମାଧାନ :
∆ OABର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟର ସ୍ଥାନାଙ୍କ ଠ(0, 0), A(2a, 0) ଓ B (0, 2b) ।
A ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (2a, 0) ଏହା x-ଅକ୍ଷ ଉପରେ ଅବସ୍ଥିତ OA = 2a ଏକକ ।
B ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (0, 2b) ଏହା y-ଅକ୍ଷ ଉପରେ ଅବସ୍ଥିତ OB = 2b ଏକକ !
⇒ କର୍ଣ୍ଣ AB = \sqrt{(0-2a)^2 +(2b-0)^2} = \sqrt{4a^2 +4b^2} = 2\sqrt{a^2 +b^2}
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -7
ମନେକର କର୍ପୂର ମଧ୍ୟବିନ୍ଦୁ P ।
AP = BP = \frac{\sqrt{a^2 +b^2}}{2} = \sqrt{a^2 +b^2}
∴ P ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{2a+0}{2}, \frac{0+2b}{2}) = (a,b)
∴ OP = \sqrt{a^2 +b^2}
∴ AP = BP = OP ଏଥରୁ ପ୍ରମାଣିତ ଯେ, କର୍ପୂର ମଧ୍ୟବିନ୍ଦୁ P ସମକୋଣୀ ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁମାନଙ୍କଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 18.
ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି ସାହାଯ୍ୟରେ ପ୍ରମାଣ କର ଯେ, ଏକ ସାମାନ୍ତରିକ ଚିତ୍ରର କର୍ଣ୍ଣଦ୍ଵୟ ପରସ୍ପରକୁ ସମଦ୍ଵିଖଣ୍ଡ କରନ୍ତି ।
ସମାଧାନ :
ପ୍ରଦତ୍ତ ଚିତ୍ରରେ OABC ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର । ମୂଳବିନ୍ଦୁ Oର ସ୍ଥାନାଙ୍କ (0, 0) ।
ମନେକର OA = a ∴ A ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (a, 0) ।
C ବିନ୍ଦୁରୁ OA ପ୍ରତି CD ଲମ୍ବ ଅଙ୍କନ କରାଯାଉ ।
ମନେକର CD = b ଏବଂ OD = p ।
∴ C ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (p, b) ଏବଂ B ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (a + p, b) ।
ବର୍ତ୍ତମାନ OABC ସାମାନ୍ତରିକ ଚିତ୍ରର ଶୀର୍ଷବିନ୍ଦୁଗୁଡ଼ିକର ସ୍ଥାନାଙ୍କ O (0, 0), A (a, 0), B (a + p, b), C (p, b) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -8
∴ କର୍ଣ୍ଣ AC ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{a+p}{2}, \frac{o+b}{2}) = (\frac{a+p}{2}, \frac{b}{2})
କର୍ଣ୍ଣ OBର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{a+p+0}{2}, \frac{b+0}{2}) = (\frac{a+p}{2}, \frac{b}{2})
∴ ସାମାନ୍ତରିକ ଚିତ୍ରର କର୍ଣ୍ଣଦ୍ଵୟର ମଧ୍ୟବିନ୍ଦୁରୁ ସ୍ଥାନଙ୍କ ଅଭିନ୍ନ ।
ଅର୍ଥାତ୍ କର୍ଣ୍ଣଦ୍ଵୟ ପରସ୍ପରକୁ ସମଦ୍ବିଖଣ୍ଡ କରନ୍ତି । (ପ୍ରମାଣିତ)

Question 19.
ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି ସାହାଯ୍ୟରେ ଦର୍ଶାଅ ଯେ, ଗୋଟିଏ ଆୟତଚିତ୍ରର କଣ୍ଠଦ୍ଵୟ ସର୍ବସମ ଏବଂ ପରସ୍ପରକୁ
ସମାଧାନ :
ପାର୍ଶ୍ୱସ୍ଥ ଚିତ୍ରରେ OABC ଏକ ଆୟତଚିତ୍ର ।
OABC ଆୟତଚିତ୍ରର ଶୀର୍ଷବିନ୍ଦୁ ଚାରୋଟିର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ
O (0, 0), A (0, a), B (a, b), C (0, b) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -9
କର୍ଣ୍ଣ AC = \sqrt{(0 – a)^2 + (b – 0)^2} = \sqrt{a^2 +b^2}
କର୍ଣ୍ଣ OB = \sqrt{(a-0)^2+(b-0)^2}= \sqrt{a^2 +b^2}
∴ AC = OB
କର୍ଣ୍ଣ ACର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{0+a}{2}, \frac{b+0}{2}) = (\frac{a}{2}, \frac{b}{2})
କର୍ଣ୍ଣ OB ର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (\frac{a+0}{2}, \frac{b+0}{2}) = (\frac{a}{2}, \frac{b}{2})
∴ OABC ଆୟତଚିତ୍ରର କର୍ଣ୍ଣଦ୍ଵୟ ସର୍ବସମ ଏବଂ ପରସ୍ପରକୁ ସମଦ୍ଵିଖଣ୍ଡ କରନ୍ତି ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(a)

Question 20.
ପାର୍ଶ୍ୱସ୍ଥ ଚିତ୍ରରେ ABC ଏକ ସମବାହୁ ତ୍ରିଭୁଜ । A ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (a, 0) ହେଲେ,
(i) ଅନ୍ୟ ଶୀର୍ଷବିନ୍ଦୁଦ୍ଵୟର ସ୍ଥାନାଙ୍କ ନିର୍ଣ୍ଣୟ କର ।
(ii) ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
(iii) BE ମଧ୍ଯମାର ଦୈର୍ଘ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
(iv) G ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ନିର୍ଣ୍ଣୟ କର ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -10
ସମାଧାନ :
ଦତ୍ତ ଚିତ୍ରରେ ABC ଏକ ସମବାହୁ ତ୍ରିଭୁଜ । A ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (a, 0) ।
∴ CO ⊥ AB ⇒ ABର ମଧ୍ୟବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (0, 0) ।
ମନେକର \frac{a+x}{2}=0 ⇒ x = -a
\frac{0+y}{2}=0 ⇒ y = 0
∴ Bବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (-a, 0) ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -11
ସମବାହୁ ତ୍ରିଭୁଜର ବାହୁର ଦୈର୍ଘ୍ୟ (AB)
\sqrt{{a-(-a)}^2 +(0-0)^2} = √(2a)² = = √4a² = 2a
∴ ସମବାହୁ ତ୍ରିଭୁଜର ଉଚ୍ଚତା (OC)
= ବାହୁର ଦୈର୍ଘ୍ୟ × \frac{\sqrt{3}}{2} = 2a × \frac{\sqrt{3}}{2} = √3a
∴ C ବିନ୍ଦୁଟି y ଅକ୍ଷ ଉପରେ ଅବସ୍ଥିତ ହେତୁ ଏହାର y ସ୍ଥାନାଙ୍କ 0 ।
C ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (0, √3a) ।
(i) ସମବାହୁ ତ୍ରିଭୁଜର ଅନ୍ୟ ଶୀର୍ଷବିନ୍ଦୁ ଦ୍ବୟର ସ୍ଥାନାଙ୍କ B (-a, 0) ଏବଂ C (0, √3a) ।
(ii) ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ = 2a
(iii)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -12
(iv)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(b) -13

The Flower-School Question Answer Class 10 English Chapter 7 BSE Odisha

Odisha State Board BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School Textbook Exercise Questions and Answers.

Class 10th English Chapter 7 The Flower-School Question Answers BSE Odisha

The Flower-School Class 10 Questions and Answers

H. Let’s Understand The Poem:
Read the poem silently and answer the following questions.
(କବିତାଟିକୁ ନୀରବରେ ପଢ଼ ଓ ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

Question 1.
What is the poem about?
(କବିତାଟି କେଉଁ ବିଷୟରେ ଲିଖ୍ ?)
Answer:
The poem is about the flower children.

Question 2.
What rumbles in the sky?
(ଆକାଶରେ କ’ଣ ଘଡ଼ଘଡ଼ିର ପ୍ରଚଣ୍ଡ ଧ୍ବନି ସୃଷ୍ଟି କରୁଛି ?)
Answer:
The storm-clouds rumble in the sky.

Question 3.
What comes marching over the heath?
(ବ୍ୟାପକ ତୃଣ ଅଞ୍ଚଳ ଉପରେ କ’ଣ ମାଡ଼ି ଆସୁଛି ?)
Answer:
The moist (slightly wet) east wind comes marching over the heath.

Question 4.
Which word tells that the east wind is not dry?
(କେଉଁ ଶବ୍ଦ ଦର୍ଶାଉଛି ଯେ ପୂବେଇ ପବନ ଶୁଖା ନୁହେଁ ?)
Answer:
The word ‘moist’ tells that the east wind is not dry.

Question 5.
How does the moist east wind come?
(ଆର୍ଦ୍ର ପୂବେଇ ପବନ କିଭଳି ଆସେ/ ଆସୁଛି ?)
Answer:
The moist east wind comes marching.

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 6.
How do the crowds of flowers come out?
(ଫୁଲଗୁଡ଼ିକର ଦଳ କିପରି ବାହାରି ଆସୁଛି ?)
Answer:
The crowds of flowers come out suddenly from nowhere.

Question 7.
Where do the flowers dance and how?
(ଫୁଲସବୁ କେଉଁଠାରେ ନାଚୁଛି ଓ କିପରି ?)
Answer:
The flowers dance upon (ନୃତ୍ୟ କରୁଛି) the grass in wild excitement.

Question 8.
Which season is described in the poem ?
(ବର୍ଷସାରା ଫୁଲଗୁଡ଼ିକ କେଉଁଠାରେ ଥାଆନ୍ତି ?)
Answer:
In the poem, the rainy season is described.

Question 9.
Where are the flowers all year round?
( ବର୍ଷସାରା ଫୁଲଗୁଡ଼ିକ କେଉଁଠାରେ ଥାଆନ୍ତି ?)
Answer:
The flowers are at their school all year round.

Question 10.
Who are the flowers compared to?
(ଫୁଲଗୁଡ଼ିକୁ କାହା ସହିତ ତୁଳନା କରାଯାଇଛି ?)
Answer:
The flowers are compared to little children.

Question 11.
How do they do their lessons? Why?
(ସେମାନେ (ଫୁଲସବୁ) କିପରି ସେମାନଙ୍କର ପାଠପଢ଼ା କରନ୍ତି ? କାହିଁକି ?)
Answer:
The flowers do their lessons shutting their doors (ସେମାନଙ୍କର ଦ୍ଵାର ବନ୍ଦ କରି). Because this is their school hour and the time to come out hasn’t come yet.

Question 12.
Who does ‘their master’ refer to?
(‘ସେମାନଙ୍କର ମାଲିକ’ କାହାକୁ ସୂଚିତ କରୁଛି ?)
Answer:
‘Their master’ refers to their teacher.

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 13.
Their master is strict or lenient? Which line in the poem tells us so?
( ସେମାନଙ୍କର ଗୁରୁ ବା ଶିକ୍ଷକ କଠୋର ବା ସରଳ ? କବିତାରେ କେଉଁ ଧାଡ଼ି ଏହା ଦର୍ଶାଉଛି ?)
Answer:
Their master is strict. The line in the poem telling us so is, “Their master makes them stand in a corner”.

Question 14.
When are they made to stand in a corner?
(ସେମାନଙ୍କୁ (ଫୁଲଗୁଡ଼ିକୁ) କେତେବେଳେ ଗୋଟିଏ କଣରେ ଠିଆ କରାଇ ଦିଆଯାଏ ?)
Answer:
They are made to stand in a comer when they play before school is over.

Question 15.
When do the flowers have their holidays?
(ଫୁଲସବୁ କେତେବେଳେ ସେମାନଙ୍କ ଛୁଟିଦିନ ପାଆନ୍ତି ?)
Answer:
When the rains come, the flowers have their holidays.

Question 16.
What changes take place in nature when the rain comes?
(ଯେତେବେଳେ ବର୍ଷା ଆସେ ପ୍ରକୃତିରେ କେଉଁସବୁ ପରିବର୍ତ୍ତନ ହୋଇଥାଏ ?)
Answer:
When the rain comes, branches clash together in the forest, the thunderclouds clap their giant hands, and the leaves rustle in the wild wind.

Question 17.
How do the flower children enjoy their holidays?
(ଫୁଲରୂପୀ ପିଲାମାନେ କିପରି ସେମାନଙ୍କର ଛୁଟି ଉପଭୋଗ କରନ୍ତି ?)
Answer:
The flower children enjoy their holidays by coming out in the rain dressed (ବର୍ଷା) in pink, yellow, and white.

Question 18.
Who is the speaker sharing his thoughts with?
(କବି କାହା ସହିତ ନିଜର ଭାବନା ବାଣ୍ଟିଛନ୍ତି ?)
Answer:
The speaker is sharing his thoughts with his mother.

Question 19.
Where is the home of the flower children?
(ଫୁଲରୂପୀ ପିଲାମାନଙ୍କର ଘର କେଉଁଠାରେ ?)
Answer:
The home of the flower children is in the sky.

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 20.
Why are they eager to go to the sky ?
(ସେମାନେ (ଫୁଲସବୁ) କାହିଁକି ଆକାଶକୁ ଯିବା ପାଇଁ ଉତ୍ସୁକ/ବ୍ୟାକୁଳ ?)
Answer:
They are eager (ଉତ୍ସୁକ/ବ୍ୟାକୁଳ) to go to the sky as their mother in the sky calls them.

Question 21.
What does the speaker guess?
(ବକ୍ତା କ’ଣ ଅନୁମାନ କରିଛନ୍ତି ?)
Answer:
The speaker guesses to whom the flower children are raising their arms.

Question 22.
Why are the flower children raising their arms?
(ଫୁଲରୂପୀ ପିଲାମାନେ କାହିଁକି ସେମାନଙ୍କର ହାତ ଉଠାଉଛନ୍ତି ?)
Answer:
The flower children are raising (ଟେକୁଛନ୍ତି) their arms (ହାତ) to their mother living in the sky to embrace them (ସେମାନଙ୍କୁ କୋଳେଇ ନେବାପାଇଁ).

Question 23.
Why are they in a hurry?
(ସେମାନେ କାହିଁକି ଏତେ ଚଞ୍ଚଳ/ବ୍ୟଗ୍ର ହୋଇ ପଡ଼ିଛନ୍ତି ?)
Answer:
They are in a hurry as their mother in the sky calls them.

Question 24.
Which line tells that the speaker also longs for his mother?
(କବିତାରେ କେଉଁ ଧାଡ଼ି ଦର୍ଶାଉଛି ଯେ ବକ୍ତା ମଧ୍ୟ ତାଙ୍କ ମାଆଙ୍କ ପାଖକୁ ଯିବାକୁ ବ୍ୟାକୁଳ ହୋଇପଡ଼ିଛନ୍ତି ?)
Answer:
The line that tells that the speaker also longs for (wishes) his mother is “they have their mother as I have my own”.

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

I. Let’s Appreciate The Poem:
(A)

Question 1.
Why do you think the flowers dance upon the grass in wild glee?
(ତୁମେ କାହିଁକି ଭାବୁଛ ଫୁଲଗୁଡ଼ିକ ଘାସ ଉପରେ ପ୍ରବଳ ଉତ୍ତେଜନାରେ ନାଚୁଛନ୍ତି ?)
Answer:
I think the flowers dance upon the gears that come down with them. During storms, branches clash together in the forest and the leaves rustle in the wild wind. The moist east wind comes marching over the uncultivated land also.

Question 2.
Do you think the speaker feels that holidays are funnier than school days?
(ତୁମେ ଭାବୁଛ କି ସ୍କୁଲ ଦିନ ଅପେକ୍ଷା ଛୁଟିଦିନସବୁ ଅଧ୍ବକ କୌତୂହଳପ୍ରଦ ବୋଲି ବକ୍ତା ଅନୁଭବ କରୁଛନ୍ତି ?)
Answer:
Yes, I think so.

Question 3.
Why does the speaker think that the flowers go to school underground?
(ବକ୍ତା କାହିଁକି ଭାବୁଛନ୍ତି ଯେ ଫୁଲଗୁଡ଼ିକ ଭୂପୃଷ୍ଠ ତଳେ ଥ‌ିବା ବିଦ୍ୟାଳୟକୁ ଯାଆନ୍ତି ?)
Answer:
The speaker thinks that the flowers go to school underground as he notices them only when the monsoon rain comes and he does not notice the flowers anywhere in other seasons. So he thinks they must have been in school underground all year round.

Question 4.
How does the speaker describe the storm?
(ବକ୍ତା କିଭଳି ଝଡ଼କୁ ବର୍ଣ୍ଣନା କରିଛନ୍ତି ?)
Answer:
According to the speaker the storm-clouds rumble (make a long deep sound) as if they clap their giant hands and showers come down with them. During storms, branches clash together in the forest and the leaves rustle in the wild wind. The moist east wind comes marching over the uncultivated land also.

Question 5.
Do you think that the speaker’s description of the flowers and their school has any reality in real life?
ତୁମେ କ’ଣ ଭାବୁଛ ବକ୍ତାଙ୍କର ଫୁଲ ଓ ସେମାନଙ୍କର ବିଦ୍ୟାଳୟ ସମ୍ପର୍କିତ ବର୍ଣ୍ଣନାର ଜୀବନର ବାସ୍ତବତା ସହ କିଛି ସମ୍ପର୍କ ଅଛି ?)
Answer:
Yes, I think that the speaker’s description of the flowers and their school has a reality in real life.

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 6.
Which elements of nature celebrate holidays with the flower children and how?
(ପ୍ରକୃତିର କେଉଁସବୁ ଉପାଦାନ ଫୁଲରୂପୀ ପିଲାମାନଙ୍କ ସହିତ ଛୁଟି ପାଳନ କରୁଛନ୍ତି ଓ କିପରି ?)
Answer:
Elements of nature like (ପରି) branches of trees in the forest, their leaves, and thunderclouds celebrate holidays with the flower children. With the blowing of wild wind, branches clash together in the forest and leaves make rustling sounds, and the thunderclouds rumble as if they clap with their big hands. In this way, they celebrate their holiday.

Question 7.
Identify the lines and phrases in the poem which indicate that the flower children have a strict system of schooling.
(କବିତାରେ ଫୁଲରୂପୀ ପିଲାମାନଙ୍କର କଠୋର ବିଦ୍ୟାଳୟ ବ୍ୟବସ୍ଥା ଥ‌ିବା କଥା ସୂଚିତ କରୁଥିବା ଧାଡ଼ି ଓ ବାକ୍ୟାଶଗୁଡ଼ିକୁ ଚିହ୍ନିତ କର ।)
Answer:
The lines and phrases in the poem which indicate that the flower children have a strict system of schooling are: “They do their lessons with doors shut, and if they want to come out to play before it is time, their master makes them stand in a comer. ”

(B)
Question 1.
The poet presents a lively description of nature during the monsoon
showers in June. The objects of nature seem to behave like human beings. Such a device in poetry is known as personification. (କବି ଜୁନ୍ ମାସର ମୌସୁମୀ ବର୍ଷା ସମୟର ପ୍ରକୃତିର ଏକ ଜୀବନ୍ତ ବିବରଣୀ ଉପସ୍ଥାପିତ କରିଛନ୍ତି । ପ୍ରକୃତିର ପଦାର୍ଥଗୁଡ଼ିକ ମନୁଷ୍ୟ ଭଳି ବ୍ୟବହାର କରୁଥ‌ିବାର ଜଣାଯାଉଛି । କବିତାରେ ଏଭଳି ଅଳଙ୍କାର କୁ personification ବା ‘ବ୍ୟକ୍ତିତ୍ବ ଆରୋପଣ’ କୁହାଯାଏ ।)
Personification is a device in which a thing or an idea or an animal is given human qualities and described as a living thing. (Personification ବା ବ୍ୟକ୍ତିତ୍ବ ଆରୋପଣ ହେଉଛି ଏକ ଅଳଙ୍କାର ଯେଉଁଥରେ ଏକ ପଦାର୍ଥ ବା ଏକ ଧାରଣା କିମ୍ବା ଏକ ପ୍ରାଣୀ ମାନବୀୟ ଗୁଣାବଳୀ ପ୍ରଦତ୍ତ ହୋଇଥାଏ ଏବଂ ଏକ ଜୀବନ୍ତ ପଦାର୍ଥ ଭଳି ବର୍ଣ୍ଣନା କରାଯାଇଥାଏ ।)
Pick out as many such examples as you can from the poem. One is done for you.
(ତୁମେ କବିତା ଯେତେ ପାରୁଛ ସେହିଭଳି ଉଦାହରଣ ବାହାର କର । ଗୋଟିଏ ତୁମ ପାଇଁ କରିଦିଆଯାଇଛି ।)
Answer:

  • Storm clouds rumble in the sky.
  • June showers come down.
  • The moist east wind comes marching.
  • Crowds of flowers come out and dance upon the grass in wild glee.
  • Branches clash together in the forest.
  • The leaves rustle in the wild wind.
  • The thunderclouds clap their giant hands.
  • The flower children rush out in dresses of pink and yellow and white.

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 2.
Poets use words and pictures to bring out comparisons between persons, ideas or objects with similar quality or appearance. (ସମାନ ଗୁଣ ବା ଚେହେରା ଥ‌ିବା ବ୍ୟକ୍ତି, ଧାରଣା ବା ପଦାର୍ଥଗୁଡ଼ିକ ମଧ୍ୟରେ ତୁଳନା କରିବା ପାଇଁ କବିମାନେ ଶବ୍ଦ ଛବି ବ୍ୟବହାର କରିଥା’ନ୍ତି ।)
What are the things below compared to in the poem you just read?
(ତୁମେ ଏବେ ପଢ଼ିଥିବା କବିତାରେ ନିମ୍ନଲିଖୁ ଜିନିଷଗୁଡ଼ିକୁ କାହା ସହ ତୁଳନା କରାଯାଇଛି ?)
One is done for you.(ଗୋଟିଏ ତୁମପାଇଁ କରି ଦିଆଯାଇଛି ।)

  • the bamboo — bagpipes
  • the flowers —
  • the underground —
  • the rains —
  • pink, yellow and white —
  • the sky —
  • the stars —

Answer.
the bamboo — bagpipes
the flowers — children
the underground — the flower-school
the rains — living animals
pink, yellow and white — colorful dresses
the sky — a living place like the earth
the stars — flower’s mothers

Question 3.
The theme of a poem is the main idea of the poem. The theme may not always be stated, but can be hinted at indirectly. Which of the following ideas supports the theme of ‘The Flower — School’? Tick your choice.
(କବିତାର ବିଷୟବସ୍ତୁ ହେଉଛି କବିତାର ମୁଖ୍ୟ ଧାରଣାସମୂହ । ବିଷୟବସ୍ତୁ ସର୍ବଦା ଉଲ୍ଲେଖ ହୋଇନଥାଏ, ମାତ୍ର ପରୋକ୍ଷରେ ସୂଚିତ କରାଯାଇଥାଏ । ‘The Flower – School’ ବା ‘ଫୁଲ-ବିଦ୍ୟାଳୟ’ର ବିଷୟବସ୍ତୁକୁ ନିମ୍ନଲିଖ୍ କେଉଁ ଧାରଣା ସମୂହ ସମର୍ଥନ କରୁଛି ।)

  • love for wildlife
  • appreciation of nature
  • praise of God
  • featuring imagination of a flower school
  • dislike for the strict school system
  • lauding the schoolmaster
  • longing for the mother

Answer:

  • appreciation of nature
  • praise of God
  • lauding the schoolmaster

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 4.
The poet’s longing for his deceased mother creates _______ in the reader’s mind.
(କବିଙ୍କଦ୍ୱାରା ନିଜର ମୃତ ମାଆଙ୍କୁ ସ୍ମରଣ ପାଠକଙ୍କ ମନରେ ______________ସୃଷ୍ଟି କରିଛି ।)
(a) pity (b) panic
(c) pathos (d) pride
Answer:
(c) pathos (ପାଥୋସ୍)

J. Let’s Do The Activities:

Role-play
The students play the roles of ‘Small Boy’ and ‘Flower Child’ in pair and practise the dialogue naturally. They are to be invited to the front of the class to play the role. They change their role after the first round. (ଛାତ୍ରଛାତ୍ରୀମାନେ ଦୁଇଜଣିଆ ଦଳ ହୋଇ ‘ଛୋଟ ବାଳକ’ ଏବଂ ‘ଫୁଲରୂପୀ ପିଲା’ ଭୂମିକାରେ ଅଭିନୟ କରିବେ ଏବଂ ସଂଳାପଗୁଡ଼ିକ ସ୍ଵାଭାବିକ ଭାବେ ଅଭ୍ୟାସ କରିବେ । ସେମାନଙ୍କୁ ଅଭିନୟ କରିବା ପାଇଁ ବିଦ୍ୟାଳୟର ଆଗ ବେଞ୍ଚକୁ ଡକାଯିବ । ପ୍ରଥମ ରାଉଣ୍ଡ ପରେ ସେମାନେ ସେମାନଙ୍କ ଭୂମିକା ପରିବର୍ତ୍ତନ କରିବେ ।)

Small Boy        : Flower Child. Flower Child, where are you going?
Flower Child    : I’m going to school.
Small Boy        : School! You’re going to school! Which school do you go to?
Flower Child    :I go to school underground.
Small Boy        : Really? Who teaches you there?
Flower Child    : Our master.
Small Boy        : Nice. Does anybody disturb you there?
Flower Child    : No, not at all. We do our lessons with doors shut.
Small Boy        : How is your master?
Flower Child    : He’s very strict. He makes us stand in a corner when we play before school is over.
Small Boy        : Thank God. Do you always study? Don’t you have any holidays to enjoy?
Flower Child    : Of course. We have. We have holidays when the rains come.
Small Boy        : What do you do then?
Flower Child    : We wearcoIourfu1 dresses and come out to enjoy the beautiful nature.
Small Boy        : Where is your home?
Flower Child    : It is in the sky. My mother lives there. I am eager to go there.
Small Boy        : Oh really? My mother also lives there.

Question 2.
Listen and Correct:
The teacher reads aloud the following sentences with some intentional factual errors. The students listen and correct them saying: “Excuse me, Sir/Ma’m. I think it is ___________ not, but it is _____________.”
(ନିମ୍ନଲିଖ୍ ଉଦ୍ଦେଶ୍ୟମୂଳକ ଭାବେ ତଥ୍ୟଗତ ତୁଟି ସମ୍ବଳିତ ବାକ୍ୟଗୁଡ଼ିକୁ ଶିକ୍ଷକ ଉଚ୍ଚ ସ୍ୱରରେ ପଢ଼ିବେ । ପିଲାମାନେ ତାହାକୁ ଶୁଣିବେ ଏବଂ ସଂଶୋଧନ କରି କହିବେ ।)
Sentences (with errors):
(i) When the storm clouds rumble on the earth, June showers come down.
(ii) The moist west wind comes marching over the heath.
(iii) The wind blows the bagpipes among the pine trees.
(iv) The crowds of flowers dance upon the bamboos in a very excited way.
(v) The flowers go to school on the top of a hill.
(vi) The flower children do their lessons with their classroom doors open.
(vii) Their teacher makes them stand under a tree.
(viii) They have their holidays at the advent of summer.
(ix) Their home is in the sea.
(x) The flowers are eager to go to their school.
Answer:
(i) Excuse me, Sir/Ma’m. I think it is not ‘the storm clouds rumble on the earth’, but it is ‘the storm clouds rumble in the sky’.
(ii) Excuse me, Sir/Ma’m. I think it is not ‘the moist west wind comes marching’, but it is ‘the moist east wind comes marching’.
(iii) Excuse me, Sir/Ma’m. I think it is not ‘the pine trees’, but it is ‘the bamboos’.
(iv) Excuse me, Sir/Ma’m. I think it is not ‘the bamboos’, but it is ‘the grass’.
(v) Excuse me, Sir/Ma’m. I think it is not ‘on the top of a hill’, but it is ‘underground’.
(vi) Excuse me, Sir/Ma’m. I think it is not ‘classroom doors open’, but it is ‘classroom doors shut’.
(vii) Excuse me, Sir/Ma’m. I think it is not ‘stand under a tree’, but it is ‘stand in a comer’.
(viii) Excuse me, Sir/Ma’m. I think it is not ‘at the advent of summer’, but it is ‘at the advent of rain’.
(ix) Excuse me, Sir/Ma’m. I think it is not ‘in the sea’, but it is ‘in the sky’.
(x) Excuse me, Sir/Ma’m. I think it is not ‘go to their school’, but it is ‘go to their home’.

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 3.
Writing
1. Given below is the summary of the poem “The Flower-School”. Fill in the blanks with suitable words to complete the summary. You may take the help of the “HELP BOX” given below. (ନିମ୍ନରେ “The Flower-School” ବା ‘ଫୁଲ-ବିଦ୍ୟାଳୟ’ର ସାରାଂଶ ଦିଆଯାଇଛି । ଶୂନ୍ୟସ୍ଥାନରେ ଉପଯୁକ୍ତ ଶବ୍ଦ ପୂରଣ କରି ସାରାଂଶକୁ ସମ୍ପୂର୍ଣ୍ଣ କର । ତୁମେ ନିମ୍ନରେ ପ୍ରଦତ୍ତ ‘HELP BOX’ ବା ‘ସହାୟକ ବାକ୍ସ’ର ସାହାଯ୍ୟ ନେଇପାର ।)

After the first shower of June, “hen the (a) _________ wind approaches blowing its (b) _________ to herald the advance of (c) ___________, the (d) _________ bloom and (e) ________ upon the grass in (f) ________ happiness. The poet thinks that before the arrival of spring, the flowers go to a school (g) ____________ and learn their lesson. They have their holidays only when it rains and they come out rushing in colorful dresses. The sky is their home towards which they raise their (h) _________ because their (i) _________ lives there and they are always in a (j) _______ to go home.

HELP BOX

mother     arms              hurry                 great        underground

east          bagpipes       flowers            rain           dance

Answer:
After the first shower of June, when the (a) east wind approaches blowing its (b) bagpipes to herald (ଘୋଷଣା କରିବା ପାଇଁ) the advance of (c) rain, the (d) flowers bloom and (e) dance upon the grass in (f) great happiness. The poet thinks that before the arrival of spring, the flowers go to a school (g) underground and learn their lesson. They have their holidays only when it rains and they come out rushing in colorful dresses. The sky is their home towards which they raise their (h) arms because their (i) mother lives there and they are always in a (j) hurry (ତରବର) to go home.

Question 2.
Imagine that you are a child of the Flower-School. Write a letter to your mother describing your experience and feelings in your school. You may begin your letter as follows : (ମନେକର ତୁମେ Flower-Schoolର ଜଣେ ଶିଶୁ । ବିଦ୍ୟାଳୟରେ ତୁମର ଅଭିଜ୍ଞତା ଓ ଅନୁଭୂତିକୁ ବର୍ଣନା କରି ମା’ଙ୍କ ପାଖକୁ ଗୋଟିଏ ଚିଠି ଲେଖ । ତୁମେ ତୁମର ଚିଠି ନିମ୍ନ ପ୍ରକାରେ ଆରମ୍ଭ କରିପାର )

                                 Flower School
Date_______

Dear Mother,

How are you? You’ll be glad to know that I go to school every day. Do you know how and where our school is? The school is situated under-ground. _____________________________________________

_____________________________________________We make a lot of fun. But our class teacher is
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________

I am looking forward to the holidays.
I miss you a lot, Mom.

                                                                                                                                                             Yours lovingly,
(Name)

Answer:

Flower School
Date: 3rd June 2020

Dear Mother,

How are you? You’ll be glad to know that I go to school every day. Do you know how and where our school is? The school is situated underground (ଭୂତଳ). We do our lessons here. We make a lot of fun. But our class teacher is very strict ( ଅତି କଠୋର ). When we play before school is over, he makes us stand in a corner. When rains come. we enjoy our holidays a lot. We came out dressed in pink and yellow and white and joyfully dance up the grass. But, mother I long for you and am eager to go home.
lam looking forward to the holidays.
I miss you a lot, Mom.

Yours lovingly,
Pinky.

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

BSE Odisha 10th Class English The Flower-School Important Questions and Answers

Very Short & Objective Questions With Answers
Answer The Following Questions In A Word Or A Phrase.

Question 1.
Where do the storm-clouds rumble?
Answer:
in the sky

Question 2.
What showers come down with the storm clouds?
Answer:
June showers

Question 3.
Who blows bagpipes?
Answer:
the moist east wind

Question 4.
Where does it blow its bagpipes?
Answer:
among the bamboos

Question 5.
Where do the crowds of flowers dance?
Answer:
upon the grass

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 6.
Whom does the wet east wind march over?
Answer:
the heath

Question 7.
Where is the school the flower go to?
Answer:
underground

Question 8.
How do flowers do their lessons?
Answer:
shutting their doors

Question 9.
Who makes the flowers stand in a corner?
Answer:
their master

Question 10.
When do the flowers have their holidays?
Answer:
with the coming of rains

Question 11.
What do branches do in the forest?
Answer:
clash (hit) together

Question 12.
What rustles in the wild wind?
Answer:
the leaves of the trees

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 13.
Who claps their giant hands?
Answer:
the thunder-clouds

Question 14.
Where is the home of flower children?
Answer:
in the sky

Question 15.
How do flower children raise their hands?
Answer:
eagerly (ବ୍ୟାକୁଳ ଭାବରେ)

Fill In The Blanks With Right Words.

1. “The Flower School” is written by ____________.
Answer:
Rabindranath Tagore

2. ____________ bring rain in June.
Answer:
Storm-clouds

3. The moist east wind marches over ____________.
Answer:
the heath

4. ____________ blows its bagpipes.
Answer:
The moist east wind

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

5. The moist east wind blows its bagpipes among ____________.
Answer:
the bamboos

6. The moist east wind blows its bagpipes by ____________.
Answer:
blowing hard like the sound flowing from bagpipes

7. When the rains come, ____________ come out suddenly.
Answer:
crowds of flowers

8. In the phrase “crowds of flowers”, the word ‘crowds’ means ____________.
Answer:
a large number of

9. “Crowds of flowers come out.” Here ‘come out’ means ____________.
Answer:
bloom or appear

10. Crowds of flowers come out of ____________.
Answer:
from unknown places

11. Crowds of flowers dance upon ____________.
Answer:
the grass

12. Crowds of flowers dance upon the grass in ____________.
Answer:
wild glee

13. The flowers go to school ____________.
Answer:
underground

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

14. The flowers of their lessons ____________.
Answer:
with doors shut

15. The actual master of the flowers is ____________.
Answer:
God

Multiple Choice Questions(Mcqs) With Answers
Pick out the correct alternative.

Question 1.
‘’The wind comes marching”, What is the figure of speech used?
(A) Simile
(B) Metaphor
(C) Personification
(D) Alliteration
Answer:
(C) Personification

Question 2.
What do the flowers do in their holidays?
(A) They blossom
(B) They wither away
(C) They remain under the cover of the earth
(D) They become fragrant
Answer:
(A) They blossom

Question 3.
Who is the poet of the poem ‘The flower school’?
(A) Rudyard Kipling
(B) Humayun Kabir
(C) R.L. Stevenson
(D) Rabindranath Tagore
Answer:
(D) Rabindranath Tagore

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 4.
Fill-in-the-Blank
The meaning of the word Rumble is ____________.
(A) a musical sound
(B) a loud resonating sound
(C) very high sound
(D) great excitement
Answer:
(B) a loud resonating sound

Question 5.
What is one word for ‘A large open area’
(A) Hall
(B) Heath
(C) Expanse
(D) Area
Answer:
(B) Heath

Question 6.
Which one below is an example of Alliteration in the poem The flower school?
(A) Bagpipes-Bamboos
(B) Storm-Sky
(C) Grass-Glee
(D) Thunder-Clouds
Answer:
(A) Bagpipes-Bamboos

Question 7.
What is the meaning of Glee?
(A) Surprised
(B) Excitement
(C) Restless
(D) Happiness
Answer:
(B) Excitement

Question 8.
‘They do their lessons with doors shut’ Who does ‘They’ refer to?
(A) Trees
(B) Clouds
(C) Leaves
(D) Flowers
Answer:
(D) Flowers

Question 9.
What is something the Crowd of Flowers does not do?
(A) clap their giant hands
(B) come out of a sudden
(C) do their lessons with doors shut
(D) dance upon the grass in wild glee
Answer:
(A) clap their giant hands

Question 10.
Who is the narrator of the poem “The flower school”
(A) Mother
(B) Father
(C) Brother
(D) Child
Answer:
(D) Child

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 11.
From where do the children come out from
(A) Home
(B) Mud
(C) School underground
(D) ‘Vessel
(C) School underground

Question 12.
Which season is referred in the poem?
(A) Summer
(B) Winter
(C) Autumn
(D) Monsoon
Answer:
(D) Monsoon

Question 13.
What happens when the Monsoon arrives?
(A) The flower children die
(B) The flower children start eating food
(C) The flower children get their holidays
(D) The flower children take a bath
Answer:
(C) The flower children get their holidays

Question 14.
According to you who is the master of the Flower children?
(A) mother
(B) poet
(C) nature
(D) birds
Answer:
(C) nature

Question 15.
Synonym of the world ‘close’
(A) open
(B) together
(C) snow
(D) shut
Answer:
(B) together

Question 16.
What represents the mother in the poem?
(A) Mother nature
(B) Rabindranath Tagore
(C) School
(D) sky
Answer:
(D) sky

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Question 17.
Which elements celebrate holidays with the flower children
(A) branches, leaves, and thunderclouds
(B) Thunder clouds, rigs, mud
(C) branches, twigs, leaves
(D) thunder clouds, twigs, mud
Answer:
(A) branches, leaves, and thunderclouds

The Flower-School Summary in English

Lead-In:
In this poem the speaker is excitedly (ପ୍ରବଳ ଆଗ୍ରହର ସହିତ) watching the monsoon rain (ମୌସୁମୀ ବର୍ଷା) and noticing (ଲକ୍ଷ୍ୟ କରୁଛନ୍ତି) small flowers that have bloomed (g© ©09) all over the place. He wonders (ଆଶ୍ଚର୍ଯ୍ୟ ପ୍ରକଟ କରିଛନ୍ତି) where they had been all the year round. He imagines (ଚିନ୍ତା|ଅନୁଭବ କରିଛନ୍ତି) they must have been in school just like him. The speaker wonders why the pretty flowers (ସୁନ୍ଦର ଫୁଲସବୁ) are so eager (ଉତ୍ସୁକ) to come out and play. He concludes (ଶେଷରେ ମତପ୍ରକାଶ କରିଛନ୍ତି) that they too need the warmth and love of their mother.

Stanzawise Explanation:
Stanza 1 (Lines 1 to 7)
When storm-clouds rumble in the sky and June showers
come down,
The moist east wind comes marching over the
heath to blow its bagpipes among the bamboo.
Then crowds of flowers come out of a sudden,
from nobody knows where, and dance upon
the grass in wild glee.

Gist: When storm clouds make a series of rumbling sounds in the sky and the heavy rain in June showers down, the soggy (wet) east wind blows over the large flat uncultivated areas with little shrubs as if blowing its bagpipes among the bamboo. All of a sudden plenty of flowers come out invisible and dance upon the grass in great delight or thrill.
ଅନୁବାଦ : ଯେତେବେଳେ ଆକାଶରେ ଝଡ଼ଜନିତ ବାଦଲ ଘଡ଼ଘଡ଼ି ଶବ୍ଦ କରେ ଓ ଜୁନ୍ ମାସର ବର୍ଷା ବର୍ଷ ଉଠେ, ଆର୍ଦ୍ରତାପୂର୍ଣ୍ଣ ପୂବେଇ ପବନ ଗୁଳ୍ମପୂର୍ଣ ବ୍ୟାପକ ପତିତ ଜମି ଉପରେ ବାଉଁଶ ବୃକ୍ଷ ମଧ୍ଯରେ ଭେରୀ ବଜାଇବାପାଇଁ ପ୍ରବାହିତ ହୋଇଆସେ । ହଠାତ୍ କେଉଁ ଅଜଣା ବା ଅଦୃଶ୍ୟ ସ୍ଥାନରୁ ଫୁଲସବୁ ଫୁଟି ବାହାରି ଆସନ୍ତି ଓ ଅତି ଆନନ୍ଦ ଓ ରୋମାଞ୍ଚରେ ସବୁଜ ଘାସ ଉପରେ ନୃତ୍ୟ କରନ୍ତି ।

Stanza 2 (Lines 8 to 13)
Mother, I really think the flowers go to school
underground.
They do their lessons with doors shut, and if
they want to come out to play before it is time,
their master makes them stand in a corner.
When the rains come they have their holidays.

Gist: The poet addresses or calls his mother and tells her that these flowers go to school underground. The flowers shutting (closing) the doors, do their lessons. Like little children at school, their master makes them stand in a corner if the flowers wish to come out to play in the gentle breeze before time. With the onset (beginning) of rain, these flowers enjoy their holidays.
ଅନୁବାଦ : କବି ଏହି ସମୟରେ ତାଙ୍କ ମାଆଙ୍କୁ ଡାକିଛନ୍ତି ଓ ସେ ତାଙ୍କୁ (ନିଜ ମା’ଙ୍କୁ) କହିଛନ୍ତି ଯେ ଏହି ଫୁଲସବୁ ଭୂପୃଷ୍ଠ ନିମ୍ନରେ ଥ‌ିବା ବିଦ୍ୟାଳୟକୁ ଯା’ନ୍ତି । ଫୁଲସବୁ ଦ୍ୱାର ବନ୍ଦ କରି ଭିତରେ ସେମାନଙ୍କ ପାଠ ପଢ଼ନ୍ତି । ଯଦି ସେମାନେ ସମୟ ପୂର୍ବରୁ ବାହାରକୁ ଆସି ମୁକ୍ତ ପବନରେ ଖେଳିବାକୁ ଚାହାନ୍ତି, ସେମାନଙ୍କର ଶିକ୍ଷକ (ମାଲିକ) ସେମାନଙ୍କୁ ଗୋଟିଏ କୋଣରେ ଠିଆ କରାଇ ଦିଅନ୍ତି । ବର୍ଷା ଆସିବାମାତ୍ରେ ଏହି ଫୁଲସବୁ ସେମାନଙ୍କର ଛୁଟି ସମୟକୁ ଖୁବ୍ ଉପଭୋଗ କରନ୍ତି ।

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Stanza3. (Lines 14 to 17)
Branches clash together in the forest and the
leaves rustle in the wild wind, the thunder-clouds
clap their giant hands and the flower children
rush Out in dresses of pink and yellow and white.

Gist: Branches of trees strike hard with one another in the forest. The leaves rustle in the hard wind. With this, the thunder-clouds appear to clap their huge hands and the flower children come out quickly as if dressed in pink, yellow, and white colors.
ଅନୁବାଦ : ବୃକ୍ଷଗୁଡ଼ିକର ଡାଳ ବର୍ଷାରେ ଜଙ୍ଗଲରେ ପରସ୍ପର ମଧ୍ୟରେ ଘର୍ଷଣ ହୋଇ ବାଡ଼େଇ ହୁଏ । ପ୍ରଚଣ୍ଡ ପବନରେ ପତ୍ରସବୁ ଖସ୍ଖସ୍ ହୁଏ । ଏଥ୍ ସହିତ ଘଡ଼ଘଡ଼ି ନାଦ କରୁଥ‌ିବା ବାଦଲଖଣ୍ଡଗୁଡ଼ିକ ସେମାନଙ୍କର ବିଶାଳ ହାତରେ ତାଳି ମାରୁଥିବାର ପ୍ରତୀୟମାନ ହୁଏ । ଫୁଲରୂପୀ ପିଲାମାନେ ଗୋଲାପୀ, ହଳଦିଆ ଓ ଧବଳ ରଙ୍ଗର ପୋଷାକ ପରିଧାନ କରି ଶୀଘ୍ର ବାହାରି ଆସନ୍ତି ।

Stanza 4 (Lines 18 to 23)
Do you know, Mother, their home is in the sky,
where the stars are?
Haven’t you seen how eager they are to get there?
Don’t you know why they are in such a hurry?
Of course, I can guess to whom they raise their arms;
they have their mother as I have my own.

Gist: The narrator asks his mother if she knows that the home of these flower children is in the star-studded sky. He asks his mother if she can’t see how eagerly they wish to return to the sky. Certainly, he (the poet or the narrator) can guess that these flower children are raising their arms to their mother living in the sky, their mother as he (the poet) himself does.
ଅନୁବାଦ : କବି ତାଙ୍କ ମା’ଙ୍କୁ ପଚାରିଛନ୍ତି ଯେ ସେ କ’ଣ ଜାଣନ୍ତି ଯେ ଏହି ଫୁଲରୂପୀ ପିଲାମାନଙ୍କର ଘର ହେଉଛି ନକ୍ଷତ୍ର ବା ତାରାଖଚିତ ଆକାଶ । ସେ ପୁନଶ୍ଚ ତାଙ୍କ ମା’ଙ୍କୁ ପ୍ରଶ୍ନ କରିଛନ୍ତି ଯେ ସେ (ତାଙ୍କ ମା’) କ’ଣ ଜାଣିପାରୁ ନାହାନ୍ତି ଏହି ଫୁଲରୂପୀ ପିଲାମାନେ କିଭଳି ଆକାଶକୁ ଫେରିଯିବାକୁ ଏତେ ବ୍ୟଗ୍ର ବା ଉତ୍ସୁକ । ପ୍ରକୃତରେ କବି ଅନୁମାନ କରିପାରୁଛନ୍ତି ଯେ ସେ ଯେପରି ନିଜ ମା’ଙ୍କ ପାଖକୁ ଯିବା ପାଇଁ ନିଜ ହାତ ଦୁଇଟି ବଢ଼ାଇ ଦିଅନ୍ତି, ଠିକ୍ ସେଇଭଳି ଏହି ଫୁଲରୂପୀ ପିଲାମାନେ ଏବେ ଆକାଶରେ ଥ‌ିବା ସେମାନଙ୍କ ମା’ ଆଡ଼କୁ ସେମାନଙ୍କର ହାତ ଉଠାଇ ଦେଇଛନ୍ତି ।

About The Poet:
Rabindranath Tagore (1861-1941) was born in Kolkata on 7th May 1861. He was a great poet, painter, patriot (ଦେଶପ୍ରେମୀ), playwright (ନାଟ୍ୟକାର), novelist (ଔପନ୍ୟାସିକ), storyteller, philosopher and educationist (ଶିକ୍ଷାବିତ୍). The common theme (ସାଧାରଣ ଥିମ୍ |) in his writing is nature. Nature left a deep-seated impression (ଗଭୀର ଆସନ) on him. He was enchanted (ବିମୋହିତ ହୋଇ ପଡ଼ୁଥିଲେ) by nature. It inculcated (ଭର୍ତ୍ତି କରିଥିଲା) a sense of freedom in him. It is also the inner voice (ଅନ୍ତଃସ୍ୱର) of his poetry. He became the first Asian to win the Nobel Prize for Literature in 1913 for his collection of poems, “Gitanjali”. He is also well known for his contribution (ଅବଦାନ) to art and music.

BSE Odisha 10th Class English Solutions Chapter 7 The Flower-School

Notes And Glossary:
rumble — make series of loud sounds (ଉଚ୍ଚ ସ୍ବରରେ ଗର୍ଜନ କରେ)
showers — rainfall lasting for a short time (କ୍ଷଣସ୍ଥାୟୀ ବର୍ଷା)
moist — slightly wet (ସାମାନ୍ୟ ଆର୍ଦ୍ର)
heath — large open uncultivated land
with shrub (ଗୁଳ୍ମପୂର୍ଣ୍ଣ ବିଶାଳ ଅଣକର୍ଷିତ ଭୂମି)
bagpipes — a musical tool (ବିଶାଳ ଭେରୀ)
in wild glee — in great excitement (ଅତି ଉତ୍ତେଜନା ବା ପ୍ରବଳ ଆନନ୍ଦରେ)
master — here, the teacher (ଶିକ୍ଷକ ବା ମାଲିକ)
clash — strike or hit against one another (ବାଡ଼େଇ ହୁଏ)
rustle — make sounds of blowing (ଖସ୍‌ଖସ୍ ହୁଏ (ପତ୍ରସବୁ))
giant — very large (ବିଶାଳ)
of course — certainly (ନିଶ୍ଚିତରୂପେ ବା ଅବଶ୍ୟ )
thunder-clouds clap — clapping of thunder clouds (ଘଡ଼ଘଡ଼ିର ଧ୍ଵନି)

BSE Odisha 10th Class English Detailed Text:

The Solitary Reaper Question Answer Class 10 English Chapter 3 BSE Odisha

Odisha State Board BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper Textbook Exercise Questions and Answers.

Class 10th English Chapter 3 The Solitary Reaper Question Answers BSE Odisha

The Solitary Reaper Class 10 Questions and Answers

G. Let’s Understand The Poem:

Answer the following questions orally. You may refer to the text with your word knowledge to locate the facts/information required.
ଭାବେ ଦିଅ । ଆବଶ୍ୟକ ତଥ୍ୟ ଚିହ୍ନଟ ବା ନିରୂପଣ ନିମନ୍ତେ ତୁମେ ତୁମ ଶବ୍ଦଜ୍ଞାନ ସହିତ ପାଠ୍ୟବିଷୟର ମଧ୍ୟ ସାହାଯ୍ୟ ନେଇପାର ।)

Question 1.
What is the central idea ((ମୁଖ୍ୟଶ ବା ସାରକଥା)of the poem? (Tick the correct answer.)
(a) Reapers can sing like birds.
(b) Sweet music appeals (ମଧୁର ସଙ୍ଗୀତ) to all.
(c) Beautiful experiences have long-lasting effects.
(d) Rich harvest (ଭଲ ଫସଲ | ଅମଳ) makes the reaper happy.
Answer:
(b) Sweet music appeals to all.

Question 2.
The setting (background — ପୃଷ୍ଠଭୂମି) of the poem is ________
(a) the Arabian deserts
(b) the British Isles
(c) the Hebrides Islands
(d) the mountain regions of Scotland
Answer:
(d) the mountain regions of Scotland

Question 3.
Who are the people described in the poem?
( କବିତାରେ କେଉଁ ବ୍ୟକ୍ତିବିଶେଷଙ୍କ ବିଷୟରେ ବର୍ଣ୍ଣନା କରାଯାଇଛି ?)
Answer:
In the poem the people passing by the poet (କବିଙ୍କ ପାଖ ଦେଇ ଅତିକ୍ରମ କରୁଥିବା କବି ଡ) are described.

Question 4.
Who does the expression “Highland Lass” refer to? Why does he describe her as “Yon solitary Highland Lass”?
(“Highland Lass” a mizig qo କରୁଛି ? ସେ (କବି) କାହିଁକି ତାକୁ (ଶସ୍ୟକଟାଳିକୁ) “Yon solitary Highland Lass” ବୋଲି ଅଭିହିତ କରିଛନ୍ତି ?)
Answer:
The expression “Highland Lass (maiden — ବାଳିକା)” refers to (implies -ସୂଚିତ କରେ |) the solitary reaper. He (the poet) describes her ((he girl or the reaper) as “Yon solitary Highland Lass” because she is all alone in the mountain regions.

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 5.
What is the girl doing?
(ଝିଅଟି କ’ଣ କରୁଛି ?)
Answer:
The girl is singing by herself (alone — ଏକାକୀ) while reaping the grain.

Question 6.
Who does the poet say “Stop here or gently pass’? Why does he say so?
(କବି କାହାକୁ କହିଛନ୍ତି “Stop here or gently pass” ? ସେ ଏହା କାହିଁକି କହିଛନ୍ତି ?)
Answer:
The poet says to the passers-by (ବାଟୋଇ ବା ପଥଚାରୀ) “Stop here or gently pass”. He says so to enjoy the reaper’s sweet song thoroughly.

Question 7.
Pick out the words which tell that the girl does not have anyone by her side.
(କବିତାରୁ ସେହି ଶବ୍ଦସମୂହ ବାଛ ଯାହା ଦର୍ଶାଉଛି ଯେ ଝିଅଟି ପାଖରେ ଆଉ କେହି ନାହାନ୍ତି ।)
Answer:
The words which tell that the girl doesn’t have anyone by her side (near her) are “Yon solitary Highland Lass” and “by herself”.

Question 8.
What is the tone of her song — happy, sad, soothing or sympathetic?
(ଝିଅଟିର ଗୀତର ଭାବ କ’ଣ – ସୁଖ, ଦୁଃଖ, ଆରାମ ବା ସହାନୁଭୂତିମୂଳକ ?)
Answer:
The tone (feeling) of her song is sadness (ବିଷାଦ).

Question 9.
Overflowing with sound – Explain.
(‘Overflowing with sound’ – (ଶବ୍ଦ ସହିତ ପ୍ରବାହିତ ହେଉଛି | )
Answer:
The expression “Overflowing with sound” means the free-flowing voice of the maiden’s song that spreads all over the valley.

Question 10.
The solitary reaper’s song reminds the poet of other singers. Who are they?
(ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ଗୀତ କବିଙ୍କୁ ଅନ୍ୟ ଗାୟିକାମାନଙ୍କ ବିଷୟରେ ମନେ ପକାଇ ଦେଉଛି । ସେମାନେ କିଏ ?)
Answer:
The solitary reaper’s song reminds (causes to remember – ମନେରଖିବାକୁ କାରଣ କରେ |) the poet of other singers. They are ‘cuckoo’ and ‘nightingale’.

Question 11.
Who sings welcome notes? Where? For whom? What for?
( ଯିଏ ସ୍ୱାଗତ ନୋଟ୍ ଗାଇଥାଏ ? ଗାନ କରିଥାଏ ? କେଉଁଠାରେ ? କାହା ପାଇଁ ? କେଉଁଥ‌ିପାଇଁ ?)
Answer:
The nightingale (ନାଇଟିଙ୍ଗଲେ |) sings welcome notes (songs) in the Arabian deserts for the band of extremely (ଅତ୍ୟନ୍ତ) tired travellers to give them the thrill of her melodious music (ପଥଶ୍ରାନ୍ତ ବାଟୋଇମାନଙ୍କୁ ମଧୁର ଗୀତର ରୋମାଞ୍ଚ ପ୍ରଦାନ କରିବା ନିମନ୍ତେ).

Question 12.
Whose voice is thrilling?
(କାହାର ସ୍ମର ରୋମାଞ୍ଚକର ଅଟେ ?)
Answer:
The cuckoo’s voice is thrilling (very exciting or joyful).

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 13.
Where does it sing? When?
(ଏହା (କୋଇଲି) କେଉଁଠାରେ ଗୀତ ଗାଏ ? କେତେବେଳେ ? )
Answer:
It (the cuckoo) sings from the farthest (remotest – ସୁଦୂର) )Hebrides in springtime.

Question 14.
Who does Wordsworth compare the farmer girl with? Why?
( କିଏ କରେWordsworth କୃଷକ ଝିଅକୁ କାହା ସହିତ ତୁଳନା କରିଛନ୍ତି ? କାହିଁକି ?)
Answer:
The poet Wordsworth compares the farmer girl with the nightingale and the cuckoo. Because the girl or the maiden or the reaper sings melodiously (very sweetly – ବହୁତ ମଧୁର) like the nightingale or the cuckoo.

Question 15.
The peasant girl’s song is not intelligible to the poet because.
Tick the right answer.) (କୃଷକ ଝିଅର ଗୀତ କବିଙ୍କ ପାଇଁ ଅସ୍ପଷ୍ଟ ବା ଅବୋଧ ଅଟେ କାରଣ ……) (ଠିକ୍ ଉତ୍ତର ପାଖରେ ଏ ଚିହ୍ନ ଦିଅ ।)
• her song is in a dialect he does not understand.
• he is far away from hearing the words of the song.
• her voice is not clear as she is humming the words.
• her voice is too soft for him to get.
Answer:
her song is in a dialect (ତାଙ୍କର ଗୀତ ଏକ ଉପଭାଷାରେ ଅଛି) he does not understand. (✓)

Question 16.
What does the phrase ‘humble lay’ mean?
(‘ନମ୍ର ଲେ’ ବାକ୍ୟର ଅର୍ଥ କ’ଣ?)
Answer:
The phrase ‘humble lay’ means the song of a regular ordinary life story.

Question 17.
The expression ‘plaintive numbers’ refers to sad music. Pick out another phrase in the poem carrying the same meaning.
(‘plaintive numbers’ ଉକ୍ତି ଦୁଃଖ ସଙ୍ଗୀତକୁ ସୂଚିତ କରୁଛି । ଏହି ଅର୍ଥ ବହନ କରୁଥ‌ିବା ଅନ୍ୟ କବିତାର ଏକ ବାକ୍ୟଶ ବାହାର କର ।)
Answer:
The expression ‘plaintive numbers’ refers to sad music. Another phrase in the poem carrying (implying -ଇଙ୍ଗିତ କରିବା) the same meaning is “a melancholy (sad) strain (song)”.

Question 18.
What does the poet mean to say “As if her song could have no ending”?
(“As if her song could have no ending” ବୋଲି କବି କେଉଁ ଅର୍ଥରେ କହିଛନ୍ତି ? )
Tick the most appropriate answer below.
(ନିମ୍ନରେ ଥ‌ିବା ଉପଯୁକ୍ତ ଉତ୍ତର ପାଖରେ I ଚିହ୍ନ ଦିଅ ।)
• Her song is too long to end.
• She keeps on singing and seems not to end.
• The poet does not want the song to end.
• The song contains an everlasting universal theme that recycles.
Answer:
The song contains (ଗୀତରର ଅଛି ) an everlasting (ଅନନ୍ତ) universal theme (ସର୍ବଭାରତୀୟ ଥିମ୍ |) which recycles. (✓)

Question 19.
The poet listens ‘motionless and still’ because _______________. (ନିଶ୍ଚଳ ଓ ନୀରବ ହୋଇ କବି ଶୁଣୁଛନ୍ତି କାରଣ ______________)
(Tick the right answer.) (ଠିକ୍ ଉତ୍ତରରେ ✓ ଚିହ୍ନ ଦିଅ ।)
• the rich melodious voice of the singer holds him mesmerized and spellbound.
• he was tired after walking uphill.
• he wanted to learn the words and rhythm of the song.
• he is rooted to the spot by the girl’s beauty.
Answer:
the rich melodious voice of the singer holds him mesmerised and spellbound (ବିମୋହିତ ଓ ଅଭିଭୂତ କରି ଦେଇଛି). (✓)

Question 20.
How did the song affect the narrator?
(ଗୀତଟି କିପରି କବିଙ୍କୁ ପ୍ରଭାବିତ କରିଥିଲା ?)
Answer:
The poet continues to hum (ଗୁଣୁଗୁଣୁ କରି ଗାଉଛନ୍ତି)the melodious song of the reaper even after a long period of time had passed since he first listened to the song.

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 21.
In stanza 1 and stanza 2, four words and phrases have been used to show that the girl working in the fields is without anyone by her. Pick out these words and phrases.
(ପଡ୍‌-୧ ଓ ପଡ୍‌କ୍ସି-୨ରେ ଚାରୋଟି ଶବ୍ଦ ଓ ବାକ୍ୟାଶ ବ୍ୟବହୃତ ହୋଇଛି ଯାହାକି ଝିଅଟି କ୍ଷେତରେ ନିଃସଙ୍ଗ ଭାବେ କାର୍ଯ୍ୟ କରୁଥ‌ିବାର ଦର୍ଶାଉଛି । ସେହି ଶବ୍ଦ ବା ବାକ୍ୟଶଗୁଡ଼ିକୁ ବାହାର କର/ଚୟନ କର ।)
Answer:
Four words or phrases in stanza 1 and stanza 2 have been used to show that the girl working in the fields is without anyone by her. These words or phrases are: ‘Single in the field’, ‘Yon solitary Highland Lass’, ‘Reaping and singing by herself and the word ‘Alone’.

Question 22.
The theme of the solitary reaper’s song contains sadness. What other words are used in place of ‘sad’ ?
(ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ଗୀତର ସାରମର୍ମ ଦୁଃଖଭିଭିକ । ‘Sad’ ବଦଳରେ ଅନ୍ୟ କେଉଁ ଶବ୍ଦଗୁଡ଼ିକ ବ୍ୟବହୃତ ହୋଇଛି ?)
Answer:
The theme of the ‘solitary reaper’s song contains sadness. The other words used in place of ‘sad’ are ‘melancholy’ (ବିଷାଦଗ୍ରସ୍ତ), ‘plaintive’ (କରୁଣ), ‘sorrow’ and ‘pain’.

Question 23.
What are the two synonyms for the ‘young girl’?
(‘ଯୁବତୀ’ ପାଇଁ ଦୁଇଟି ସମକକ୍ଷ କ’ଣ?)
Answer:
The two synonyms (ପ୍ରତିଶବ୍ଦ) for the ‘young girl’ are ‘Lass’ and ‘Maiden’.

Question 24.
Three other words are used to mean ‘song’. What are they?
(‘song’ ବା ଗୀତକୁ ସୂଚିତ କରୁଥ‌ିବା ଅନ୍ୟ ତିନୋଟି ଶବ୍ଦ ବ୍ୟବହୃତ ହୋଇଛି । ସେଗୁଡ଼ିକ କ’ଣ ?)
Answer:
The three other words to mean ‘song’ are ‘strain’, ‘numbers’, and ‘lay’.

Question 25.
A melancholy strain’ in stanza 2 means ‘sad song’. Find out another phrase in stanza 5 with a similar meaning.
(ପଡ୍‌-୨ରେ ‘A melancholy strain’ର ଅର୍ଥ ହେଉଛି ‘ଦୁଃଖ ଗୀତ’ । ପଙ୍‌-୫ରେ ଏହି ଅର୍ଥ ବହନ କରୁଥିବା ଅନ୍ୟ ଏକ ବାକ୍ୟଶ ବାହାର କର ।)
Answer:
‘A melancholy strain’ in stanza 2 means ‘sad song’. Another phrase in stanza 5 with a similar meaning is ‘plaintive numbers’.

Question 26.
Which word in stanza 5 expresses the poet’s guess?
( ପଙ୍‌କ୍ତି-୫ରେ କେଉଁ ଶବ୍ଦ କବିଙ୍କର ଅନୁମାନକୁ ପ୍ରକାଶ କରୁଛି ?)
Answer:
The word ‘perhaps’ (ବୋଧହୁଏ) in stanza 5 expresses the poet’s guess.

F. Let’s Appreciate The Poem:

Question 1.
Describe what picture on the valley and the farm worker come to your mind as you read the poem.
(ତୁମେ କବିତାକୁ ପଢ଼ିବାବେଳେ ଉପତ୍ୟକା ଓ କୃଷିକ୍ଷେତ୍ରରେ କାର୍ଯ୍ୟରତ ଶ୍ରମଜୀବୀର କେଉଁ ଛବି ତୁମ ମନକୁ ଆସୁଛି ବର୍ଣ୍ଣନା କର ।)
Answer:
As I read the poem. the background of the poem which is set in a mountain valley and a solitary reaper girl working there comes to my mind. The valley spreads across the mountains. The green lush alley (ସବୁଜିମା ଭରା ଗଳିକନ୍ଦି) looks feasting (ଭୋଜିଭାତ) to the eyes of the on lookers (ନିରୀକ୍ଷଣକାରୀ). In the field of the valley, the sight of the solitary (lonely) reaper comes to the notice of the poet. The lonely girl while cutting and binding the grain is found singing a gloomy (ବିଷାଦମୟ ଗୀତ) song. The girl presents the picture of a fresh inspiring (ପ୍ରେରଣାଦାୟୀ) and enjoyable (ଉପଭୋଗ୍ୟ) commodity (ଦ୍ରବ୍ୟ) in the widespread solitariness (ଏକାକୀପଣ) of the dull (ନିସ୍ତେଜ) railey.

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 2.
Why do you think Wordsworth has chosen the song of the nightingale and cuckoo for comparison with the solitary reaper’s song?
(ତୁମେ କାହିଁକି ଭାବୁଛ Wordsworth ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ଗୀତ ସହ ତୁଳନା କରିବା ପାଇଁ ବୁଲ୍‌ବୁଲ୍ ଓ କୋଇଲିର ଗୀତକୁ ଚୟନ କରିଛନ୍ତି ?)
Answer:
Obviously (ଅବଶ୍ୟ କବି) the poet William Wordsworth is absolutely (thoroughly – ସମ୍ପୂର୍ଣ୍ଣ ଭାବରେ) )captivated (ବିମୋହିତ) by the hunting melody (ଶିକାର ମେଲୋଡି) of the reaper’s song though he hasn’t been able to understand the theme. The maiden’s song reminds (ମନେ ପକାଇ ଦେଲା) him of the songs of the nightingale and the cuckoo. The poet’s choosing the song(s) of these two birds stands crystal clear ( ନିର୍ମଳ) to us. The tired travelers of the Arabian deserts listening to the soulful (sweet – ମଧୁର) music of the nightingale and the people getting immense joy (ଅସୀମ ଆନନ୍ଦ) from the note (song) of the cuckoo flowing (ବହିୟାଉଛି କୋକିଲ) from the far-off Hebrides in spring color of the poet’s imagination (କଳ୍ପନା). The songs of the reaper as well as (and) those of the nightingales and the cuckoos though (ଯଦିଓ) unintelligible (not fit to be understood – ବୁଝିବା ଯୋଗ୍ୟ ନୁହେଁ) have the source (ଉତ୍ସ) of perennial joy (ଚିରନ୍ତନ ଆନନ୍ଦ) in the routine dull life (ନିତିଦିନିଆ ନୀରସ ଜୀବନରେ) of human beings.

Question 3.
Whose song is sweeter according to the poet – the nightingale and the cuckoos or the solitary reaper’s? or the solitary reapers?
(କବିଙ୍କ ଅନୁସାରେ |ମତରେ କାହାର ଗୀତ ମଧୁରତର – ବୁଲ୍‌ବୁଲ୍ ବା କୋଇଲିର ବା ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ?)
Answer:
According to the poet, the solitary reaper’s song is sweeter.

I. Let’s Write:

Question 1.
The poet cannot understand the words of the songq vet he raised several possibilities about its theme. In the diagram below are some of the possibilities. Read the stanzas-S and 6, and find out the phrases that match each. Work In pairs and complete the diagram writing the correct phrases in the blanks. One is done for you. (କବି ଗୀତର ଶବ୍ଦଗୁଡ଼ିକୁ ବୁଝିପାରୁନାହାନ୍ତି, ତଥାପି ସେ ଗୀତର ସାରମର୍ମ ବିଷୟରେ ଅନେକ ସମ୍ଭାବନାର ପ୍ରଶ୍ନ ଉଠାଇଛନ୍ତି । ନିମ୍ନ ଅଙ୍କିତ diagramରେ କେତେକ ସମ୍ଭାବନା ପ୍ରଦାନ କରାଯାଇଛି । ପଙ୍‌-୫ ଓ ୬ ପଢ଼ ଏବଂ ପ୍ରତ୍ୟେକ ସହ ଖାପ ଖାଉଥ‌ିବା ବାକ୍ୟଶଗୁଡ଼ିକୁ ଚୟନ କର, ଦୁଇ ଦୁଇଜଣ ହୋଇ କାର୍ଯ୍ୟ କର ଏବଂ diagramର ଶୂନ୍ୟସ୍ଥାନଗୁଡ଼ିକରେ ସଠିକ୍ ବାକ୍ୟାଶ ଲେଖୁ ପୂରଣ କର ।)

The poet cannot understand the words of the song 1

Answer:

The poet cannot understand the words of the song 2

Question 2.
In stanza-3 and 4, the poet compares the solitary reaper’s song with that of the nightingale and the cuckoo. On the basis of your reading the poem and your imagination, complete the table below with required information/facts. Work in groups of 4. Then check your findings with others in a brief class discussion. discussion. (ପଙ୍‌କ୍ତି – ୩ ଓ ୪ରେ କବି ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ଗୀତକୁ ବୁଲ୍‌ବୁଲ୍‌ର ଗୀତ ଓ କୋଇଲିର ଗୀତ ସହିତ ତୁଳନା କରିଛନ୍ତି । ତୁମର କବିତା ପଠନ ଓ କଳ୍ପନାକୁ ଆଧାର କରି ନିମ୍ନଲିଖ୍ ସାରଣୀ (table)କୁ ଆବଶ୍ୟକୀୟ ତଥ୍ୟ ସହ ପୂରଣ କର । ୪ ଜଣିଆ ଦଳରେ କାର୍ଯ୍ୟ କର । ତୁମର ତଥ୍ୟ ସହ ଅନ୍ୟମାନଙ୍କ ତଥ୍ୟ ସଂକ୍ଷିପ୍ତ ଆଲୋଚନା ମାଧ୍ୟମରେ ତୁଳନା କର ।)

Singer Place Listener Impact on the listener
Solitary reaper Scottish Highland the poet holds him spellbound
Nightingale
Cuckoo

Now write one paragraph for each sub-table using the facts/information available hereunder. One is done for you.
(ଏବେ ପ୍ରତି ଉପ-ସାରଣୀ ପାଇଁ ସେଥ୍ରେ ଉପଲବ୍ଧ ତଥ୍ୟଗୁଡ଼ିକୁ ନେଇ ଗୋଟିଏ ଗୋଟିଏ ଅନୁଚ୍ଛେଦ ଲେଖ । ଗୋଟିଏ ତୁମପାଇଁ କରି ଦିଆଯାଇଛି ।)

The solitary reaper was singing a melodious song as she reaped crops in the deep valley of the Scottish Highlands. The poet chanced to see (ଦେଖିବା ପାଇଁ ମନ ବଳାଇଲେ) and hear her. The tone and the tune enchanted (ଟ୍ୟୁନ୍ କବିଙ୍କୁ ମନ୍ତ୍ରମୁଗ୍ଧ କରିଦେଲା) the poet. It held him mesmerized and spellbound. The poet stood motionless and still as he listened to the song.

Answer:
On the Nightingale (ରାତି ଅଧରେ) -As the poet listened to the song of the reaper more and more, the picture of nightingale crosses his mind (ମନକୁ ଆସିଛି). When the travelers are already exhausted (very tired- ଅତି କ୍ଳାନ୍ତ) from the long journey. rest themselves in the cool shades of sorne oasis (ମରୂଦ୍ୟାନର ଶୀତଳ ଛାୟାରେ ବିଶ୍ରାମ ନିଅନ୍ତି )they chance to listen (ହଠାତ୍ ଶୁଣିବାକୁ ପାଆନ୍ତି) to the captivating music (ପୁଲକିତ ସଙ୍ଗୀତ) of the nightingale and they forget the weariness (ଅବସାଦ) of the long travel.

On the Cuckoo (କୁକୁଡା ଉପରେ) -The sweet and soulful music of the reaper’s song continued to echo (ପ୍ରତିଧ୍ଵନିତ ହେବାକୁ ଲାଗିଲା) in the poet’s heart. The cuckoo, the harbinger (announcer – ଘୋଷକ) of spring appeared to tickle (ବସନ୍ତର ଝଙ୍କାର ଉଠିଲା) the poet’s imagination. The bird’s thrilling music overflowed, breaking the silence(ନିସ୍ତବ୍ଧତାକୁ ଭାଙ୍ଗି) of the group of islands that lie to the north-west of Scotland.

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 3.
Imagine that you are the poet, William Wordsworth. Just after hearing the solitary reaper, you will continue on your walk and reach home. Try to describe your experience to your younger brother and what you saw and felt.
(ମନେକର ତୁମେ ହେଉଛି କବି William Wordsworth । (ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ଗୀତ ଶୁଣିବା ପରେ, ତୁମେ ତୁମର ଚାଲିବା ବଜାୟ ରଖ୍ ଘରେ ପହଞ୍ଚିଯିବ । ତୁମେ ଯାହା ଦେଖୁଲ ଓ ଅନୁଭବ କଲ ସେ ସମ୍ପର୍କରେ ତୁମର ଅନୁଭୂତି ତୁମର ସାନଭାଇ ନିକଟରେ ବର୍ଣ୍ଣନା କର ।)
Answer:
Dear Preeti, just now I was walking in the valley. I saw a solitary reaper working alone in the field. While working he was singing to herself. I was so mesmerized by her singing that I stopped there for a moment and listened to her song. The reaper was singing while cutting and binding the grain. She was singing in a very tragic (ମର୍ମନ୍ତୁଦ) voice. But her voice was extremely melodious (ଅତ୍ୟନ୍ତ ମଧୁର). The more I listened to the song, the more thrill I felt in my mind and heart. Even (ଏପରିକି) the musical notes (ସଙ୍ଗୀତ ସମ୍ବନ୍ଧୀୟ) of the nightingale and the cuckoo appeared to have lacked (ଅଭାବ) the depth (ଗଭୀରତା) of feeling. As I climbed up the hill and stood motionless, the reaper’s song made me spellbound (ମୋତେ ଅପାର ଆନନ୍ଦରେ ଅଭିଭୂତ କରିଦେଲା). The song continues to trasport (ଚିରନ୍ତନ ଆନନ୍ଦର ଆବେଗକୁ ମୋତେ କ୍ରମାଗତ ଦେଇଯାଉଛି) bliss (ଅନାବିଳ ଆନନ୍ଦ).

Question 4.
‘The Solitary Reaper’ is a superb panorama of events that slowly and silently glides from one to the other. Given below a glimpse of the poet’s lofty thoughts occurring in the poem. But they miss their sequence. Can you reorder them as they occur in the poem?
Write (a) / (b) / (c) / (d / (e) in the boxes to show the order. (‘The Solitary Reaper’ କବିତା ହେଉଛି ଘଟଣାସମୂହର ଏକ ଉଚ୍ଚକୋଟୀର ଚଳନ୍ତି ପ୍ରବାହ ଯେଉଁଗୁଡ଼ିକ ଧୀରେ ଧୀରେ ଗୋଟିଏରୁ ଅନ୍ୟ ଗୋଟିଏକୁ ଗତି କରୁଛି । ନିମ୍ନରେ କବିଙ୍କର କବିତାରେ ପ୍ରକାଶ ପାଇଥିବା ଚମତ୍କାର ଭାବନାଗୁଡ଼ିକର ଏକ କ୍ଷୁଦ୍ର ପ୍ରତିଛବି ପ୍ରଦାନ କରାଯାଇଛି । କିନ୍ତୁ ସେଗୁଡ଼ିକ କ୍ରମ ଅନୁସାରେ ନାହାନ୍ତି । କବିତାର ଘଟଣାକ୍ରମ ଅନୁସାରେ ସଜାଇ ଲେଖ । କ୍ରମ ନିର୍ଦ୍ଦେଶ ପାଇଁ ବାକ୍ସଗୁଡ଼ିକରେ (a) | (b) | (c) | (d) / (e) ଲେଖ ।)
(a) The poet’s guess (ଅନୁମାନ ) is that the solitary reaper’s song contains a theme of sorrow, loss or pain.
(b) Touching tone and melody of the song holds the poet mesmerised and spell bound.
(c) Poet walks up the hill carrying the maiden’s song in his heart and head.
(d) The young farm worker sings to herself (ନିଜକୁ ନିଜେ ଗୀତ ଗାଉଛି) as (ଯେପରି) she is reaping (ଅମଳ) the corn.
(e) Wordsworth compares the girl’s song with the songs of the nightingale and the cuckoo.
Answer:
(d)/(b)/(e)/(a)/(c)

Question 5.
Given below isa description similar to your experience. But some words/phrases are missing in it. Complete the description using appropriate words/phrases from the from the HELP BOX. (ତୁମର ଅନୁଭୂତି ସହ ସମାନ ଏକ ବର୍ଣ୍ଣନା ଦିଆଯାଇଛି । HELP BOXରୁ ସଠିକ୍ ଶବ୍ଦ ବା ବାକ୍ୟାଶଗୁଡ଼ିକୁ ବାଛି ନିମ୍ନ ପ୍ରଦତ୍ତ ବର୍ଣ୍ଣନାକୁ ପୂରଣ କର ।)

“Just now, J was walking, I saw a _______ in the field. She was ________ as she worked. I was so affected ________ that I ___________. She had, which seemed to _______ was a sad one, and I could not. But its ________ and melancholy sound. and its ________ reminded me of the song and. After some time I walked ___________, of the young _____ with me.”

HELP BOX

Beauty up the hill singing to herself
a nightingale in the valley stopped and listened
a cuckoo by her singing till the whole valley
the song a beautiful voice understand the words
woman’s song young farm worker touched me greatly
plaintive tone carrying the memory

Answer:
“Just now, I was walking in the valley, I saw a young farm worker (ଶ୍ରମିକ କାର୍ଯ୍ୟରତ ଯୁବତୀ) in the field. She was singing to herself as she worked. I was so affected
(ଏତେ ପ୍ରଭାବିତ ହୋଇ ଯାଇଥୁଲି)her singing that (ତାହା) I stopped and listened.
She had a beautiful voice, which seemed to fill the whole valley (ସମ୍ପୂର୍ଣ୍ଣ ଉପତ୍ୟକାକୁ ପ୍ରତିଧ୍ଵନିତ କରୁଥିବାର ଜଣାପଡୁଥିଲା ). The song was a sad one, and I could not understand the words.
But its plaintive tone (ବିଷାଦମୟ ସ୍ବର) and melancholy sound touched me greatly, and its beauty reminded me of the song of a nightingale and a cuckoo. After some time I walked the hill carrying the memory (ସ୍ମୃତି ବହନ କରି), of the young woman’s song with me.”

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 6.
Write answers to all the questions under “G. Let us understand the poem”.

BSE Odisha 10th Class English The Solitary Reaper Important Questions and Answers

Very Short A Objective Questions With Answers:
A. Answer The Following Questions In Word Or A Phrase.

Question 1.
Who was singing in the field?
Answer:
the solitary reaper/the Highland Lass

Question 2.
Where was the field?
Answer:
in the regions of the high mountains

Question 3.
What was the girl?
Answer:
a solitary reaper

Question 4.
What was the girl doing in the field?
Answer:
singing to herself while cutting the grain

Question 5.
What request does the poet make to the passers-by?
Answer:
to stop there or pass slowly

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 6.
What was overflowing with the sound of the song?
Answer:
the profound or deep valley

Question 7.
What was the vale like?
Answer:
deep and widespread

Question 8.
What sort of song was the reaper singing?
Answer:
a sad or melancholic song

Question 9.
Whose welcome notes are greeted by tired travelers?
Answer:
the nightingale’s

Question 10.
Where is the shady haunt?
Answer:
among the Arabian deserts

Question 11.
When does the cuckoo sing?
Answer:
in springtime

Question 12.
Whose song was more appealing than those of the nightingales and cuckoos?
Answer:
the reaper’s song

Question 13.
Why couldn’t the poet understand the girl’s song?
Answer:
the language of the song being unfamiliar (ଅଜଣା ଥ‌ିବାରୁ )

Question 14.
How did the poet listen to the reaper’s song?
Answer:
standing (ଠିଆ ହୋଇ) still and motionless

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 15.
What was the note of the reaper’s music?
Answer:
sad or sorrowful

B. Fill In The Blanks With Right Words:

1. The solitary reaper was busy cutting the grain in ___________.
Answer:
the field

2. The word ‘Highland’ means ___________.
Answer:
high mountain regions

3. The reaper was cutting and ___________.
Answer:
binding the grain or crops

4. ___________is resounded with the girl’s song.
Answer:
The profound valley /vale

Question 5.
The poet appeals to ___________ to stop there on gently pass.
Answer:
the passers-by

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 6.
The solitary reaper was singing a ___________.
Answer:
melancholic or sad song

Question 7.
___________are the weary bands.
Answer:
The tired travelers

Question 8.
The cuckoo’s song broke ___________.
Answer:
the silence of the seas

Question 9.
The travelers rest in ___________.
Answer:
shady or cool haunts

Question 10.
___________ gives comfort to tired travelers.
Answer:
The nightingale & sweet note

Question 11.
The sad songs of the girl flow from ___________.
Answer:
old, unhappy, and familiar matters

Question 12.
More welcome notes come from ___________.
Answer:
the nightingale

Question 13.
The reaper’s song may have ___________.
Answer:
different themes

Question 14.
Words like ‘maiden’ and ‘lass’ stand for ___________.
Answer:
the reaper

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 15.
While at work the reaper held ___________ in her hand.
Answer:
a sickle

Multiple Choice Questions (Mcqs) With Answers
Pick out the correct alternative.

Question 1.
The poem “Solitary Reaper” is written by ___________.
(A) William Wordsworth
(B) John Keats
(C) Shakespeare
(D) R.N. Tagore
Answer:
(A) William Wordsworth

Question 2.
William Wordsworth is/was famous as one of the poets.
(A) greatest romantic
(B) greatest Nature
(C) greatest social
(D) greatest traditional
Answer:
(B) greatest Nature

Question 3.
Wordsworth belongs to the age in English literature.
(A) medieval
(B) Victorian
(C) romantic
(D) ancient
Answer:
(C) romantic

Question 4.
In the line/expression “Yon Solitary Highland Lass !“. the word ‘Yon’ means.
(A) young
(B) overhear
(C) there
(D) over there
Answer:
(D) over there

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 5.
The narrator / The poet requests the passers-by to stop there or.
(A) pass noiselessly
(B) pass quickly
(C) go ahead
(D) gently pass
Answer:
(D) gently pass

Question 6.
The phrase “Highland Lass” means the girl living in the high mountain region of.
(A) England
(B) Ireland
(C) Scotland
(D) Holland
Answer:
(C) Scotland

Question 7.
The solitary reaper is reaping and by herself.
(A) binding
(B) singing
(C) harvesting
(D) sowing
Answer:
(B) singing

Question 8.
are requested by the poet to pass gently (slowly).
(A) The passers-by
(B) The travelers
(C) The farmers
(D) The villagers
Answer:
(A) The passers-by

Question 9.
In the line “Behold her single in the field”, the pronoun ‘her’ stands for.
(A) the solitary reaper
(B) the solitary girl
(C) The solitary worker
(D) The old lady
Answer:
(A) the solitary reaper

Question 10.
The deep and is filled with the sweet song of the solitary reaper.
(A) shallow valley
(B) wide valley
(C) narrow valley
(D) profound valley
Answer:
(B) wide valley

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 11.
“Deep and wide” denotes the word meant for the ‘valley’.
(A) shallow
(B) huge
(C) profound
(D) marvelous
Answer:
(C) marvelous

Question 12.
Alone cuts and binds the grain.
(A) the young farmer
(B) the old farmer
(C) the solitary worker
(D) the solitary reaper
Answer:
(D) the solitary reaper

Question 13.
The group of rest / are resting in the shady haunt.
(A) young travelers
(B) weary pilgrims
(C) rich travelers
(D) tired travelers
Answer:
(D) tired travelers

Question 14.
The soothing notes of providing (give) peace and joy to tired travelers in the desert.
(A) Cuckoo
(B) Nightingale
(C) Parrot
(D) Pegion
Answer:
(B) Nightingale

Question 15.
The Nightingale is famous for its sweetness or melodious.
(A) voice
(B) tone
(C) note
(D) song
Answer:
(A) voice

Question 16.
The melody of the breaks / is breaking the silence of the seas.
(A) nightingale
(B) cuckoo
(C) sparrow
(D) parrot
Answer:
(B) cuckoo

Question 17.
The bird cuckoo sings in.
(A) autumn
(B) winter
(C) summer
(D) spring
Answer:
(D) spring

Question 18.
The word in the poem denotes that the reaper is unmarried.
(A) yon
(B) maiden
(C) solitary
(D) single
Answer:
(B) maiden

Question 19.
The phrase “familiar matter” described in the poem means the events of.
(A) day-to-day life
(B) past life
(C) normal life
(D) material ife
Answer:
(A) day-to-day life

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Question 20.
The poet listened to the reaper’s / girl’s / farmer girl’s song still and.
(A) unmoved
(B) quiet
(C) motionless
(D) speechless
Answer:
(C) motionless

The Solitary Reaper Summary in English

Lead-In:
In the present poem, Wordsworth recalls (ୱାର୍ଡସୱର୍ଥ ସ୍ମରଣ କରେ) an experience (ଏକ ଅଭିଜ୍ଞତା)that had created a lasting impression (ଯାହା ଏକ ସ୍ଥାୟୀ ଭାବନା ସୃଷ୍ଟି କରିଥିଲା ​​|) upon his mind. The poet’s depicting (ଚିତ୍ରଣ) a solitary Highland Lass (ଏକ ନିର୍ଜନ ଉଚ୍ଚଭୂମି ଲାସ୍) and the effect (ପ୍ରଭାବ) of her sweet song on him forms the crux (major subject — ମୁଖ୍ୟ ବିଷୟ) of the poem.

Stanzawise Explanation:
Stanza 1 (Lines 1 to 4)
Behold her single in the field.
Yon solitary Highland Lass!
Reaping and singing by herself;
Stop here, or gently pass!
Gist: In the first stanza, the poet urges (calls) the by-passers to behold (to see) the solitary Highland Lass (maiden) who while reaping the corns is singing by herself (alone). The poet appeals to them (the by-passers) to stop near the girl or cross the path slowly.
ସାରାଂଶ : କବିତାର ପ୍ରଥମ ପତ୍‌ତ୍ତିରେ କବି ନିକଟରେ ଯାତାୟାତ କରୁଥିବା ପଥଚାରୀମାନଙ୍କୁ ଉଚ୍ଚ ପର୍ବତଶ୍ରେଣୀରେ ଶସ୍ୟ କାଟୁଥ‌ିବା ଓ ମନକୁ ମନ ଗୀତ ଗାଉଥିବା ଏକ ଝିଅକୁ ଲକ୍ଷ୍ୟ କରିବାପାଇଁ ଆଗ୍ରହ ପ୍ରକାଶ କରିଛନ୍ତି । ଝିଅଟି ପାଖରେ ଅଟକିଯିବା ପାଇଁ ବା ଧୀରେ ଧୀରେ ରାସ୍ତାକୁ ଅତିକ୍ରମ କରିବାପାଇଁ ସେ (କବି) ପଥଚାରୀମାନଙ୍କୁ ନିବେଦନ କରିଛନ୍ତି ।

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Stanza 2 (Lines 5 to 8)
Alone she cuts and binds the grain,
And sings a melancholy strain;
O listen! for the Vale, profound Is overflowing with sound.
Gist: The girl or the maiden is found cutting and binding the grain. While doing so, she keeps on singing a sad number (song). The poet calls upon (urges) the passers-by to listen to that melodious song. The sprawling (widespread) valley of the mountain region is overflowed with the sweetness of the song.
ସାରାଂଶ : ଉଚ୍ଚ ପର୍ବତାଞ୍ଚଳରେ କୁମାରୀ ବା ଝିଅଟି ଶସ୍ୟ କାଟି ବିଡ଼ା ବାନ୍ଧୁଥ‌ିବାର ଦେଖାଯାଉଛି । ଏହା କରୁଥିଲାବେଳେ ସେ ଏକ ଦୁଃଖ ବା ବିଷାଦଭରା ଗୀତ ଗାଇ ଚାଲିଛି । ସେହି ମଧୁର ଗୀତକୁ ଶୁଣିବା ନିମିତ୍ତ ସେ (କବି) ପଥଚାରୀମାନଙ୍କୁ ଆହ୍ବାନ ଦେଇଛନ୍ତି । ଝିଅର ଗୀତ ସମ୍ପୂର୍ଣ୍ଣ ପର୍ବଶ୍ରେଣୀର ବିସ୍ତୃତ ଉପତ୍ୟକାକୁ ବିମୋହିତ କରିଦେଇଛି ।

Stanza 3 (Lines 9 to 12)
No Nightingale did ever chaunt.
More welcome notes to weary bands
Of travelers in some shady haunt,
Among Arabian sands;
Gist: As the poet says, the hunting melody of the reaper’s song easily surpassed the music of Nightingale, which captivates the band of tired travelers taking a rest in some oasis in the Arabian desert.
ସାରାଂଶ : କବିଙ୍କ କହିବା ଅନୁସାରେ ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ମଧୁର ମର୍ମସ୍ପର୍ଶୀ ଗୀତ ସହଜରେ ନାଇଟିଙ୍ଗଲ୍ ବା ବୁଲ୍‌ବୁଲ୍ ପକ୍ଷୀର ସଙ୍ଗୀତକୁ ଟପିଯାଇଥିଲା, ଯାହା ଆରବ ମରୁଭୂମିର ମରୂଦ୍ୟାନରେ ବିଶ୍ରାମରତ କ୍ଳାନ୍ତ ପଥକଗୋଷ୍ଠୀକୁ ବିମୋହିତ କରିଥାଏ ।

Stanza 4 (Lines 13 to 16)
A voice so thrilling ne ’er was heard
In springtime from the Cuckoo-bird
Breaking the silence of the seas
Among the farthest Hebrides.
Gist: The melodious voice of the solitary reaper was so appealing. Even the cuckoo’s song in spring lacks the intensity of thrill. The reaper’s breathtaking (very sweet) voice seemed to have broken the silence of the cluster (group) of islands off the north Atlantic coast of Scotland.
ସାରାଂଶ : ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ମଧୁର ସ୍ଵର ଅତ୍ୟନ୍ତ ମୁଗ୍‌ଧକାରୀ ଥିଲା । ଏପରିକି ବସନ୍ତରେ କୋଇଲିର କୁହୁତାନରେ ଏଭଳି ରୋମାଞ୍ଚକର ତୀବ୍ରତା ନ ଥାଏ । ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ମାଦକଭରା ସ୍ଵର ସ୍କଟ୍‌ଲାଣ୍ଡର ସୁଦୂର ଉତ୍ତର ଆଟଲାଣ୍ଟିକ୍ କୂଳରେ ଅବସ୍ଥିତ ଦ୍ଵୀପପୁଞ୍ଜର ନୀରବତାକୁ ଭଙ୍ଗ କରୁଥିବାର ପ୍ରତୀୟମାନ ହେଉଛି ।

Stanza 5 (Lines 17 to 20)
Will no one tell me what she sings?-
Perhaps the plaintive numbers flow
For old, unhappy, far-off things,
And battles long ago;
Gist: The poet fails to understand the theme of the song. Perhaps sad songs arising out of old, unhappy things of the past or very ancient battles could have formed the theme of the reaper’s song.
ସାରାଂଶ : ନିର୍ଜନ ଶସ୍ୟକଟାଳିର ବିଷାଦଭରା ଓ ହୃଦୟସ୍ପର୍ଶୀ ଗୀତର ସାରକଥା ବୁଝିବାରେ କବି ଅସଫଳ ହୋଇଛନ୍ତି । ବୋଧହୁଏ ଅତୀତର କୌଣସି ଦୁଃଖଦାୟକ ଘଟଣା ବା ସଙ୍ଘଟିତ ଯୁଦ୍ଧ ଝିଅଟିର ଗୀତର ସାରକଥା ହୋଇଥାଇପାରେ ବୋଲି ସେ ମନେ କରୁଛନ୍ତି ।

Stanza 6 (Lines 21 to 24)
Or is it some more humble lay,
Familiar matter of today?
Some natural sorrow, loss, or pain,
That has been, and maybe again?
Gist: The theme of the song could be the song of the everyday stories of life or the usual things of the day. Maybe matters of natural sorrow or loss or pain had produced the gist or theme of the song.
ସାରାଂଶ : ଗୀତର ସାରମର୍ମ ହୁଏତ ସବୁଦିନିଆ ଜୀବନ କାହାଣୀର ଗୀତ ବା ସାଧାରଣ ଘଟଣାବଳୀର ସନ୍ଦେଶ ହୋଇପାରେ । ଲାଗୁଥିଲା ସ୍ଵାଭାବିକ ଦୁଃଖ ବା କ୍ଷତି ବା ଯନ୍ତ୍ରଣା ଗୀତର ମାର୍ମିକ ଭାବ ହୋଇଥାଇପାରେ ।

Stanza 7 (Lines 25 to 28)
Whate ’er the theme, the Maiden sang
As if her song could have no ending;
I saw her singing at her work,
And o ’er the sickle bending; –
Gist: The theme of the song may be diverse (different). But the maiden’s (reaper’s) song seemed to have no end. The poet beheld (saw) the girl singing at her work bending low with a sickle (billhook) in her hand.
ସାରାଂଶ : ଗୀତର ସାରମର୍ମ ଭିନ୍ନ ଭିନ୍ନ ହୋଇପାରେ, କିନ୍ତୁ କୁମାରୀର (ଅବିବାହିତ ଶସ୍ୟକଟାଳିର) ଗୀତର ଅନ୍ତ ନ ଥ‌ିବାର ଜଣାପଡୁଥିଲା । ଝିଅଟି ହାତରେ ଦାଆ ଧରି ନଇଁପଡ଼ି ଶସ୍ୟ କାଟିବାରେ ବ୍ୟସ୍ତ ଥ‌ିବାର କବି ଦେଖ‌ିଲେ ।

BSE Odisha 10th Class English Solutions Chapter 3 The Solitary Reaper

Stanza 8 (Lines 29 to 32)
I listen ’d, motionless and still;
And, as I mounted up the hill,
The music in my heart I bore,
Long after it was heard no more.
Gist: The poet continued to listen to the song quietly and attentively as he climbed up the hill. Though several (many) days have passed, the musical song of the girl still continues to throb in his heart.
ସାରାଂଶ : ପାହାଡ଼ ଚଢ଼ୁଥିଲାବେଳେ କବି ଲଗାତର ଭାବେ ଶସ୍ୟକଟାଳିର ଗୀତକୁ ନୀରବ ଓ ଏକାଗ୍ର ଚିତ୍ତରେ ଶୁଣୁଥିଲେ । ଯଦିଓ ସେ ଏହି ଗୀତ ଶୁଣିବାର ଅନେକ ଦିନ ବିତିଯାଇଛି, ତଥାପି ଝିଅର ସଙ୍ଗୀତଭରା ଗୀତ ତାଙ୍କ ହୃଦୟକୁ ଏବେ ବି ପୁଲକିତ କରୁଛି ।

About The Poet:
William Wordsworth was born on 7th April 1770, in Cockermouth in the Lake District, England. He is regarded (ତାଙ୍କୁ ସମ୍ମାନିତ କରାଯାଏ) as a worshipper (ଉପାସକ) of nature. Love of nature is a major theme (ମୁଖ୍ୟ ଥିମ୍ |) of his poetry. He wrote about ordinary (ସାଧାରଣ) men and women in the language of the ordinary people. For him (ତାଙ୍କ ପାଇଁ), “Poetry is the spontaneous óverflow of powerful feelings (କବିତା ହେଉଛି ଶକ୍ତିଶାଳୀ ଭାବନାର ସ୍ବତଃସ୍ଫୂର୍ତ ଉପଦ୍ରବ) arising from emotions (ଆବେଗର ଉଦୟ) recollected in tranquility (ଶାନ୍ତିରେ ସ୍ମରଣ)”. He died at Rydal Mount and Gardens, United Kingdom on April 23, 1850.

Word Meaning / Glossary:
As written in the text (ପାଠ୍ୟ ବିଷୟରେ ପ୍ରଦତ୍ତ ଶବ୍ଦ ବା ବାକ୍ୟଶ ଅନୁସାରେ)

behold – look at or see (ଦେଖିବା)
yon – (old English) over there (ସେହିଠାରେ) or that (କିମ୍ବା ତାହା |)
solitary – single or alone or deserted ((ନିର୍ଜନ ବା ଏକାକିନୀ ))
Do you see the solitary house/farmer there?
Highland Lass – the girl living in the highlands (mountain regions) of Scotland (ସ୍କଟ୍‌ଲାଣ୍ଡ ଦେଶର ଉଚ୍ଚ ପର୍ବତାଞ୍ଚଳରେ ବାସ କରୁଥିବା ଝିଅ ବା କୁମାରୀ)
melancholy strain – sad or sorrowful song (ବିଷାଦଭରା ବା ଦୁଃଖପୂର୍ଣ୍ଣ ଗୀତ)
vale – valley (ଉପତ୍ୟକା) (poetic form of valley – valleyରକାବ୍ୟରୂପ )
The vale of the mountain is sprawling (ବିଚ୍ଛୁରିତ).
profound – widespread (ବ୍ୟାପକ)
We see a profound valley in the Himalayas.
did chaunt – sang or chanted (ଗୀଥିଲା ବା ଗାୟନ କରିଥିଲା)
weary – very tired or exhausted (ଅବସାଦଗ୍ରସ୍ତ)
The weary passer-by rested under a tree shade (ଛାଇ) for some time, band-group (ଗୋଷ୍ଠୀ)
The owner of the house has been killed by a band of robbers (ଡକାୟତଙ୍କ ବ୍ୟାଣ୍ଡ୍).
welcome notes – very sweet or melodious songs (ଅତି ମଧୁର ବା ମଧୁର ଗୀତ)
in some shady haunt – in some cool and sheltered oasis (କିଛି ଥଣ୍ଡା ଓ ଆଶ୍ରୟସ୍ଥଳୀରେ)
among Arabian sands – among the desert of Arabia (Middle East) (ଆରବ ମରୁଭୂମି ମଧ୍ୟରେ (ମଧ୍ୟ ପୂର୍ବ)
so thrilling- very exciting or amusing ( ଅତି ରୋମାଞ୍ଚିତ)
The cuckoo has a thrilling melody (ରୋମାଞ୍ଚକର).
ne’er – never (କେବେ ନୁହେଁ)
vate pround – a deep and wide valley (ଗଭୀର ଓ ବିସ୍ତୃତ ଉପତ୍ୟକା))
farthest Hebrides – the most remote group or cluster of islands that lies to the north-west of Scotland (ସ୍କଟଲାଣ୍ଡର ଉତ୍ତର-ପଶ୍ଚିମରେ ଅବସ୍ଥିତ ସବୁଠାରୁ ଦୁର୍ଗମ ଗୋଷ୍ଠୀ ବା ଦ୍ୱୀପପୁଞ୍ଜ)
plaintive numbers – sad songs (ଦୁଃଖର ଗୀତ)
humble lay – ordinary song (ସାଧାରଣ ଗୀତ)
sickle – a tool with a curved blade on a short handle for cutting grass, corn, etc. (ଘାସ ବା ଶସ୍ୟକଟା ଯନ୍ତ୍ର (ଦାଆ))
The old man is mowing (କାଟୁଛି) the grass with a sickle.
motionless – without movement (ବା ନିଶ୍ଚଳ ହୋଇ)
The boy stood motionless in fear.
battles long ago – the Scots were divided into clans or large clans (କ୍ଳାନ୍ସ) or large family groups or tribes.
In the past, these clans sometimes fought. Songs have been written about these battles. (ଅତୀତରେ ସ୍କଟ୍‌ସୀ (ସ୍କଟ୍‌ଲାଣ୍ଡର ଅସ୍ଵାସୀ) ଅନେକ ବିଶାଳ ଗୋଷ୍ଠୀରେ ବିଭକ୍ତ
Scotland – A country bordering England in the United Kindgom (ଯୁକ୍ତରାଜ୍ୟରେ ଇଂଲଣ୍ଡ ଦେଶ ସୀମାରେ ଅବସ୍ଥିତ ଅନ୍ୟ ଏକ ଦେଶ ) .

BSE Odisha 10th Class English Detailed Text:

A letter to God Question Answer Class 10 English Chapter 2 BSE Odisha

Odisha State Board BSE Odisha 10th Class English Solutions Chapter 2 A letter to God Textbook Exercise Questions and Answers.

Class 10th English Chapter 2 A letter to God Question Answers BSE Odisha

A letter to God Class 10 Questions and Answers

E. Let’s Understand The Text:

Question 1.
Where did Lencho live?
(ଲେଞ୍ଜୋ କେଉଁଠାରେ ବାସ କରୁଥିଲା ?)
Answer:
Lencho lived in a solitary house that sat on the top of a low hill in the valley.

Question 2.
What did he hope for?
(ସେ କ’ଣ ଆଶା କରୁଥିଲା ?)
Answer:
He hoped for a downpour or at least a good shower of rain which was the only thing the earth needed for a good harvest.

Question 3.
What did he say about the raindrops?
(ସେ ବର୍ଷାବିନ୍ଦୁଗୁଡ଼ିକ ବିଷୟରେ କ’ଣ କହିଥିଲା ?)
Answer:
Lencho remarked with excitement that the raindrops falling from the sky were new coins. The big drops were ten-cent pieces and the little ones were five-cent pieces.

Question 4.
How did the rain change?
(ବର୍ଷା କିପରି ବଦଳିଗଲା ?)
Answer:
The rain changed into a hailstorm and along with the rain very large hailstones began to fall.

Question 5.
What happened to Lencho’s corn Heads?
(ଲେଞ୍ଜୋର ଶସ୍ୟକ୍ଷେତଗୁଡ଼ିକର କ’ଣ ହେଲା ?)
Answer:
As a strong wind began to blow and along with the rain, large hailstones began to fall, Lencho’s corn fields looked white as if it was covered with salt. His corn fields were completely destroyed.

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 6.
Who did Lencho have faith in?
(ଲେଞ୍ଜୋ କାହା ଉପରେ ବିଶ୍ବାସ ରଖୁଥିଲା ?)
Answer:
Lencho had deep faith in God.

Question 7.
Who did he write a letter to?
(ସେ କାହା ପାଖକୁ ଚିଠିଟିଏ ଲେଖୁଲା ?)
Answer:
Lencho wrote a letter to God.

Question 8.
Who read the letter?
(ଚିଠିଟିକୁ କିଏ ପଢ଼ିଲା ?)
Answer:
The postman and then the postmaster read the letter.

Question 9.
What did the postmaster do?
(ପୋଷ୍ଟମାଷ୍ଟର କ’ଣ କଲେ ?)
Answer:
The postmaster thought of writing a letter in order not to shake Lencho’s faith in God. But after opening the letter he knew that Lencho asked for hundred pesos from God. So he collected some money from his employees and several friends and contributed a part of his salary. Then he sent the money in an envelope addressed to Lencho. He was able to send only a little more than half.

Question 10.
Was Lencho surprised to find a letter for him with money in it?
(ଲେଞ୍ଜୋ ତା’ ପାଇଁ ଟଙ୍କା ଥ‌ିବା ଚିଠିଟିଏ ଦେଖୁ ଆଶ୍ଚର୍ଯ୍ୟ ହେଲା କି ?)
Answer:
No, Lencho was not surprised to find a letter for him with money in it.

Question 11.
What was Lencho’s reaction after getting the letter?
(ଚିଠିଟି ପାଇବା ପରେ ଲେଞ୍ଜୋର ପ୍ରତିକ୍ରିୟା କ’ଣ ଥିଲା ?)
Answer:
After getting the letter Lencho did not show the slightest surprise upon seeing the money in it. But he became angry while he counted the money because he got seventy pesos only whereas he asked for 100 pesos. He knew it well that neither God could have made a mistake nor could have denied what he had asked for. He began to suspect the integrity of the post office employees. In his next letter, he requested God to send him the rest money not through mail as the post office employees were a bunch of crooks.

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

F. Let’s Read Between The Lines:

(a)
(i) Why did Lencho keep on looking at the sky throughout the morning?
(ଲେଞ୍ଜୋ କାହିଁକି ସାରା ସକାଳ ଆକାଶକୁ ଚାହିଁ ରହିଥିଲା ?)
Answer:
There was no rain for some days and the earth needed a downpour or at least a shower for a good harvest. So Lencho kept on looking towards the northeast sky and waited for rain throughout the morning and waited for rain.

(ii) Why was the field white after the storm?
(ଝଡ଼ ପରେ କ୍ଷେତ କାହିଁକି ଧଳା ହୋଇଗଲା ?)
Answer:
Due to a hailstorm for an hour, large hailstones fell on the field along with rain. So after the storm, the field was white as if it was covered with salt.

(iii) Why did Lencho say the raindrops were like ‘new coins’?
(ଲେଞ୍ଜୋ କାହିଁକି ବର୍ଷାବିନ୍ଦୁଗୁଡ଼ିକ ନୂଆ ମୁଦ୍ରା ଭଳି ବୋଲି କହିଥିଲା ?)
Answer:
Just as Lencho predicted big raindrops began to fall from the sky. Seeing it Lencho became extremely happy having a hope of a good harvest. So he. said the raindrops were as valuable as new silver coins.

(iv) Why did Lencho prefer locusts to the storm?
(ଲେଞ୍ଜୋ କାହିଁକି ଝଡ଼ ଅପେକ୍ଷା ପଙ୍ଗପାଳଙ୍କୁ ପସନ୍ଦ କରିଥିଲା ?)
Answer:
The crops of Lencho were completely destroyed by hailstorms. He thought that his family would go without food that year. He knew that a plague of locusts would have left more than that. The hail had left nothing. So he preferred locusts to the storm.

(v) Did Lencho try to find out who had sent the money to him? Why/Why not?
(କିଏ ତା’ ପାଖକୁ ଟଙ୍କା ପଠାଇଥିଲା ଜାଣିବାକୁ ଲେଞ୍ଜୋ ଚେଷ୍ଟ କଲା କି ? କାହିଁକି କାହିଁକି ନୁହେଁ ?)
Answer:
Lencho didn’t try to find out who had sent the money to him. It was because he had firm faith in God and thought that God had listened to his prayer and had sent him the money.

(vi) What would be the reaction of the post office employees when they read the second letter.
Answer:
When the post office employees read the second letter, the postmaster showed his curiosity to read the letter as what Lencho had written.

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

(b) Read the passage from the text and answer the questions that follow :
(ପାଠ୍ୟବିଷୟରୁ ଅନୁଚ୍ଛେଦଟିକୁ ପଢ଼ ଓ ତଳେ ଥ‌ିବା ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)
All through the night _______________________ act of charity.
(Five paragraphs)

(i) Who does Lencho have complete faith in?
(ଲେଞ୍ଜୋ କାହା ଉପରେ ପୂର୍ଣ୍ଣ ବିଶ୍ଵାସ ରଖିଛି ? )
Answer:
Lencho has complete faith in God.

(ii) ‘Lencho was an ox of a man’ – What does the line mean?
(‘ଲେ ବଳଦ ଭଳି ମଣିଷଟିଏ ଥିଲା’ – ଏହି ଧାଡ଼ିଟି କ’ଣ ବୁଝାଉଛି ? )
Answer:
The line ‘Lencho was an ox of a man’ means though Lencho was a man, he had to work hard in the field like an ox. The writer uses such a metaphor.

(iii) What was the postmaster like?
(ପୋଷ୍ଟମାଷ୍ଟର କିଭଳି ଥିଲେ ?)
Answer:
The postmaster was a fat man and had a friendly and pleasant personality. He was very kind and helpful also. As he was a charitable person he came forward to help Lencho by collecting and contributing money from his salary.

(iv) Why did the postmaster send money to Lencho?
(ପୋଷ୍ଟମାଷ୍ଟର ଲେଞ୍ଚୋ ପାଖକୁ କାହିଁକି ଟଙ୍କା ପଠାଇଲେ ?)
Answer:
When the postmaster saw the letter addressed to God, he was greatly surprised to think about the writer’s firm faith in God. In order not to shake the writer’s faith in God, he decided to answer the letter. He did not want Lencho to face the unfortunate results of his crop loss. He managed to send him seventy pesos to let him overcome his loss of crops.

(v) What does the expression ‘an act of charity’ mean?
(‘ଏକ ବଦାନ୍ୟତାପୂର୍ଣ୍ଣ କାର୍ଯ୍ୟ’ ଶବ୍ଦପୁଞ୍ଜର ଅର୍ଥ କ’ଣ ?)
Answer:
The expression ‘an act of charity’ means an act of showing kindness and generosity towards people who are in distress.

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

G. Let’s Learn Some New Words:

(i) Look at the following sentence from the story.
(ଗଳ୍ପର ନିମ୍ନଲିଖତ ବାକ୍ୟଟିକୁ ଦେଖ ।)
But suddenly a strong wind began to blow and along with the rain very large hailstones began to fall.

  • What are hailstones?
    (କୁଆପଥର କ’ଣ ?)
    ‘Hailstones’ are small balls of ice that fall like rain.(‘କୁଆପଥର’ ହେଉଛି ବରଫର ଛୋଟ ବରଫପେଣ୍ଡୁ ଯାହା ବର୍ଷାପରି ପଡ଼େ ।)
  • What is a hailstorm?
    (କୁଆପଥର ଝଡ଼ କ’ଣ ?)
    A storm in which hailstones fall is a ‘hailstorm’. We know that a storm is bad weather with strong winds, rain, thunder and lightning. (ଯେଉଁ ଝଡ଼ରେ କୁଆପଥର ପଡ଼େ ତାହା ହେଉଛି ‘କୁଆପଥର ଝଡ଼’ । ଆମେ ଜାଣୁ ଯେ ଝଡ଼ ହେଉଛି ଏକ ଖରାପ ପାଗ ଯେଉଁଥରେ ପ୍ରବଳ ପବନ ଓ ବର୍ଷା ସାଙ୍ଗକୁ ଘଡ଼ଘଡ଼ି ଓ ବିଜୁଳି ମାରୁଥାଏ ।)

There are different names in different parts of the world for storms, depending on their nature. (ଝଡ଼ ପାଇଁ ପୃଥ‌ିବୀର ବିଭିନ୍ନ ସ୍ଥାନରେ ସେମାନଙ୍କ ପ୍ରକୃତି ଅନୁସାରେ ଭିନ୍ନ ଭିନ୍ନ ନାମ ରହିଛି ।). Try to match the names in the box with their descriptions below, and fill in the blanks. (ନିମ୍ନସ୍ଥ ବର୍ଣ୍ଣନା ସହ ମେଳ ଖାଉଥ‌ିବା ନାଁଗୁଡ଼ିକୁ କୋଠରି ଭିତରୁ ବାଛିବାକୁ ଚେଷ୍ଟା କର ଏବଂ ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।)

gale, whirlwind, cyclone, hurricane, tornado, typhoon

1. A violent tropical storm in which strong winds move in a circle: __ __c __ __ __
2. An extremely strong wind: __ a __ __
3. A violent tropical storm with very strong winds: __ __ p __ __ __
4. A violent storm with strong winds, especially in the Western Atlantic Ocean: __ __ r __ __ __ __ __ __
5. A violent storm whose center is a cloud in the shape of a funnel: __ __ __ n __ __ __
6. A very strong wind that moves very fast in a spinning movement and causes a lot of damage: __ __ __ __ l __ __ __ __

Answers:
1. cyclone
2. gale
3. typhoon
4. hurricane
5. tornado
6. whirlwind

ii) Mark how the word ‘hope’ is used in these sentences from the story.
(ଗଳ୍ପର ଏହି ବାକ୍ୟଗୁଡ଼ିକରେ ‘hope’ ଶବ୍ଦଟି କିପରି ବ୍ୟବହୃତ ହୋଇଛି ଲକ୍ଷ୍ୟ କର ।)

(a) I hope it (the hailstorm) passes quickly. (ମୁଁ ଆଶା କରୁଛି ଏହା (କୁଆପଥର) ଶୀଘ୍ର ଅତିକ୍ରମ କରିବ |)
(b) There was a single hope : help from God. (ଏକମାତ୍ର ଆଶା ଥିଲା – ଈଶ୍ବରଙ୍କଠାରୁ ସାହାଯ୍ୟ)

In sentence ‘a’, hope is used as a verb which means you wish for something to happen. (ବାକ୍ୟ ‘a’ରେ ‘hope’ ଏକ verb(କ୍ରିୟା)ରୂପେ ବ୍ୟବହୃତ ହୋଇଛି ଯାହାର ଅର୍ଥ ତୁମେ କିଛି ଘଟିବ ବୋଲି ଆଶା କରୁଛ ।)
In sentence ‘b’ it is a noun meaning a chance for something to happen. (ବାକ୍ୟ ‘b’ରେ ଏହା ଏକ ବିଶେଷ୍ୟ ଅଟେ ଯାହାର ଅର୍ଥ କିଛି ଘଟିବାର ଏକ ସମ୍ଭାବନାକୁ ବୁଝାଉଛି ।)

Difference between Noun and Verb
NOUN                                                                                                  VERB
(i) a naming word (ନାମବାଚକ ଶବ୍ଦ)                                                           (i) an action word (କାର୍ଯ୍ୟବାଚକ ଶବ୍ଦ)
(ii) used as a subject or object (କର୍ତ୍ତା ବା କର୍ମରୂପେ ବ୍ୟବହୃତ)                         (ii) used as tense (କାଳରୂପେ ବ୍ୟବହୃତ)
Example:
The book (ବହିଟି) is our best friend.                                                       Did you book for Angul ?(ବୁକ୍ କରିଛନ୍ତି କି?)
They drink water (ଜଳ) .                                                                         They water (ସେମାନେ ଜଳ ଦିଅନ୍ତି |) plants.
I write on a paper (କାଗଜ) .                                                                     We paper (ଆମେ କାଗଜ କାନ୍ଥ) walls.
This is my little finger (ଆଙ୍ଗୁଠି ).                                                               Don’t finger (ଆଙ୍ଗୁଠି କରନ୍ତୁ) the mobile
Where is the bottle (ବୋତଲ) ?                                                                Mother bottled the pickle. (ବୋତଲରେ ପୂରାଇ ରଖୁ)

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Match the sentences in Column A with the meanings of ‘hope’ in Column B.
(Aସ୍ତମ୍ଭରେ ଥ‌ିବା ବାକ୍ୟଗୁଡ଼ିକ ସହିତ Bସ୍ତମ୍ଭରେ ଥ‌ିବା ‘hope’ର ଅର୍ଥଗୁଡ଼ିକ ସହିତ ମିଳାଅ ।)

Column – A Column – B
1. Will you get the subjects you want to study in college?
I hope so.
a feeling that something good will probably happen
2. I hope you don’t mind my saying this, but I don’t like the way you are arguing. thinking that this would happen (It may or may not have happened)
3. This discovery will give new hope to HIV/AIDS sufferers. stopped believing that this good thing would happen
4. We are hoping against hope that the judges would not notice our mistakes. wanting something to happen (and thinking it quite possible)
5. I called early in the hope of speaking to her before she went to school. showing concern that what you say should not offend disturb or  the other person : a way of being polite
6. Just when everybody had given up hope, the fisherman came back seven days after the cyclone. wishing for something to happen, although this is very unlikely.

Answer:

Column – A Column – B
1. Will you get the subjects you want to study in college? I hope (verb) so. (ମୁଁ ଆଶା କରୁଛି) wanting something to happen (and thinking it quite possible)
2. I hope (verb) you don’t mind my saying this, but I don’t like the way you are arguing. showing concern that what you say should not offend or disturb the other person: a way of being polite.
3. This discovery (ଆବିଷ୍କାର |) will give new hope to Hl V/AIDS sufferers. a feeling that something good will probably happen.
4. We are hoping against hope (noun) that the judges would not notice our mistakes. wishing for something to happen although this is very unlikely.
5. I called early in the hope (noun) of speaking to her before she went to school. thinking that this would happen (It may or may not have happened)
6. Just when everybody had given up hope (noun) the fisherman came back seven days after the cyclone. stopped believing that this good thing would happen.

Now read the story and make a list of words used as verbs and nouns. Make sentences of your own using them as verbs and nouns in your words.
(ଏବେ ଗପଟିକୁ ପଢ଼ ଏବଂ କ୍ରିୟା ଓ ବିଶେଷଭାବେ ବ୍ୟବହୃତ ହୋଇଥିବା ଶବ୍ଦଗୁଡ଼ିକର ଏକ ତାଲିକା କର । ସେଗୁଡ଼ିକୁ କ୍ରିୟା ଓ ବିଶେଷ୍ୟଭାବେ ବ୍ୟବହାର କରି ନିଜ ଭାଷାରେ ବାକ୍ୟ ଗଠନ କର ।)
Answer:
The list of words used as verbs and nouns in the story: are need, water, reply, work, drop, return, rain, help, place, and comment.

need         (N)
(V)
We take money from the bank as the need arises.
I need a rented house to stay in in Bhubaneswar.
water        (N)
(V)
Water is a basic need for human beings.
We should water our plants in the afternoon.
reply         (N)
(V)
He did not give reply to my letter.
He is able to reply to all letters.
work         (N)
(V)
Work is worship.
He works in a private company
 drop          (N)
(V)
Every drop of water is useful for us.
He dropped the glass which he hold in his hand
return       (N)
(V)
He has already submitted his income tax return.
He returned home after his work was over.
rain            (N)
(V)
The rain continued for two hours.
It rained heavily yesterday.
help           (N)
(V)
I need your help.
He helps me whenever I am in need.
place         (N)
(V)
Cuttack is a suitable place to live in.
He placed the telephone on a table.
comment  (N)
(V)
No unfavorable comment should be made about others.
The minister refused to comment on the rumor of his resignation

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

H. Let’s Listen And Speak:
(ଦୁଇ ଜଣ ଲେଖାଏଁ ବାପା ଏବଂ ପୁଅର ଅଭିନୟ କର । ଆବଶ୍ୟକ ହେଲେ ଏହାକୁ ମା’ ଏବଂ ଝିଅ ଅଭିନୟକୁ ବଦଳାଇ ପାର ।)

Dad: It’s quite late. Go to bed.
Son: I’ve got the final examination tomorrow.
Dad: It’s almost 12.30 at the night. No more argument. Off to bed.
Son: I have a lot to learn yet.
Dad: It’s essential to get your concepts clear.
Son: But I need to make sure that I know everything that’s required.
Dad: Isn’t it necessary to read the books to get the concepts clear?
Son: All right, father.

Read the text and prepare a dialogue like the one given above and play the roles. (ପାଠ୍ୟ ବିଷୟଟିକୁ ପଢ଼ ଏବଂ ଉପରେ ପ୍ରଦତ୍ତ ବାର୍ତ୍ତାଳାପ ଭଳି ବାର୍ତ୍ତାଳାପଟିଏ ପ୍ରସ୍ତୁତ କରି ଅଭିନୟ କର ।)

Answer:
Daughter: Mummy, I am going to bed because it is 11.30 p.m.
Mother: Have you finished your home task?
Daughter: Yes, Mum. There was little homework for today. Besides, I completed a part of it at school in my leisure hour.
Mother: Yesterday I met your English teacher at the market. He told that some of your grammar concepts are not clear. He advised giving emphasis on it.
Daughter: I am trying my best to get the concept clear with the help of my teacher.
Mother: All right. You can go now.
Daughter: Good night! Mummy.

I. Lets Learn Language:

(i) Relative Clauses (Adjective Clause) (ବିଶେଷଣ ଖଣ୍ଡବାକ୍ୟ):
Look at the following sentence : (ନିମ୍ନଲିଖ୍ ବାକ୍ୟଟିକୁ ଦେଖ )

Throughout the morning Lencho – who knew his fields intimately (ଯିଏ ନିଜ କ୍ଷେତଗୁଡ଼ିକୁ ଭଲ ଭାବରେ ଜାଣିଥିଲା) looked at the sky.
This sentence may also be written as: (ଏହି ବାକ୍ୟଟିକୁ ଏପରି ମଧ୍ୟ ଲେଖାଯାଇପାରେ )
All morning Lencho, who knew his fields intimately, looked at the sky.
The underlined parts of the sentences provide us with more information about Lencho and the woman. We call it a Relative Clause. Mark that they begin with a relative pronoun, who. Other common relative pronouns are whom, which, that, and whose.
(ରେଖାଙ୍କିତ ଅଂଶଟି ଆମକୁ ଲେଞ୍ଜୋ ବିଷୟରେ ଅଧ‌ିକ ସୂଚନା ପ୍ରଦାନ କରୁଛି । ଏହାକୁ ଆମେ “Relative clause” (ବିଶେଷଣ ଉପବାକ୍ୟ) କହୁ । ଏହା ଏକ ସର୍ବନାମ whoରେ ଆରମ୍ଭ ହୋଇଥିବା ଲକ୍ଷ୍ୟ କର । who, whom, which, that, whose ଇତ୍ୟାଦି ଅନ୍ୟାନ୍ୟ Relative Pronoun ଅଟନ୍ତି ।)

The relative clause in the above sentence is called a
non-defining relative clause because we already know the identity of the person described. We don’t need the information in the relative clause to pick the person out of a larger set. (ଉପରିଲିଖ୍ ବାକ୍ୟରେ ଥ‌ିବା Relative clauseକୁ non-defining Relative clause କୁହାଯାଏ କାରଣ ଆମେ ବର୍ଣ୍ଣନା କରାଯାଇଥିବା ଲୋକଟିର ପରିଚୟ ବିଷୟରେ ପୂର୍ବରୁ ଜାଣିସାରିଛୁ । ଏକ ବୃହତ୍ ପରିସରରୁ ସେ ବ୍ୟକ୍ତିକୁ ଜାଣିବା ପାଇଁ relative clauseରେ ଥ‌ିବା ସୂଚନା ଆମର ଆବଶ୍ୟକ ନାହିଁ ।)

A.NON-DEFINING RELATIVE CLAUSE
Non-defining Relative Clause usually has a comma preceding and following it. Some writers use a dash (-) instead (as in the story). If the relative clause comes at the end, we just put a full stop. Non-defining Relative clause 1 666 comma (,) ବିରାମ ଚିହ୍ନଟି ରହେ । କିଛି ଲେଖକ comma ପରିବର୍ତ୍ତେ dash (–) ଚିହ୍ନ ବ୍ୟବହାର କରିଥା’ନ୍ତି, ଯେପରିକି ଉକ୍ତ ଗଳ୍ପଟିରେ ଲେଖକ ବ୍ୟବହାର କରିଛନ୍ତି । ଯଦି Relative clauseଟି ଶେଷରେ ରହେ ତେବେ ଆମେ ଏକ full stop (ପୂର୍ଷଚ୍ଛେଦ) ଦେଇଥାଉ ।)

A Non-defining Relative clause does not identify its antecedent (noun phrase used before the relative pronoun). (ଯଥା – whose, whom, what, when ଇତ୍ୟାଦିର ଠିକ୍ ପୂର୍ବରୁ 4 noun phrase antecedentକୁ ଚିହ୍ନଟ କରେ ନାହିଁ ।)
It (NDRC) only gives extra or additional information about its antecedent. (NDRC କେବଳ ନିଜର antecedent ବିଷୟରେ ଅତିରିକ୍ତ ସୂଚନା ଦେଇଥାଏ ।)

A clause (ଏକ ଧାରା |) is either a simple sentence or a part of a bigger sentence having subject and predicate. (ଗୋଟିଏ ଖଣ୍ଡବାକ୍ୟ ହୁଏତ ଗୋଟିଏ ସରଳବାକ୍ୟ ବା ଏକ ବୃହତ୍ ବାକ୍ୟର ଏକ ଅଂଶ)

Join the pairs of sentences given below using a relative pronoun.(ନିମ୍ନରେ ପ୍ରଦତ୍ତ ବାକ୍ୟ ଯୁଗ୍ମକୁ Relative pronoun ବ୍ୟବହାର କରି ସଂଯୋଗ କର ।)

  1. My mother is going to host a T.V. show on cooking.
    She cooks very well.
  2. Our institution is highly popular.
    It works for public welfare.
  3. Satish scored a goal at the last minute.
    He was fortunate.
  4. Mother Teresa is revered as a saint.
    She served mankind.
  5. I often go to Mumbai.
    Mumbai is the commercial capital of India.
  6. These sportspersons are going to meet the President.
    Their performance has been excellent.

Answer:

  1. My mother, who cooks very well, is going to host ( ଉପସ୍ଥାପନ କରିବେ ) a T.V. show on cooking.
  2. Our institution, which works for public welfare (ଯାହା ଜନ କଲ୍ୟାଣ ପାଇଁ କାର୍ଯ୍ୟ କରେ |), is highly popular.
  3. Satish, who was fortunate (ଭାଗ୍ୟବାନ ), scored a goal in the last minute.
  4. Mother Teresa, who served mankind (ମାନବଜାତିର ସେବା କରିଥିଲେ), is revered as a saint.
  5. I often go to Mumbai, which is the commercial (ବ୍ୟବସାୟିକ )capital of India.
  6. These sports-persons, whose performance has been excellent (କାର୍ଯ୍ୟଦକ୍ଷତା), are going to meet the President.

Sometimes the relative pronoun in a relative clause remains ‘hidden’. For example, look at the first sentence of the story.(ବେଳେବେଳେ Relative clauseରେ Relative pronoun ଲୁକ୍‌କାୟିତ (hidden) ରହେ । ଉଦାହରଣସ୍ୱରୂପ, ଗଳ୍ପର ପ୍ରଥମ ବାକ୍ୟଟିକୁ ଦେଖ ।)

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

a) The house — the only one in the entire valley — sat on the Crest of a low hill.
We can rewrite the sentence as (without any change in the meaning)
(ଏହି ବାକ୍ୟର କୌଣସି ଅର୍ଥ ପରିବର୍ତ୍ତନ ନ କରି ଆମେ ଏପରି ଲେଖିପାରିବା )
The house—which was the only one in the entire valley—sat on the crest of a low hill.
In the original sentence of the text (a) the relative pronoun ‘which’ and the verb ‘was’ are not present (hidden). ପାଠ୍ୟରେ ଥ‌ିବା ମୂଳବାକ୍ୟ (a)66 Relative pronoun ‘which’ କ୍ରିୟା ‘was’ ଉପସ୍ଥିତ ନାହିଁ (ଲୁକ୍‌କାୟିତ ଅଛି) ।)

(ii) Using negatives (ନାସ୍ତିବାଚକ) for emphasis (ଗୁରୁତ୍ଵ ପ୍ରଦାନ) :
We know that sentences with words such as ‘no’, ‘not’, or ‘nothing’ show the absence of something, or contradict something. (ଆମେ ଜାଣୁ ଯେ ‘no’, ‘not’ ବା ‘nothing’ ଭଳି ଶବ୍ଦଗୁଡ଼ିକ ଥ‌ିବା ବାକ୍ୟଗୁଡ଼ିକ କୌଣସି ଜିନିଷର ଅନୁପସ୍ଥିତି ବା ବିରୋଧ କରୁଥ‌ିବା ବିଷୟରେ ସୂଚାଏ ।)
For example — (From the text) (ପାଠ୍ୟବିଷୟରୁ ଉଦାହରଣସ୍ୱରୂପ )

  • This year we will have no corn. (The crops have failed.) (ଫସଲ ହେବ ନାହିଁ)
  • (b) The hail has left nothing. (Absence of a crop) (ଶସ୍ୟ ନ ଥ‌ିବା ଅର୍ଥରେ)
  • (c) These aren’t raindrops falling from the sky, they are new coins.
    (Contradicts the common idea of what the drops of water falling from the sky are.) (ଆକାଶକୁ ସାଧାରଣତଃ ବର୍ଷାବିନ୍ଦୁ ପଡ଼ିବାର ବିପରୀତ ଅର୍ଥରେ )
    But sometimes negative words are used to emphasize an idea. Look at the sentences from story :
    (କିନ୍ତୁ ବେଳେବେଳେ ନାସ୍ତିସୂଚକ ଶବ୍ଦସମୂହ କୌଣସି ଧାରଣାକୁ ଗୁରୁତ୍ୱ ଦେବାକୁ ବ୍ୟବହୃତ ହୋଇଥାଏ । ଗଳ୍ପର ଏହି ବାକ୍ୟଗୁଡ଼ିକୁ ଦେଖ )
  • (d) Lencho …. had donc nothing else but see the sky towards the north-east. (He had done only this.) (ସେ କେବଳ ଏହା କରିଥିଲା)
  • (e) The man went out for no other reason than to have the pleasure of feeling the rain on his body. (He had only this reason.) (ତା’ ପାଖରେ କେବଳ ଏହି କାରଣ ଥିଲା)
  • (f) Lencho showed not the slightest suprise on seeing the money. (He showed no surprise at all.) (ସେ ଆଦୌ ଆଶ୍ଚର୍ଯ୍ୟର ଭାବନା ଦେଖାଇଲା ନାହିଁ )

Now look back at the example ‘c’. Mark that the contradiction in fact serves to emphasize the value or usefulness of the rain to the farmer.
(ଏବେ ଉଦାହରଣ ‘c’କୁ ଦେଖ । ଲକ୍ଷ୍ୟକର ଯେ ତଥ୍ୟର ବିରୁଦ୍ଧତା କୃଷକ ନିମନ୍ତେ ବର୍ଷାର ଆବଶ୍ୟକତା ଉପରେ ଗୁରୁତ୍ଵ ଦେବାରେ ସହାୟକ ହୋଇଛି ।)
Find sentences in the story with negative words, which express the following ideas emphatically : (ନାସ୍ତିସୂଚକ ଶବ୍ଦଥ‌ିବା ବାକ୍ୟଗୁଡ଼ିକ ଗଳ୍ପରୁ ବାଛ ଯାହାକି ନିମ୍ନୋକ୍ତ ଧାରଣାକୁ ଦୃଢ଼ଭାବେ ପ୍ରକାଶ କରୁଥ‌ିବ ।)

(1) The trees lost all their Leaves.
(2) The letter was addressed to God himself.
(3) The postman saw this address for the first time in his life.
Answer:
(1) Not a leaf (ପତ୍ର) remained on the trees.
(2) It was nothing less than a letter to God.
(3) Never in his career as a postman had he known that address.

J. Let’s Write:
Report Writing (ବିବରଣୀ ଲିଖନ)

Read the newspaper report given below. (ନିମ୍ନରେ ପ୍ରଦତ୍ତ ସମ୍ବାଦପତ୍ର ବିବରଣୀ ପଢ଼ ।)
Note the information given at different points. (ବିଭିନ୍ନ ସ୍ଥଳରେ ପ୍ରଦତ୍ତ ସୂଚନାଗୁଡ଼ିକୁ ଟିପି ରଖ ।)

Title – Rath Yatra pulls in lakhs of devotees to Puri
Source – The Times of India.
Place and date – Puri 22nd June
Topic sentence and introduction – All roads led to Purl on Thursday with lakhs of devotees flocking to the town to participate in the grand Rath Yatra festival. A wave of euphoria swept across the beach town with the reigning
Details of deities the Yatra – commencing their nine-day ‘outing’ from Jagannath shrine to Gundicha temple in a boisterous procession.
Information on the terror threat and stampede –  However, the threat of possible terror attack and the death of two pilgrims allegedly in a stampede cast a shadow on the festivities. There were intelligence reports that some subversive groups might and create trouble during Rath Yatra.
Further details of action taken by the authorities –  “Security was tightened compared to the previous years” the DGP said. “We also appeal to the people to remain alert.” the  DGP added.

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Word Notes:

led to Puri – ପୁରୀକୁ ନେଇଗଲା |
Lakhs of devotees – ଲକ୍ଷ ଲକ୍ଷ ଭକ୍ତ
Flocking – ପ୍ରବାହିତ
grand – ଗ୍ରାଣ୍ଡ୍
Wave of euphoria – ଇଉଫୋରିଆର ତରଙ୍ଗ |
swept across – ଅତିକ୍ରମ କଲା |
beach – ବେଳାଭୂମି
shrine – ଶ୍ରୀକ୍ଷେତ୍ର
with the reigning deities – ଶାସକ ଦେବତାମାନଙ୍କ ସହିତ |
outing – ବାହାଘର
commencing – ଆରମ୍ଭ
allegedly – ଅଭିଯୋଗ ହୋଇଛି
in a boisterous procession – ଏକ ଶୋଭାଯାତ୍ରାରେ |
pilgirm – ତୀର୍ଥଯାତ୍ରୀ
appealed – ଆବେଦନ କରିଥ

1) Now imagine that you are a reporter for ‘The Indian Express’. You have received the news about the recent cyclone that hit Odisha. Using the guidelines given in the box above, complete a similar report for the newspaper.
(ଏବେ କଳ୍ପନା କର ଯେ ତୁମେ ‘The Indian Express’ର ଜଣେ ସାମ୍ବାଦିକ । ଓଡ଼ିଶାକୁ ଅଳ୍ପଦିନ ତଳେ କ୍ଷତିଗ୍ରସ୍ତ କରିଥିବା ବାତ୍ୟା ବିଷୟରେ ତୁମେ ଏକ ସମ୍ବାଦ ପାଇଲ । ଉପରେ କୋଠରିରେ ପ୍ରଦତ୍ତ ନିର୍ଦେଶଗୁଡ଼ିକୁ ବ୍ୟବହାର କରି ସମ୍ବାଦପତ୍ର ନିମନ୍ତେ ଏକାପରି ଖବର ପ୍ରସ୍ତୁତ କର ।).
Answer:
The Super Cyclone brings about the unthinkable tragedy
The Indian Express,
Cuttack, 05 November
The super cyclone of the 29th of October in the district brought about an unthinkable tragedy to the people and their property. A large number of trees were uprooted. Thatched houses were blown away and the mud-built houses of the poor people collapsed. Crops were completely destroyed. Livestock died in numbers. Above all, it caused a great loss of lives and properties.
However, government and voluntary organizations geared up to provide relief commodities. Medicines were supplied to the affected people. Helps from all quarters flew continuously.

K.Let’s Know More (Vocabulary):
A. Related words.

Noun Verb Adjective Adverb
intimacy intimate intimate intimately
preparation prepare preparatory/ prepared
prediction predict predictable predictably
destruction destroy destructive destructively
sadness sadden sad sadly
promise promise promising
approach approach approachable
expose expose exposed
Amiability amiable amiably
charity charitable charitably
surprise surprise surprised/surprising surprisingly
denial deny deniable
instruction instruct instructive instructively
confidence confide confident confidently
immediacy immediate immediately
hunger hungry hungrily
expression express express expressly
heart hearty heartily
obligation oblige obliging obligingly
destruction destroy destructive
resolution resolve
correspondence correspond corresponding correspondingly

 

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God
B. A single word for the group of words:
1. a long and narrow area of land between the hills                        – valley
2. the highest part of the hill                                                            – crest
3. scattered over the area                                                                 – dotted with
4. the cutting and gathering of the corn                                          – harvest
5. a heavy fall of rain                                                                         – downpour
6. a short period of light rain                                                            – shower
7. a light meal was taken in the evening                                           – supper
8. a heavy meal was taken at night                                                   – dinner
9. say that something will happen but are not sure                          – predict
10. very big                                                                                        – huge
11. come nearer to someone                                                             – approach
12. cover with a piece of cloth                                                           – drape
13. small balls of ice that fall from the sky                                         – hailstones
14. look like somebody or something                                                – resemble
15. leave oneself expose to open danger                                           – expose
16. that has been turned into ice                                                        -frozen
17. the spiritual past of someone that is believed to continue existing after death – the soul
18. a large number of unpleasant animals or insects                         – plague
19. an insect like a grasshopper                                                         – locust
20. without others, being alone                                                         – solitary
21. unhappy or worried                                                                      – upset
22. a serious lack of food which caused death                                   – hunger
23. the feeling of inside about what is right or wrong                       – conscience
24. a box which is meant for dropping letters                                    – mailbox
25. with great pleasure                                                                       – heartily
26. friendly and pleasant                                                                    – amiable
27. the act of writing letters to someone                                           – correspondence
28. make a remark or criticize                                                             – a comment
29. a firm decision                                                                               – resolution
30. good feelings among people                                                        – goodwill
31. a dishonest person                                                                        – crook
32. being kind and helping people                                                     – charity
33. a feeling of happiness                                                                   – contentment
34. strong faith or trust                                                                       – confidence
35. a man in charge of a post office                                                    – postmaster
36. a man whose job is to collect and deliver letters, parcels, money orders, etc. – a postman
37. a person who has been employed                                                – an employee
38. a very short period of time                                                            – the moment
39. a few, not many                                                                             – several
40. ideas or feelings showing through words or actions                     – expression
41. very bad weather with heavy rain, wind, thunder, and lightning   – storm
42. a storm in which hailstones fall                                                      – hailstorm
43. a violent storm in the tropical region                                             – typhoon
44. an extremely strong wind                                                               – gale
45. a tropical storm in which strong winds move in circles                  – cyclone
46. strong winds in Western Atlantic Ocean                                        – hurricane
47. a funnel-shaped strong winds                                                        – Tornado
48. a strong wind that moves in a spinning movement                       – a whirlwind

C. Opposite words (antonyms) of the following words:

low — high
always — never
intimate — distant, formal
smaller — larger
approach — withdraw
sweet — sour
satisfied — dissatisfied
drape — uncover
strong — weak
cover — uncover
upset (y) — Console
sadness — happiness
solitary — busy, sociable
remember — forget
alive — dead
amiable — unpleasant, unfriendly
goodwill — ill-will, hostility
impossible — possible
usual — unusual
contentment — unhappiness, displeasure
count — uncount
open — close
faith — disbelief, disloyalty, distrust
near — far
pass — fail
ripe — immature, green
good — bad, evil
older — younger
begin (y) — end
fresh — stale
pleasure — pain
regard — disregard
drop — rise, lift
large — small
upset (adj) — calm
destroy — build, create, construct
sorrowful — cheerful
loss — profit
die — live
inside — outside
serious — careless
several — few
able — unable/disable
huge — tiny
confidence — doubt, distrust
public-private
willing — unwilling
resemble — vary, differ, contrast
quickly — slowly
expose — cover, protect, conceal
remain — leave, depart
trouble (n) — luck, pleasure, peace
deny — admit, accept

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

BSE Odisha 10th Class English The Solitary Reaper Important Questions and Answers

Very Short & Objective Questions With Answers:
Answer The Following Questions In A Word Or A Phrase:

Question 1.
Which country did G.L. Fuentes belong to?
Answer:
Mexico

Question 2.
At what age did G.L. Fuentes start writing?
Answer:
15

Question 3.
What was the only one in the entire valley?
Answer:
Lencho’s house

Question 4.
What was flowing by the low hill?
Answer:
the river

Question 5.
What was dotted with the flowers?
Answer:
the field of ripe corn

Question 6.
When did Lencho look at the sky in the northeast direction?
Answer:
in the morning

Question 7.
‘Now we’re really going to get some water, woman.’ Who is the ‘woman’ here?
Answer:
Lencho’s wife

Question 8.
Who was working in the field?
Answer:
the older boys

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 9.
How long were the little boys playing?
Answer:
till dinner

Question 10.
How was the air when it started raining?
Answer:
fresh and sweet

Question 11.
What did Lencho consider the raindrops?
Answer:
new coins

Question 12.
What began to fall along with the rain?
Answer:
very large hailstones

Question 13.
What did the large hailstones resemble?
Answer:
new silver coins

Question 14.
What did the boys collect when hailstones began to fall?
Answer:
the frozen pearls

Question 15.
“I hope it passed quickly.” What does ‘it’ refer to?
Answer:
the hailstones

Question 16.
How long did the hailstorm continue?
Answer:
for an hour

Question 17.
What was totally destroyed due to a hailstorm?
Answer:
corn

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 18.
What were Lencho’s sons filled with after a hailstorm?
Answer:
sadness

Question 19.
What was that night like?
Answer:
sorrowful

Question 20.
“All our work, for nothing.” To whom did Lencho say so?
Answer:
to his sons

Question 21.
What was the single hope in that solitary home?
Answer:
help from God

Question 22.
What can see everything, even what is deep in one’s conscience?
Answer:
God’s eyes

Question 23.
How was Lencho working in the fields?
Answer:
like an animals

Question 24.
How much money did Lencho need?
Answer:
hundred pesos

Question 25.
What did Lencho write on the envelope?
Answer:
To Gode

Question 26.
Who dropped the letter into the mailbox?
Answer:
Lencho

Question 27.
The postman went to his boss laughing heartily. Who is the ‘boss’ here?
Answer:
the postmaster

Question 28.
Who had in his career never known that address?
Answer:
the postman

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 29.
What was the postmaster like?
Answer:
a fat amiable person

Question 30.
“What a faith !” who said this?
Answer:
the postmaster

Question 31.
Where did the postmaster tap the letter?
Answer:
on his desk

Question 32.
Who opened Lencho’s letter to God?
Answer:
the postmaster

Question 33.
What did the postmaster stick to?
Answer:
his resolution

Question 34.
How much money did the postmaster himself give?
Answer:
a part of his salary

Question 35.
What did the letter to Lencho contain?
Answer:
only a single word as a signature: God

Question 36.
Who handed the letter to Lencho the following Sunday?
Answer:
the postman

Question 37.
Who was experiencing the contentment of a man who had performed a good deed?
Answer:
the postmaster

Question 38.
Who had unflinching faith in God?
Answer:
Lencho

Question 39.
What did Lencho ask for near the window of the post office?
Answer:
paper and ink

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 40.
Where did Lencho write his second letter to God?
Answer:
on the public writing table.

Fill In The Blanks With Right Words:

1. The lonely house of Lencho sat on ___________.
Answer:
the crest of low hill

2. The ripe corn field always promised ___________.
Answer:
a good harvest

3. The only thing the earth needed was a ___________.
Answer:
a downpour or at least a shower

4. Lencho had predicted big drops of rain during ___________.
Answer:
the meal

5. Lencho compared raindrops with ___________.
Answer:
new coins

6. The big raindrops are ___________ cent pieces.
Answer:
10

7. ___________ began to blow suddenly.
Answer:
A strong wind

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

8. Along with the rain ___________ began to fall.
Answer:
large hailstones

9. Hailstones resembled ___________.
Answer:
new silver coins

10. Lencho’s cornfield looked white as if covered with ___________.
Answer:
salt

11. ___________ were gone from the plants.
Answer:
Flowers

12. Lencho’s soul was filled with ___________.
Answer:
sadness

13. The children went out to collect ___________ in the rain.
Answer:
frozen pearls

14. With a ___________ Lencho regarded the field of ripe corn with its flowers draped in a curtain of rain.
Answer:
satisfied expression

15. Lencho’s family lived in the ___________ house in the middle of the valley.
Answer:
solitary

16. ___________ was a single hope of Lencho.
Answer:
With the help of God

17. ‘Don’t be so upset’ Lencho said this to his ___________.
Answer:
family

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

18. Lencho thought only of one hope. That is ___________.
Answer:
the help of God

19. God sees what is deep in one’s ___________.
Answer:
conscience

20. Lencho was ___________.
Answer:
anoxia man

21. Lencho began to write a letter to God at ___________ on the following Sunday.
Answer:
daybreak

22. The synonym of ‘upset’ is ___________.
Answer:
disturbed

23. The antonym of ‘drape’ is ___________.
Answer:
reveal/unwrap

24. People say no one dies of ___________.
Answer:
hunger

25. According to Lencho a plague of ___________ would have left more than this.
Answer:
locusts

26. Lencho carried his first letter to ___________.
Answer:
town

27. Lencho asked God ___________ pesos in his first letter.
Answer:
100

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

28. Lencho wrote ___________ on the envelope of the first letter.
Answer:
To God

29. After placing a stamp on the envelope, he dropped it into ___________.
Answer:
the mailbox

30. One of the employees refers to ___________.
Answer:
the postman

31. The postmaster was a ___________ fellow.
Answer:
amiable

32. The postmaster immediately turned ___________ after tapping the letter on his desk.
Answer:
serious

33. What faith! It is a ___________ on Lencho.
Answer:
comment

34. Synonym of ‘Correspondence’ is ___________.
Answer:
accord/equivalence

35. Lencho demanded hundred pesos in order to ___________ his field again.
Answer:
sow

36. Small balls of ice that fall to the ground with rain is called ___________.
Answer:
hailstones

37. ___________ read the first letter of Lencho.
Answer:
Postmaster

38. The antonymn of ‘amiable’ is ___________.
Answer:
unfriendly/disgraceful

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

39. To answer the letter the postmaster needed ___________ than goodwill, ink, and paper.
Answer:
something more

40. The Postmaker stuck to his ____________.
Answer:
resolution

MULTIPLE CHOICE QUESTIONS (MCQS) WITH ANSWERS
Pick out the correct alternative.

Question 1.
Who was Lencho?
(A) A postman
(B) A postmaster
(C) A farmer
(D) A laborer
Answer:
(C) A farmer

Question 2.
Lencho hoped Lorraine because ___________.
(A) the weather was rather hot.
(B) the crop in his field badly needed water
(C) there was a drought in Mexico
(D) he would sow seeds in his field
Answer:
(B) the crop in his field badly needed water

Question 3.
After the destruction caused by the hailstorm. Lencho was worried about ___________.
(A) his children
(B) the shortage of food for the entire year
(C) the crops destroyed by the hailstorms
(D) the pleasant days to come in future
Answer:
(B) the shortage of food for the entire year

Question 4.
Who was deeply moved by Lencho’s faith in God?
(A) the postman
(B) the postmaster
(C) the children
(D) the woman
Answer:
(B) the postmaster

Question 5.
Lencho lived on the crest of a ___________.
(A) mountain
(B) hill
(C) hillock
(D) plateau
Answer:
(C) hillock

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 6.
Lencho did not try to find out the real sender of the money because ___________.
(A) he was dreaming about the happy days ahead
(B) he had deep faith in God
(C) he was heartbroken at the loss of crops
(D) he was very ungrateful
Answer:
(B) he had deep faith in God

Question 7.
When Lencho counted the money, he became angry and had all doubts about ___________.
(A) the postmaster
(B) the postman
(C) his wife and his sons
(D) the people working in the post office.
Answer:
(D) the people working in the post office.

Question 8.
Lencho blamed the post office employees because ___________.
(A) they did not help his family
(B) he had great faith in God
(C) the post office employees had stolen a part of the money.
(D) they didn’t deliver the letter on time.
Answer:
(C) the post office employees had stolen a part of the money

Question 9.
The postmaster along with the post office employees sent Lencho the money because ___________.
(A) they were related to each other
(B) it was an act of kindness and selflessness
(C) the postmaster was a rich man
(D) he was the victim of the hailstorm
Answer:
(B) it was an act of kindness and selflessness

Question 10.
In the story ‘ The rain turned into hailstorm’ implies ___________.
(A) the conflict between the postman and the postmaster
(B) the conflict between man and nature
(C) the conflict between his wife and children
(D) the conflict between the storm and the crops
Answer:
(B) the conflict between man and nature

Question 11.
From the height of the low hill, one could see ____________.
(A) the stream and the brook
(B) the river and the cornfield
(C) the garden with trees in the valley
(D) the hut and the trees
Answer:
(B) the river and the cornfield

Question 12.
Lencho looked at the sky towards the ____________.
(A) south-west
(B) south-east
(C) north-west
(D) north-east
Answer:
(D) north-east

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 13.
Lencho thought that the only thing that the earth needed was ___________.
(A) a downpour
(B) a shower
(C) a storm
(D) a hailstorm
Answer:
(A) a downpour

Question 14.
The older boys were working in ___________.
(A) the woman
(B) crops
(C) Lencho
(D) raindrops
Answer:
(D) raindrops

Question 15.
The smaller boys were playing ____________.
(A) in the playground
(B) in the meadow
(C) near the house
(D) in the park
Answer:
(C) near the house

Question 16.
Big drops of rain began to fall during ___________.
(A) the dinner
(B) the supper
(C) the lunch
(D) the breakfast
Answer:
(B) the supper

Question 17.
Huge mountains of clouds could be seen approaching ___________.
(A) in the south-east
(B) in the north-west
(C) in the south-west
(D) in the north-east
Answer:
(D) in the north-east

Question 18.
When big drops of rain began to fall, Lencho went out to have the pleasure of feeling the rain ___________.
(A) on his head
(B) on his hand
(C) on his body
(D) on his legs
Answer:
(C) on his body

Question 19.
Lencho said that the big raindrops were like ____________ cent pieces.
(A) five
(B) ten
(C) twenty
(D) fifty
Answer:
(B) ten

Question 20.
After the hailstorm. Lencho’s cornfield was covered with __________________
(A) clouds
(B) locusts
(C) hailstones
(D) salt
Answer:
(C) hailstones

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 21.
The expression ‘frozen pearls’ refers to ___________.
(A) locusts
(B) crops
(C) hailstones
(D) raindrops
Answer:
(C) hailstones

Question 22.
The hailstones looked like __________.
(A) old gold coins
(B) new copper coins
(C) new silver coins
(D) old bronze coins
Answer:
(C) new silver coins

Question 23.
‘All our work for nothing’ said by year said by ___________.
(A) Lencho
(B) the woman
(C) the small boys
(D) the older boys
Answer:
(C) the small boys

Question 24.
We’ll go hungry __________.
(A) fruitless
(B) the postman
(C) flowerless
(D) the boys
Answer:
(A) fruitless

Question 25.
Lencho’s field looked as if it was covered with ____________.
(A) cotton
(B) snow
(C) foam
(D) salt
Answer:
(D) salt

Question 26.
The hail rained for __________.
(A) an hour
(B) two hours
(C) three hours
(D) four hours
Answer:
(A) an hour

Question 27.
The hailstorm made the plants ___________.
(A) the dinner
(B) the supper
(C) the lunch
(D) the breakfast
Answer:
(B) the supper

Question 28.
What time did Lencho write the letter to God?
(A) at sunset
(B) at dusk
(C) at daybreak
(D) at night
Answer:
(C) at daybreak

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 29.
Lencho preferred ___________ to hailstorm.
(A) downpour
(B) ox
(C) mosquitoes
(D) locusts
Answer:
(D) locusts

Question 30.
God can see everything even what is ___________.
(A) heart
(B) conscience
(C) mind
(D) personality
Answer:
(B) conscience

Vocabulary
Do As Directed:

Question 1.
They are dishonest people. (Substitute a single word for the underlined words)
Answer:
crooks

Question 2.
He expressed his satisfaction. (Substitute a single word for the underlined word)
Answer:
contentment

Question 3.
Scattered over an area (Write a single word)
Answer:
dotted with

Question 4.
Lencho lived in a solitary house. The underlined word means __________.
Answer:
lonely

Question 5.
Mr. Kar is working in an ____________ organization. (Fill in the blank with a word opposite to public)
Answer:
private

Question 6.
Dr. Goutam Maharana is a friendly and pleasant young man. (Substitute a single word for the underlined part)
Answer:
amiable

Question 7.
Lencho said, “Don’t be upset.” Here ‘upset’ means __________. (Fill up the blanks)
Answer:
disturbed

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 8.
In spring we __________the swimming pool. (Fill in the blank with a word opposite to ‘cover’)
Answer:
uncover

Question 9.
Aditya expressed his __________ to see the evils. (Fill in the blank with a word opposite to ‘pleasure’)
Answer:
displeasure

Question 10.
The ____________ of the meteorologists came true. (Fill in the blank with a word related to ‘predict’)
Answer:
prediction

Question 11.
The bombs caused a lot of __________. (Fill in the blank with a word related to ‘destroy’)
Answer:
destruction

Question 12.
With a satisfied __________, he looked at his cornfield. (Fill in the blank with the noun form of ‘express’)
Answer:
expression

Question 13.
He can foretell the future. (Substitute a single word for the underlined portion)
Answer:
predict

Question 14.
But he stuck to his resolution. Here ‘resolution’ means __________. (Fill up the blanks)
Answer:
firm decision

Question 15.
Nobody likes __________ people. (Fill in the blank with a word opposite to ‘honest’)
Answer:
dishonest

Question 16.
God sees everything, even what is deep in one’s inner sense of right or wrong. A single word for the underlined expression will be ___________.
Answer:
conscience

SUBJECTIVE QUESTIONS WITH ANSWERS
Answer the following question in about 50 words.

Question 1.
Who was Lencho? What was he like?
Answer:
Lencho was a farmer who lived with his family in a solitary house on the top of a low hill. He was very poor and innocent. He was very hardworking. He knew his fields well and worked there like an animal. No doubt he was literate and he was able to read and write. Besides he had firm faith in God.

Question 2.
Why did Lencho keep on looking at the sky throughout the morning?
Answer:
Lencho was a poor farmer who knew his com fields closely. He could see his fields of ripe com full of flowers that always promised a good harvest. He hoped for a downpour or at least a shower of rain for his fields of ripe corn. So Lencho kept on looking at the sky towards the northeast throughout the morning.

Question 3.
How was Lencho’s cornfield destroyed?
(Or)
What was the effect of the hailstorm on Lencho’s cornfield?
Answer:
Just as Lencho had predicted, big drops of rain began to fall at night. His joy knew no bounds as the water was badly necessary for his field. But soon a strong wind began to blow and along with the rain very large hailstones began to fall. It continued for an hour. The field was completely covered with hailstones. As a result, the com was totally destroyed.

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 4.
What did Lencho predict? How was it materialized?
Answer:
Once Lencho felt the necessity of a shower of rain for his com field in order to have a good harvest. So one morning he kept on looking at the sky towards the northeast anticipating a downpour or a shower. He was sure about it in the afternoon and informed his wife all about it. His imagination materialized when he was taking his meal. Big drops of rain began to fall and in the northeast huge mountains of clouds could be seen approaching.

Question 5.
Why did Lencho say that the raindrops were like new coins?
Answer:
Lencho’s ripe corn field was badly in need of rain and he had been anxiously waiting for the raindrops for a long time, as a shower of rain had great importance for him. In the northeast huge mountains of clouds approached and then the air became fresh and sweet. Lencho went out for no other reason than to have the pleasure of feeling the rain on his body. When he returned home, he exclaimed that those raindrops were like new silver coins.

Question 6.
Why didn’t Lencho’s happiness last long?
Answer:
It began to rain at night just as Lencho had predicted before. He was delighted to see it. He came out and felt the pleasure of rain on his own body. But due to the irony of his fate, his happiness didn’t last long. Soon large hailstones began to fall along with a strong wind. The hail rained in the valley for an hour which snatched away all his happiness.

Question 7.
Why did Lencho prefer locusts to the storm?
Answer:
Locusts are a kind of insects that eat crops and vegetables. They fly in large numbers and eat up some crops and then leave the fields. But the storm came and spoiled the whole of the ripe corps in the Lencho’s corn field. So Lencho preferred locusts to the storm because the locusts would not have caused so much damage to the crops as the storm had done.

Question 8.
Why did Lencho write a letter to God?
(Or)
Why did Lencho write a letter to God? What did he pray to him?
Answer:
Lencho had deep faith in Almighty God. He was sure that God would certainly help him during his distress. When all his com fields were destroyed by the hailstones, Lencho’s heart was filled with sorrow. He became helpless and depended on God who might save his family from starvation and hunger throughout the year. Since he had a single hope: hope from God, he wrote a letter to Him asking for a hundred pesos to sow his field again and feed his family until the next crop came.

Question 9.
What was the only hope in everybody’s heart?
Answer:
The only hope in everybody’s heart was that God would certainly help them as God sees everything, even what is deep in one’s conscience. So the following Sunday at daybreak Lencho wrote a letter to God praying to Him to send him a hundred pesos so that he could sow his field again and feed his family until the next crop came.

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 10.
What did the employees feel after receiving Lencho’s letter to God?
Answer:
The postman laughed heartily when he saw the letter addressed to God. Career as a postman had he known that address. The postmaster also broke out laughing. Soon he became serious and tapped the letter on his desk. He was astonished to see Lencho’s deep faith in God.

Question 11.
WhatdidLenchowritetoGodinhis first letter?
Answer:
Lencho’s com field was totally destroyed by the hailstorm. So he thought that his family would go hungry that year. He hoped that God would help him. He made a humble prayer to God to help him otherwise his family would go without food that year. He also wrote that he needed a hundred pesos to sow his field again and to live until the next crop came.

Question 12.
What was Lencho’s reaction after the hailstorm?
Answer:
After the hailstorm, Lencho was disheartened to see his com field which was completely destroyed. He thought of himself as well as of his family members. He realized that his family would go hungry that year. He expressed it before his sons. He also knew it well that no one could help them there. He consoled his family as he had deep faith in God.

Question 13.
What did Lencho write to God in his last letter?
Answer:
In his last letter Lencho wrote to God that of the money that he asked for, only seventy pesos reached him. He requested God to send him the rest amount since he needed it very much. He also asked Him to send it to him not through the mail as the post office employees were a bunch of crooks.

Question 14.
What was Lencho’s reaction when he received the letter from the post office?
(Or)
What was Lencho’s reaction to the reply to his letter?
Answer:
Lencho’s expectation of getting a reply from God came true when he received the letter from the postman. He was not at all surprised to see it as he had firm faith in God. He confidently opened it and counted the money. Soon he got angry as it was less than a hundred pesos. He never suspected God rather he suspected the integrity of the employees of the post office.

Question 15.
What would have been the reaction of the postal employees if they had read Lencho’s second letter?
Answer:
Lencho’s second letter to God contained a statement of his dissatisfaction with the postal employees. The second letter written by Lencho was the accusation of cheating on the money from the parcel by them. If they had read the second letter sent by Lencho, they would have strongly blamed him and would have considered him ungrateful.

Question 16.
Why did the postmaster send money to Lencho?
Answer:
The postmaster was an amiable person, who praised Lencho for his deep faith in God. He decided to answer the letter. He went through the letter and found that his family would go hungry that year and needed a hundred pesos in order to sow his fields again. When he knew that it needed not merely goodwill, but financial help, he gave a part of his salary and collected money from several of his employees and some more money from his friends as an act of charity.

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Question 17.
Why did Lencho’s field look white after the storm?
Answer:
Suddenly a strong wind began to blow along with a heavy shower of rain. Soon very large hailstones began falling from the sky. The rain continued for an hour. Unfortunately, the strong wind turned into a hailstorm. The hailstones rained for an hour and fell on Lencho’s house, the garden, the hillside, the corn field, and the entire valley. So his cornfield became full of hailstones and looked white as if it was covered with salt.

Question 18.
Why did Lencho call the post office employees a bunch of crooks’?
Answer:
Lencho was very sure that God could neither make a mistake nor deny him what he had asked for. His confidence in God was so deep that he expected to get a hundred pesos positively. When he received the letter from God, he eagerly counted the money. He found that there were only seventy pesos in the envelope. He suspected that the post office employees were dishonest persons, who must have taken the remaining part of the money. For this reason, he called the post office employees a bunch of crooks.

Question 19.
How does the writer describe two kinds of conflicts in the story ‘A Letter to God’?
Answer:
There are two kinds of conflict described in the story. Firstly Lencho’s com field was destroyed by a hailstorm and thereby Lencho became helpless. It is the conflict between Nature and man. Even though the postmaster and other employees collected money and managed to send seventy pesos to Lencho, as an act of kindness, Lencho blamed them for taking away a part of his money. This statement reflects that Lencho didn’t have faith in man. So it is the conflict between man and man.

A letter to God Summary in English

Lead-In:
It is believed that faith can move mountains. People live in faith. Particularly people have faith in God. People endure misfortunes because of their faith for the commencement of a better time. Faith keeps us alive and makes us optimistic. Here, in the present context, Lencho is a farmer who has firm faith in God. He writes a letter to God during his distress and asks for some financial assistance. Though it doesn’t reach its destination, still the poor farmer gets a part of his request. He couldn’t disbelieve it.

Paragraph-wise Explanation:
Para: The house-the only one in the entire valley – sat on the crest of a low hill. From this height, one would see the river and the field of ripe corn dotted with the flowers that always promised a good harvest. The only thing the earth needed was a downpour or at least a shower. Throughout the morning Lencho – who knew his fields intimately- had done nothing else but see the sky towards the northeast.
ଅନୁବାଦ : ସମଗ୍ର ଉପତ୍ୟକାରେ ଥ‌ିବା ଏକମାତ୍ର ଘରଟି ଗୋଟିଏ ନିମ୍ନ ପାହାଡ଼ର ଉପରିଭାଗରେ ଥିଲା । ଏହି ଉଚ୍ଚତାରୁ ଯେ କେହି ନଦୀ ଏବଂ ବିସ୍ତୀର୍ଣ୍ଣ ଅଞ୍ଚଳ ପରିବ୍ୟାପ୍ତ ଫୁଲ ଉଡ଼ାଉଥ‌ିବା ପାଚିଲା ଶସ୍ୟକ୍ଷେତ୍ର ଦେଖପାରିବ ଯାହାକି ସବୁବେଳେ ଭଲ ଫସଲ ଅମଳ ଦେବାର ଆଶା ସଞ୍ଚାର ଅସରାଏ କେବଳ ପୃଥ‌ିବୀ ମୂଷଳ ଧାରାରେ ବୃଷ୍ଟି କିମ୍ବା ଅତି ଭାବରେ ଜାଣିଥୁଲା ବର୍ଷା ଦରକାର କରୁଥିଲା । ଲେଞ୍ଚେ ଯିଏକି ତା’ର କ୍ଷେତକୁ ଭଲ କମ୍‌ରେ କରାଇଥାଏ । ସାରା ସକାଳ କେବଳ ଉତ୍ତର-ପୂର୍ବ ଆକାଶକୁ ଚାହିଁବା ବ୍ୟତୀତ ଅନ୍ୟ କିଛି ହିଁ କରି ନଥିଲା ।

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Para: “Now we’re really going to get some water, woman. ” The woman who was preparing supper, replied, “Yes, God willing “. The older boys were working in the field, while the smaller ones were playing near the house until the woman called to them all, “Come for dinner”. It was during the meal that, just as Lencho had predicted, big drops of rain began to fall. In the northeast, huge mountains of clouds could be seen approaching. The air was fresh and sweet. The man went out for no other reason than to have the pleasure of feeling the rain on his body, and when he returned he exclaimed, “These aren’t raindrops falling from the sky, they are new coins. The big drops are ten cent pieces and the little ones are fives.”
ଅନୁବାଦ : ‘‘ହେ ନାରୀ, ଏବେ ବାସ୍ତବରେ ଆମେ କିଛି ପାଣି ପାଇବାକୁ ଯାଉଛେ !’’ ଯେଉଁ ସ୍ତ୍ରୀଲୋକଟି ସନ୍ଧ୍ୟାକାଳୀନ ଖାଦ୍ୟ ପ୍ରସ୍ତୁତ କରୁଥିଲେ ସେ ଉତ୍ତର ଦେଲେ, ‘ହଁ, ଈଶ୍ୱରଙ୍କ ଇଚ୍ଛା ।’’ ବୟସ୍କ ବାଳକମାନେ କ୍ଷେତ୍ରରେ କାମ କରୁଥିଲେ, ଯେତେବେଳେ କି ସାନ ସାନ ପିଲାମାନେ ଘର ପାଖରେ ଖେଳୁଥିଲେ । ସ୍ତ୍ରୀଲୋକଟି ସମସ୍ତଙ୍କୁ ‘ଖାଇବ ଆସ’ ବୋଲି ଡାକିଲେ । ଠିକ୍ ଯେପରି ଲେ ଭବିଷ୍ୟବାଣୀ କରିଥିଲା, ଖାଇବା ଚାଲିଥିବା ସମୟରେ ସେହିପରି ବଡ଼ ବଡ଼ ବର୍ଷା ବିନ୍ଦୁମାନ ପଢ଼ିବା ଆରମ୍ଭ କରିଦେଲା । ଉତ୍ତର-ପୂର୍ବାକାଶରେ ବିରାଟ ପର୍ବତ ଆକାରର ବାଦଲ ଖଣ୍ଡମାନ ମାଡ଼ି ଆସୁଥିବାର ଦେଖାଯାଉଥିଲା । ପବନ ଖୁବ୍ ସତେଜ ଓ ଆନନ୍ଦଦାୟକ ଥିଲା । ଲୋକଟି ଅନ୍ୟ କୌଣସି କାରଣ ପାଇଁ ନୁହେଁ, କେବଳ ତା’ ଶରୀରରେ ବର୍ଷା ସ୍ପର୍ଶର ଆନନ୍ଦ ଉପଭୋଗ କରିବାକୁ ବାହାରକୁ ଚାଲିଗଲା ଏବଂ ଫେରିବା ସମୟରେ ଆବେଗରେ କହି ପକାଇଲା, ‘‘ଆକାଶରୁ ଝରି ପଡୁଥ‌ିବା ଏଗୁଡ଼ିକ ବର୍ଷାବିନ୍ଦୁ ନୁହେଁ, ସେଗୁଡ଼ିକ ନୂତନ ମୁଦ୍ରା । ବଡ଼ ବଡ଼ ବର୍ଷାବିନ୍ଦୁଗୁଡ଼ିକ ଦଶ ସେଣ୍ଟ୍ ମୁଦ୍ରା (ଆମେରିକୀୟ ମୁଦ୍ରା) ଏବଂ ଛୋଟ ବର୍ଷାବିନ୍ଦୁଗୁଡ଼ିକ ପାଞ୍ଚ ସେଣ୍ଟ୍ ମୁଦ୍ରା ସଦୃଶ ।’’

Para: With a satisfied expression he regarded the field of ripe corn with its flower, draped in a curtain of rain. But suddenly a strong wind began to blow and along with the rain very large hailstones began to fall. These truly did resemble new silver coins. The boys, exposing themselves to the rain, ran out to collect the frozen pearls.
ଅନୁବାଦ : ସନ୍ତୋଷଜନକ ଅଭିବ୍ୟକ୍ତି ସହ ସେ ପାଚିଲା ଶସ୍ୟକ୍ଷେତ ତା’ ଫୁଲ ସହ ବର୍ଷାରୂପକ ପରଦାରେ ଆବୃତ ହୋଇଯାଇଛି ବୋଲି ଲେଞ୍ଝା ଭାବୁଥିଲା । କିନ୍ତୁ ହଠାତ୍ ପ୍ରବଳ ବେଗରେ ଶକ୍ତିଶାଳୀ ପବନ ବହିବାକୁ ଆରମ୍ଭ କଲା ଏବଂ ବର୍ଷା ସହିତ ବଡ଼ ବଡ଼ କୁଆପଥର ଖଣ୍ଡସବୁ ପଡ଼ିବା ଆରମ୍ଭ ହେଲା । ବାସ୍ତବରେ କୁଆପଥରଗୁଡ଼ିକ ନୂଆ ରୁପା ମୁଦ୍ରାଭଳି ଦେଖାଯାଉଥିଲା । ବର୍ଷାରେ ଭିଜିଭିଜି ପିଲାମାନେ ବରଫ ମୁକ୍ତାଗୁଡ଼ିକୁ ସଂଗ୍ରହ କରିବାପାଇଁ ବାହାରକୁ ଦୌଡ଼ିଲେ ।

Para: “It’s really getting bad now “, exclaimed the man, “I hope it passes quickly. ” It did not pass quickly. For an hour the hail rained on the house, the garden, the hillside, the cornfield, on the whole valley. The field was white as if covered with salt.
ଅନୁବାଦ : ଲୋକଟି ବିସ୍ମୟର ସହ କହିଲା, ‘ବର୍ତ୍ତମାନ ପ୍ରକୃତରେ କିଛି ଖରାପ ଘଟିବାକୁ ଯାଉଛି ଏବଂ ମୁଁ ଆଶା କରୁଛି ଏହା ଖୁବ୍ ଶୀଘ୍ର ଚାଲିଯିବ ।’’ କିନ୍ତୁ ଏହା ଶୀଘ୍ର ଅତିକ୍ରମ କଲା ନାହିଁ । ଦୀର୍ଘ ଏକଘଣ୍ଟା ଧରି ଘର ଉପରେ, ବଗିଚାରେ, ପାହାଡ଼ କଡ଼ରେ, ଶସ୍ୟକ୍ଷେତରେ ତଥା ସମଗ୍ର ଉପତ୍ୟକାରେ କୁଆପଥର ବୃଷ୍ଟି ହେଲା । ଶସ୍ୟକ୍ଷେତଗୁଡ଼ିକ ଲୁଣଦ୍ୱାରା ଆବୃତ ହେବାଭଳି ସମ୍ପୂର୍ଣ୍ଣ ଧଳା ଦେଖାଯାଉଥିଲା ।

Para: Not a leaf remained on the trees. The corn was totally destroyed. The flowers were gone from the plants. Lencho’s soul was filled with sadness. When the storm had passed, he stood in the middle of the field and said to his sons, “A plague of locusts would have left more than this. The hail has left nothing, this year we will have no corn.”
ଅନୁବାଦ : ଗଛଗୁଡ଼ିକରେ ପତ୍ରଟିଏ ମଧ୍ଯ ରହିଲା ନାହିଁ । ଶସ୍ୟଗୁଡ଼ିକ ସମ୍ପୂର୍ଣ୍ଣ ନଷ୍ଟ ହୋଇଗଲା । ଶସ୍ୟଗଛଗୁଡ଼ିକରୁ ଫୁଲସବୁ ଝଡ଼ି ପଡ଼ିଲା । ଲେଞ୍ଚୋର ଆତ୍ମା ଦୁଃଖରେ ଭରିଗଲା । ଯେତେବେଳେ ଝଡ଼ ଚାଲିଗଲା, ସେ କ୍ଷେତ ମଝିରେ ଛିଡ଼ା ହୋଇ ତା’ର ପୁଅମାନଙ୍କୁ କହିଲା, ‘‘ପଙ୍ଗପାଳ ଉପଦ୍ରବ କରିଥିଲେ ହୁଏତ ଏହାଠାରୁ ଅଧିକ ଛାଡ଼ି ଦେଇଥା’ନ୍ତେ । କୁଆପଥର ବୃଷ୍ଟି କିଛି ଛାଡ଼ିଯାଇ ନାହିଁ । ଏବର୍ଷ ଆମେ ଆଦୌ ଶସ୍ୟ ପାଇବା ନାହିଁ ।

Para: That night was a sorrowful one.
“All our work, for nothing. ”
“There’s no one who can help us. ”
“We’ll all go hungry this year.”
But in the hearts of all who lived in that solitary house in the middle of the valley, there was a single hope: help from God.
“Don’t be so upset, even though this seems like a total loss. Remember, no one dies of hunger.”
That’s what they say : no one dies of hunger.
ଅନୁବାଦ : ସେହି ରାତିଟି ଏକ ଦୁଃଖଦ ରାତି ଥିଲା । ‘‘ଆମର ସମସ୍ତ ଶ୍ରମ ନିରର୍ଥକ ହେଲା ।’’ ‘‘ଆମକୁ ସାହାଯ୍ୟ କରିପାରିବା ଭଳି କେହି ନାହାନ୍ତି ।’’ ‘‘ଏବର୍ଷ ଆମେ ଭୋକିଲା ରହିବା ।’’ମାତ୍ର ଉପତ୍ୟକାର ମଧ୍ୟଭାଗସ୍ଥିତ ସେହି ନିଃସଙ୍ଗ ଘରଟିରେ ବାସ କରୁଥିବା ସମସ୍ତଙ୍କ ହୃଦୟରେ ଗୋଟିଏ ମାତ୍ର
ଆଶା ଥିଲା ; ଭଗବାନ୍‌ଙ୍କଠାରୁ ସାହାଯ୍ୟ ପ୍ରାପ୍ତି ।“‘ବିଚଳିତ ହୁଅ ନାହିଁ, ଯଦିଓ ଏହା ଏକ ସମ୍ପୂର୍ଣ୍ଣ କ୍ଷତି ଭଳି ଜଣାପଡ଼ୁଛି । ମନେରଖ, କେହି ଭୋକରେ ମରନ୍ତି ନାହିଁ ।”’‘କେହି ଭୋକରେ ମରେ ନାହିଁ – ଏପରି ସେମାନେ କହନ୍ତି ।’’

Para: All through the night, Lencho thought only of one hope: the help of God, whose eyes, as he had been instructed, see everything, even what is deep in one’s conscience. Lencho was an ox of a man, working like an animal in the fields, but still, he knew how to write. The following Sunday, at daybreak, he began to write a letter which he himself would carry to town and place in the mail. It was nothing less than a letter to God.
ଅନୁବାଦ : ରାତିସାରା ଲେଞ୍ଜୋ କେବଳ ସେଇ ଗୋଟିଏ ଆଶା, ଈଶ୍ବରଙ୍କ ସାହାଯ୍ୟ କଥା ଭାବୁଥୁଲା, ଯାହାଙ୍କର ଆଖ୍ ତାଙ୍କୁ ଯେପରି କୁହାଯାଇଛି ସବୁକିଛି ଦେଖେ ଏପରିକି ଜଣକର ଗଭୀର ହୃଦୟରେ କ’ଣ ଅଛି ଦେଖାରେ । ଲେଞ୍ଜୋ କଠିନ ପରିଶ୍ରମୀ ଥିଲା ଓ ସେ ତା’ର ଜମିରେ ବଳଦ ଭଳି ଖଟୁଥିଲା, କିନ୍ତୁ ତଥାପି ସେ ଲେଖୁବା ଜାଣିଥିଲା । ତା’ ପରବର୍ତୀ ରବିବାର ସକାଳୁ ସେ ଗୋଟିଏ ଚିଠି ଲେଖିବାକୁ ଆରମ୍ଭ କଲା, ଯାହାକୁ ସେ ନିଜେ ସହରକୁ ନେଲା ଓ ଡାକରେ ପକାଇ ଆସିଲା । ଏହା ଈଶ୍ଵରଙ୍କ ନିକଟକୁ ଏକ ପତ୍ର ବ୍ୟତୀତ ଅନ୍ୟ କିଛି ନ ଥିଲା ।

Para: “God,” he wrote, “if you don’t help me, my family and I will go hungry this year. I need a hundred pesos in order to sow my field again and to live until the crop comes, because of the hailstorm…. ”
ଅନୁବାଦ : ସେ ଲେଖୁଥିଲା, ‘ହେ ଈଶ୍ଵର, ଯଦି ଆପଣ ମୋତେ ସାହାଯ୍ୟ ନ କରନ୍ତି, ତେବେ ଏବର୍ଷ ମୋ ପରିବାର ଓ ମୁଁ ଭୋକରେ ରହିବୁ । ଜମିରେ ପୁନର୍ବାର ବିହନ ବୁଣିବାପାଇଁ ଓ ଫସଲ ଅମଳ ହେବା ପର୍ଯ୍ୟନ୍ତ ବଞ୍ଚିବା ପାଇଁ ମୋର ଏକ ଶହ ପୋସୋ (ଲାଟିନ୍‌ ଆମେରିକାର ମୁଦ୍ରା) ଆବଶ୍ୟକ କାରଣ କୁଆପଥର ……”’

Para: He wrote “To God’ on the envelope, put the letter inside, and, still troubled, went to town. At the post office, he placed a stamp on the letter and dropped it into the mailbox.
ଅନୁବାଦ : ସେ ଲଫାପା ଉପରେ “To God” (ଈଶ୍ବରଙ୍କୁ) ବୋଲି ଲେଖୁଲା, ଚିଠିଟିକୁ ତା’ ଭିତରେ ରଖିଲା, ତଥାପି ଶଙ୍କାକୁଳ ମନରେ ସହରକୁ ଗଲା । ଡାକଘରେ ସେ ଚିଠି ଉପରେ ଟିକଟ ଲଗାଇଲା ଓ ଡାକବାକ୍ସ ଭିତରେ ସେଇଟିକୁ ପକାଇଦେଲା ।

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Para: One of the employees, who was a postman and also helped at the post office, went to his boss laughing heartily, and showed him the letter to God. Never in his career as a postman had he known that address. The postmaster – a fat, amiable fellow also broke out laughing, but almost immediately he turned serious and, tapping the letter on his desk, commented, “What faith! I wish I had the faith in the man who wrote this letter. Starting up a correspondence with God !”
ଅନୁବାଦ : ଜଣେ କର୍ମଚାରୀ, ଯିଏକି ଜଣେ ଡାକବାଲା ଥୁଲା ଓ ଡାକଘରେ ମଧ୍ୟ ସାହାଯ୍ୟ କରୁଥିଲା, ହସି ହସି ଉପରିସ୍ଥ ଅଧିକାରୀଙ୍କ ନିକଟକୁ ଗଲା ଓ ଈଶ୍ବରଙ୍କ ପାଖକୁ ଲେଖାଯାଇଥିବା ଚିଠିଟିକୁ ଦେଖାଇଲା । ଡାକବାଲାଭାବେ ତା’ର ଚାକିରିକାଳ ମଧ୍ୟରେ ସେ ଏପରି ଠିକଣା କେବେ ଜାଣି ନଥିଲା । ଅମାୟିକ ପ୍ରକୃତିର ମେଦବହୁଳ ପୋଷ୍ଟମାଷ୍ଟର, ମଧ୍ୟ ହସି ହସି ଫାଟି ପଡ଼ିଲେ, କିନ୍ତୁ ହଠାତ୍ ସେ ଗମ୍ଭୀର ହୋଇଗଲେ ଏବଂ ତାଙ୍କ ଡେସ୍କ ଉପରେ ଚିଠିଟିକୁ ଟିକିଏ ବାଡ଼େଇ ଦେଇ ମନ୍ତବ୍ୟ ଦେଲେ, ‘କି ବିଶ୍ଵାସ ! ଏହି ଚିଠିଟି ଲେଖୁଥ‌ିବା ଲୋକର ବିଶ୍ଵାସ ଭଳି ମୋର ବିଶ୍ବାସ ଥାଆନ୍ତି କି ! ଈଶ୍ଵରଙ୍କ ସହିତ ପତ୍ରାଳାପ ଆରମ୍ଭ କରିଛି ।’’

Para: So, in order not to shake the writer’s faith in God, the postmaster came up with an idea: answer the letter. But when he opened it, it was evident that to answer it he needed something more than goodwill, ink, and paper. But he stuck to his resolution: he asked for money from his employees, he himself gave part of his salary, and several friends of his were obliged to give something for an act of charity.
ଅନୁବାଦ : ଲେଖକଙ୍କର ଈଶ୍ଵରଙ୍କ ଉପରେ ଆସ୍ଥା ନ ତୁଟାଇବାପାଇଁ ପୋଷ୍ଟମାଷ୍ଟରଙ୍କ ମନ ମଧ୍ୟରେ ଚିଠିର ଉତ୍ତର ଲେଖୁବାପାଇଁ ବିଚାରଟିଏ ଆସିଲା । କିନ୍ତୁ ସେ ଯେତେବେଳେ ଏହାକୁ ଖୋଲିଲେ, ସେଥୁରୁ ଜଣାପଡ଼ିଲା ଯେ ଏହାର ଉତ୍ତର ଦେବାପାଇଁ ସେ ଶୁଭେଚ୍ଛା, କାଳି ଓ କାଗଜ ବ୍ୟତୀତ ଆଉ କିଛି ଅଧିକ ଦରକାର କରୁଥିଲେ । କିନ୍ତୁ ସେ ଯାହା ସଙ୍କଳ୍ପ କରିଥିଲେ ବା ଚିନ୍ତା କରିଥିଲେ ସେଇଥ୍ରେ ଅଟଳ ରହିଲେ । ସେ ତାଙ୍କର କର୍ମଚାରୀମାନଙ୍କଠାରୁ କିଛି ଟଙ୍କା ମାଗିଲେ ଓ ସେ ନିଜେ ତାଙ୍କ ଦରମାରୁ କିଛି ଅଂଶ ପ୍ରଦାନ କଲେ ଏବଂ ତାଙ୍କର ଅନେକ ବନ୍ଧୁ ଏହି ବଦାନ୍ୟପୂର୍ଣ୍ଣ କାର୍ଯ୍ୟ ପାଇଁ କିଛି ଅର୍ଥ ଦେଇ କୃତଜ୍ଞ ହେଲେ ।

Para: It was impossible for him to gather together the hundred pesos, so he was able to send the farmer only a little more than half. He put the money in an envelope addressed to Lencho and with it a letter containing only a single word as a signature: God.
ଅନୁବାଦ : ଲେଞ୍ଜୋ ମାଗିଥ୍‌ ଶହେ ପେସୋ ଏକତ୍ର ସଂଗ୍ରହ କରିବା ତାଙ୍କ ପକ୍ଷେ ଅସମ୍ଭବ ଥିଲା, ତେଣୁ ସେ କୃଷକ ନିକଟକୁ ଅଧାରୁ ଟିକିଏ ଅଧିକ ପଠାଇବାକୁ ସମର୍ଥ ହେଲେ । ଲେଞ୍ଜୋ ଠିକଣା ଲେଖାଥିବା ଲଫାପା ଭିତରେ ସେ ଟଙ୍କା ଓ ତା’ ସହ ‘ଈଶ୍ବର’ ବୋଲି ସ୍ବାକ୍ଷର ହୋଇଥିବା ଚିଠିଟିକୁ ରଖୁଲେ ।

Para: The following Sunday Lencho came a bit earlier than usual to ask if there was a letter for him. It was the postman himself who handed the letter to him while the postmaster, experiencing the contentment of a man who has performed a good deed, looked on from his office.
ଅନୁବାଦ : ତା’ପାଇଁ କିଛି ଚିଠି ଅଛି କି ନାହିଁ ଜାଣିବାପାଇଁ ଲେଞ୍ଜୋ ପରବର୍ତ୍ତୀ ରବିବାର ଟିକିଏ ସହଳ ପହଞ୍ଚଗଲା । ଡାକବାଲା ନିଜେ ଚିଠିଟିକୁ ତାକୁ ବଢ଼ାଇଦେଲେ ଓ ଜଣେ ଭଲ କାମ କରିଥିବା ବ୍ୟକ୍ତି ଭଳି ସନ୍ତୋଷ ଅନୁଭବ କରି ପୋଷ୍ଟମାଷ୍ଟର ତାଙ୍କ ଅଫିସ୍‌ରୁ ଚାହିଁ ରହିଲେ ।

Para: Lencho showed not the slightest surprise on seeing the money; such was his confidence-but that he became angry when he counted the money. God could not have made a mistake, nor could he have denied Lencho what he had requested.
ଅନୁବାଦ : ଲେଞ୍ଚୋର ଏତେ ଦୃଢ଼ ବିଶ୍ୱାସ ଥିଲା ଯେ ଟଙ୍କାକୁ ଦେଖୁ ସେ ତିଳେମାତ୍ର ବିସ୍ମିତ ହେଲା ନାହିଁ । ମାତ୍ର ଟଙ୍କାଗୁଡ଼ିକୁ ଗଣିବାବେଳେ ସେ ରାଗିଗଲା । ଈଶ୍ବର କେବେହେଲେ ଭୁଲ୍ କରିପାରିବେ ନାହିଁ କିମ୍ବା ଲେଞ୍ଜୋ ଯାହା ଅନୁରୋଧ କରିଥିଲା ତାକୁ ମଧ୍ୟ କେବେହେଲେ ପ୍ରତ୍ୟାଖ୍ୟାନ କରିପାରିବେ ନାହିଁ ।

Para: Immediately, Lencho went up to the window to ask for paper and ink. On the public writing table, he started to write, with much wrinkling of his brow, caused by the effort he had to make to express his ideas. When he finished, he went to the window to buy a stamp which he licked and then affixed to the envelope with a blow of his fist. The moment the letter fell into the mailbox the postmaster went to open it. It said: “God: Of the money that I asked for, only seventy pesos reached me. Send me the rest since I need it very much. But don’t send it to me through the mail because the post office employees are a bunch of crooks. Len cho.”
ଅନୁବାଦ : ସଙ୍ଗେ ସଙ୍ଗେ ଲେଞ୍ଜୋ କାଗଜ ଓ କାଳି ମାଗିବାପାଇଁ ଝରକା ପର୍ଯ୍ୟନ୍ତ ଗଲା । ତା’ର ଭାବନାକୁ ପରିପ୍ରକାଶ କରିବାକୁ ଭୁଲତା କୁଞ୍ଚନ ସହିତ ସର୍ବସାଧାରଣଙ୍କ ପାଇଁ ଉଦ୍ଦିଷ୍ଟ ଟେବୁଲ ଉପରେ ସେ ଲେଖୁବାକୁ ଆରମ୍ଭ କଲା । ଚିଠି ଲେଖା ଶେଷ କରିବା ପରେ ଲେଞ୍ଚେ ଝରକା ପାଖକୁ ଟିକଟଟିଏ କିଣିବାକୁ ଗଲା ଓ ଟିକେଟରେ ସେ ଛେପ ମାରି ଲଫାପା ଉପରେ ବିଧାମାରି ଲଗାଇଦେଲା । ଡାକବାକ୍ସରେ ଚିଠିଟି ପଡ଼ିବାମାତ୍ରେ ପୋଷ୍ଟମାଷ୍ଟର ତାକୁ ଖୋଲିବାକୁ ଚାଲିଗଲେ । ଏଥ‌ିରେ ଲେଖାଥିଲା : ‘ହେ ଈଶ୍ଵର, ମୁଁ ମାଗିଥିବା ଟଙ୍କା ମଧ୍ୟରୁ ମାତ୍ର ସତୁରି ପୋସୋ ମୋ ପାଖରେ ପହଞ୍ଚିଲା । ଅବଶିଷ୍ଟ ଟଙ୍କାତକ ମୋ ପାଖକୁ ପଠାନ୍ତୁ କାରଣ ଏହା ମୋ ପାଇଁ ନିତାନ୍ତ ଆବଶ୍ୟକ । ମାତ୍ର ଏହାକୁ ଡାକ ଯୋଗେ ମୋ ପାଖକୁ ପଠାନ୍ତୁ ନାହିଁ କାରଣ ଡାକଘର କର୍ମଚାରୀମାନେ ହେଉଛନ୍ତି ଦଳେ ଅସାଧୁ ଲୋକ ।’’ ଇତି ଲେଞ୍ଜୋ ।

About The Author:
Gregorio Lopez Fuentes (1895-1966) was a famous Mexican novelist (ଔପନ୍ୟାସିକ), poet and journalist. Fuentes started writing at the age of 15, when the Mexican Revolution (ବିପ୍ଳବ) began. Many of his books are related (ସମ୍ପର୍କୀୟ) to the civil conflict. His stories are exciting and humorous. Many of his works are concerned with the oppression of Americans. In 1935, he was awarded the National Prize of Arts and Science.

BSE Odisha 10th Class English Solutions Chapter 2 A letter to God

Word Meaning / Glossary:

entire (Adj) – whole
The entire class went to the picnic.
crest (N) – top/the highest part of a hill
There is a temple on the crest of a hill.
corn (N) – crops
dotted with – scattered over an area
The sea is dotted with ships.
harvest (N) – yield
promised(V)- offered
needed (V) – required
at least – to the minimum
throughout (Prep) – all through
downpour (N)- heavy rainfall
shower (N) – a spell of light rain
intimately (Adv) – closely
supper (N) – light meal taken in the evening
was preparing (V) – was cooking
God willing – if God wills
until- not till
predict – foretell the future
No one can predict birth and death,
huge – very big
I saw a huge elephant,
approaching (V)- coming fast
exclaimed (V) – became surprised
falling (V) – dropping with a satisfied expression – with a content
regarder – thought, considered
drape – cover
The ground was draped with snow.
hailstones (N) – pellets of hail
did resemble (V)- looked alike
exposing themselves to the rain – coming out in the rain
frozen pearls (NP) – very cold small, white solid bead-like substance
really (Adv) — indeed
as if(Conj) — as though
He is walking as if he is mad.
rained (V) — poured down
soul (N) — heart
locusts (N) — insects that fly in big groups and destroy crops
The farmers use pesticides to protect the crops from locusts.
sorrowful (Adj) — having sadness
He got a piece of sorrowful news.
was totally destroyed — was completely spoiled
were gone — were disappeared
we’ll all go hungry— we will all live without food
solitary(Adj)— lonely, single
Upset (Adj) — disturbed, gloomy
She is most upset about her loss of property.
seems (V) — appears
a total loss (NP)— a complete loss
all through — throughout
conscience (N) — an inner sense of right or wrong
peso (N) – the currency of several Latin American countries
Lencho received only seventy pesos from the postmaster.
all-day break — at the crack of dawn/predawn
nothing less than— like
an ox of a man — very hard-working like an ox
one of the employees – କର୍ମଚାରୀଙ୍କ ମଧ୍ୟରୁ ଜଣେ
boss (N) — senior officer
laughing heartily— laughing with an open heart
amiable (Adj)— friendly and pleasant
He is an amiable young man.
fellow (N) — person
broke out laughing— laughed suddenly
commented (V) — opined
faith (N) — belief
correspondence(N) — an act of writing letters
Man cannot make correspondence with God.
in order not to shake — not to shatter (ici
came up with an idea — had a good idea
evident (Adj) — obvious, clear
stuck to — ଅଟକି ରହିବା
goodwill (N) — welfare
resolution (N) — a firm decision
Robinson Crusoe made a resolution to leave home all the first opportunity.
asked for (V) — demanded
were obliged (V) — were forced
salary (N) — monthly allowance
charity (N) — offering, donation, financial assistance
containing (V) —having
signature (N) — sign
handed (V) — gave by hand
mail (N) — post
blow (V) — knock
contentment (N) — satisfaction
The cat purred in obvious contentment.
a good deed (NP) — a good act
slight surprise (NP) — least surprise (little— less — least)
confidence (N) — self-belief, faith, trust
wrinkling (Adj) — କୁଞ୍ଚନ
caused by (V) — created by
licked (V) — pasted on the spittle
affïxed (V) — fixed properly, attached
with a blow of his fist — a strike of his grip
crooks (N) — dishonest persons/people
Nobody likes the crooks.

BSE Odisha 10th Class English Detailed Text: