CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 14 Limit and Differentiation

Limit Of A Function:

A real number ‘l’ is called the limit of the function f(x) as x tends to ‘a’ if for every ∈ > 0, there exist δ > 0 such that |f(x) – l| < ∈ whenever |x – a| < δ
We write \(\lim _{x \rightarrow a}\) f(x) = l
Left and right hand limit:

Left hand limit of f(x) as x → a is:
\(\lim _{x \rightarrow a-}\) f(x) = \(\lim _{x \rightarrow 0}\) f(a – h)

Right hand limit of f(x) as x → a is:
\(\lim _{x \rightarrow a+}\) f(x) = \(\lim _{h \rightarrow 0}\) f(a + h)

Existance of limit:
\(\lim _{x \rightarrow a}\) f(x) exists if it is unique, irrespective of any type of approach i.e if LHL = RHL. i.e if \(\lim _{x \rightarrow a-}\) f(x) = \(\lim _{x \rightarrow a+}\) f(x)

Indeterminate forms:
The forms : \(\frac{0}{0}, \frac{\infty}{\infty}\), ∞ – ∞, 0 × ∞, 0°, ∞° and 1 are called indeterminate forms in mathematics.

CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation

Properties of limit:
CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation

Some standard limits:
CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation 1

Limit At Infinite And Infinite Limits:

(a) We write \(\lim _{x \rightarrow a}\) f(x) = ∞ if for a given m > 0, there exists δ > 0 such that |x – a| < δ ⇒ f(x) > m for large m.

(b) We write \(\lim _{x \rightarrow a}\) f(x) = -∞ if for a given m < 0, there exists δ > 0 such that |x – a| < δ ⇒ f(x) < m for large |m|.

(c) \(\lim _{x \rightarrow ∞}\) f(x) = l if for given ∈ > 0 there exists k > 0 such that x > k ⇒ |f(x) – l| < ∈ for large x.

(d) \(\lim _{x \rightarrow -∞}\) f(x) = l if for given ∈ > 0, there exists k < 0 such that x < k ⇒ |f(x) – l| < ∈ for large |k|.

(e) We write \(\lim _{x \rightarrow ∞}\) f(x) = ∞ if for m > 0 there exists k > 0 such that x > x ⇒ f(x) > m for large m.

(f) \(\lim _{x \rightarrow ∞}\) xn = \(\left\{\begin{array}{lll}
\infty & \text { if } & n>0 \\
1 & \text { if } & n=0 \\
0 & \text { if } & n<0
\end{array}\right.\)

(g) \(\lim _{n \rightarrow ∞}\) xn = \(\left\{\begin{array}{ccc}
0 & \text { if } & |x|<1 \\
1 & \text { if } & x=1 \\
\infty & \text { for } & x>1
\end{array}\right.\) does not exist for x ≤ -1.

CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation

Some useful expansions:
CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation 2

Techniques to find limit:
If \(\lim _{x \rightarrow a}\) f(x), does not take any indeterminate form then get the limit just by putting x = a(provided that the limit is finite).
If \(\lim _{x \rightarrow a}\) f(x) takes any indeterminate form then either use formula or simplify to remove the indeterminate form before finding limit.
The indeterminate form can be removed by using.

  • Factorisation
  • Rationalisation
  • Expand  formula or any other techniques.

Differentiation:

(a) Let y = f(x) is a function.
The derivative (differential coefficient) of y or f(x) with respect to x is \(\frac{d y}{d x}\) = f'(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

(b) The differentiation of y = f(x) at x = a is \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=\mathrm{a}}\) = f'(a) = \(\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)

(c) Differentiability of y = f(x) at x = a:
CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation 3

(d) Geometrical meaning of differentiation:
Geometrically f'(x) or \(\frac{d y}{d x}\) represents the slope of tangent to y = f(x) at any point P(x, y)
⇒ Slope of tangent to y = f(x) at A(x1, y1) = \(\left.\frac{d y}{d x}\right]_{\left(x_1, y_1\right)}\)

(e) Some rules of differentiation:
CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation 4

CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation

(f) Differentiation of some standard functions:
CHSE Odisha Class 11 Math Notes Chapter 14 Limit and Differentiation 5

CHSE Odisha Class 11 Math Notes Chapter 7 Linear Inequalities

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 7 Linear Inequalities will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 7 Linear Inequalities

Inequality:
A statement with symbols like >, ≥, <, ≤ is an inequality.

Different types of inequality:

(a) Numerical inequality: It is an inequality involving numbers not variables.
(b) Literal inequality: It is the inequality involving literal numbers(variable).
(c) Strict inequality: An inequality with only > or < symbols is a strict inequality.
(d) Slack inequality: An inequality with only ≥ or ≤ symbols is a slack inequality.

Linear inequality:
An inequality involving variables in the first degree is called linear inequalities.
(a) General form of inequalities:
(i) In one variable: ax + b > or ≥ or < or ≤ 0
(ii) In two variables: ax + by + c > or ≥ or < or ≤ 0.

Intervals:

  • Closed Interval: [a, b] = {x ∈ R: a ≤ x ≤ b}
  • Open Interval: (a, b) = {x ∈ R: a < x < b}
  • Semi-open or semi-closed interval:
    ⇒ [a, b) = {x ∈ R: a ≤ x < b}
    ⇒ (a, b] = {x ∈ R: a < x ≤ b}

Basic properties of inequalities:
(1) a > b, b > c ⇒ a > c
(2) a > b ⇒ a ± c > b ± c
(3) a > b

  • m > 0 ⇒ am > bm, \(\frac{a}{m}>\frac{b}{m}\)
  • m < 0 ⇒ am < bm, \(\frac{a}{m}<\frac{b}{m}\)

(4) If a > b > 0, then
a2 > b2, |a| > |b| and \(\frac{1}{a}>\frac{1}{b}\)
If a < b < 0, then
|a| > |b| and \(\frac{1}{a}>\frac{1}{b}\)

CHSE Odisha Class 11 Math Notes Chapter 7 Linear Inequalities

Graphical solution of linear inequalities in two variables:
Working rule:

Let the inequality is ax + by + c < or ≤ or > or ≥ 0

Step – 1: Consider the equation ax + by + c = 0 in place of the inequality and draw its graph (Draw a dotted line for > or < and a bold line for ≥ or ≤).
Step – 2: Take any point that does not lie on the graph, and put the coordinate in the inequality.
If you get true then the inequality is satisfied. Shade the half-plane containing that point otherwise the inequality is not satisfied. In this case shade the half plane region that does not contain the point.
Step – 3: The shaded region is the required solution.

Solution of a system of linear inequalities in two variables:

Step – 1: Draw the graph of all lines.
Step – 2: Shade the appropriate region for each inequality.
Step – 3: The common region is the required solution.

BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a)

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a)

(କ – ବିଭାଗ )

Question 1.
ନିମ୍ନ ଉକ୍ତିମାନଙ୍କ ମଧ୍ୟରୁ ଯେଉଁଟି ଠିକ୍ ତା’ ପାଖରେ T ଓ ଯେଉଁଟି ଭୁଲ ତା’ ପାଖରେ F ଲେଖ ।
(i) ଦୁଇଟି କ୍ରମିକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ ସେ ଦ୍ଵୟ ମଧ୍ୟବର୍ତ୍ତୀ ଯୁଗ୍ମସଂଖ୍ୟା ସଙ୍ଗେ ସମାନ ।
(ii) ଏକ ସମାନ୍ତର ପ୍ରଗତିରେ ଥିବା ତିନୋଟି କ୍ରମିକ ପଦର ମାଧ୍ଯମାନ ସେମାନଙ୍କର ମଧ୍ଯମପଦ ସଙ୍ଗେ ସମାନ ।
(iv) ଭିନ୍ନ ଭିନ୍ନ ଆରମ୍ଭ ବିନ୍ଦୁ ନେଇ ଦତ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ନିର୍ଣ୍ଣୟ କଲେ ଭିନ୍ନ ଭିନ୍ନ ଉତ୍ତର ମିଳିବ ।
(v) କୌଣସି ତଥ୍ୟାବଳୀର ଆରମ୍ଭ ବିନ୍ଦୁ 20 ହେଲେ ଅନ୍ତର୍ଭୁକ୍ତ ଲବ୍‌ଧାଙ୍କ 15ର ବିଚ୍ୟୁତି 5 ।
(vi) ପ୍ରଥମ n ସଂଖ୍ୟକ ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ \(\frac{n+2}{2}\)।
(vii) ପ୍ରଥମ n ସଂଖ୍ୟକ ଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 2n + 2 ।
(viii) ପ୍ରଥମ ଦଶଗୋଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 10 ।
(ix) 15 ଗୋଟି ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 17 । ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାକୁ 2 ଦ୍ୱାରା ଗୁଣି ସେମାନଙ୍କର ମାଧ୍ଯମାନ ସ୍ଥିର କଲେ ମାଧ୍ୟମାନ 8.5 ହେବ ।
(x) ପ୍ରଥମ 20ଟି ଯୁଗ୍ମ ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ, ପ୍ରଥମ 20ଟି ଗଣନ ସଂଖ୍ୟାର ମାଧମାନର ଦୁଇ ଗୁଣ ।
ଉ :
(i) ଦୁଇଟି କ୍ରମିକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ ସେ ଦ୍ଵୟ ମଧ୍ୟବର୍ତ୍ତୀ ଯୁଗ୍ମସଂଖ୍ୟା ସଙ୍ଗେ ସମାନ । (T)
(ii) ଏକ ସମାନ୍ତର ପ୍ରଗତିରେ ଥିବା ତିନୋଟି କ୍ରମିକ ପଦର ମାଧ୍ଯମାନ ସେମାନଙ୍କର ମଧ୍ଯମପଦ ସଙ୍ଗେ ସମାନ । (T)
(iv) ଭିନ୍ନ ଭିନ୍ନ ଆରମ୍ଭ ବିନ୍ଦୁ ନେଇ ଦତ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ନିର୍ଣ୍ଣୟ କଲେ ଭିନ୍ନ ଭିନ୍ନ ଉତ୍ତର ମିଳିବ । (T)
(v) କୌଣସି ତଥ୍ୟାବଳୀର ଆରମ୍ଭ ବିନ୍ଦୁ 20 ହେଲେ ଅନ୍ତର୍ଭୁକ୍ତ ଲବ୍‌ଧାଙ୍କ 15ର ବିଚ୍ୟୁତି 5 । (F)
(vi) ପ୍ରଥମ n ସଂଖ୍ୟକ ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ \(\frac{n+1}{2}\)। (T)
(vii) ପ୍ରଥମ n ସଂଖ୍ୟକ ଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 2n + 2 । (F)
(viii) ପ୍ରଥମ ଦଶଗୋଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 10 । (T)
(ix) 15 ଗୋଟି ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 17 । ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାକୁ 2 ଦ୍ୱାରା ଗୁଣି ସେମାନଙ୍କର ମାଧ୍ଯମାନ ସ୍ଥିର କଲେ ମାଧ୍ୟମାନ 8.5 ହେବ । (F)
(x) ପ୍ରଥମ 20ଟି ଯୁଗ୍ମ ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ, ପ୍ରଥମ 20ଟି ଗଣନ ସଂଖ୍ୟାର ମାଧମାନର ଦୁଇ ଗୁଣ । (F)

BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a)

ବ୍ୟାଖ୍ୟା ସହ ଉତ୍ତର:
(i) (T) (କାରଣ 3 ଓ 5ର ମାଧ୍ୟମାନ \(\frac{3+5}{2}=4\))
(ii) (T) (କାରଣ AM \(\frac{a+b}{2}\))
(iii) (T) (କାରଣ ମାଧ୍ଯମାନର ପ୍ରତିଶବ୍ଦ ହାରାହାରି ଅଟେ ।)
(iv) (F) (ସର୍ବଦା ବିଚ୍ୟୁତିର ମାଧ୍ଯମାନ ସହିତ ଆରମ୍ଭ ବିନ୍ଦୁ ଯୋଗ କରାଯାଏ, ତେଣୁ ଉତ୍ତର ସର୍ବଦା ସମାନ ହେବ ।)
(v) (F) (କାରଣ ବିଚ୍ୟୁତି = ଲବ୍‌ଧାଙ୍କ – ଆରମ୍ଭ ବିନ୍ଦୁ = 15 – 20 = – 5)
(vi) (T) (କାରଣ ପ୍ରଥମ n ସଂଖ୍ୟକ ସଂଖ୍ୟାର ସମଷ୍ଟି = \(\frac{n(n+1)}{2}\)
∴ ମାଧ୍ୟମାନ = \(\frac{n(n+1)}{2n}=\frac{n+1}{2}\))
(vii) (F) (ସୂତ୍ର ଅନୁସାରେ n + 1 ହେବ ।)
(viii) (T) (କାରଣ ପ୍ରଥମ ଦଶଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ସମଷ୍ଟି = 10², ମାଧ୍ୟମାନ = \(\frac{10²}{10}\) = 10)
(ix) (F) (କାରଣ ମାଧମାନ 2 ଗୁଣ ହେବ ।)
(x) (F) (କାରଣ, ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାରେ 2 ଗୁଣିଲେ ତା 20ଟି ଯୁଗ୍ମ ଗଣନ ସଂଖ୍ୟା ହେବ ।)

Question 2.
ପ୍ରତ୍ୟେକ ପ୍ରଶ୍ନ ପାଇଁ ପ୍ରଦତ୍ତ ସମ୍ଭାବ୍ୟ ଉତ୍ତରମାନଙ୍କ ମଧ୍ୟରୁ ଠିକ୍ ଉତ୍ତରଟି ବାଛ ।
(i) 61, 62, 68, 56, 64, 72, 69, 51, 71, 67, 70, 55, 63 ଏହି ଲବ୍ଧାଙ୍କମାନଙ୍କର ମାଧ୍ୟମାନ ନିରୂପଣ ଲାଗି ନିମ୍ନସ୍ଥ ସଂଖ୍ୟାମାନଙ୍କ ମଧ୍ୟରୁ କେଉଁଟି ଉପଯୁକ୍ତ ଆରମ୍ଭ ବିନ୍ଦୁ ହେବ ?
(A) 55
(B) 60
(C) 70
(D) 72

(ii) ପ୍ରଥମ 20ଟି ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ କେତେ ?
(A) 10
(B) 10½
(C) \(\frac{21}{20}\)
(D) 210

(iii) ପ୍ରଥମ ‘n’ ସଂଖ୍ୟକ ସଂପ୍ରସାରିତ ସ୍ଵାଭାବିକ ସଂଖ୍ୟା (Whole number)ର ମାଧ୍ଯମାନ କେତେ ?
(A) \(\frac{n-1)}{2}\)
(B) \(\frac{n}{2}\)
(C) \(\frac{n+1}{2}\)
(D) n

(iv) ପ୍ରଥମ ‘n’ ସଂଖ୍ୟକ ଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ଯମାନ କେତେ ?
(A) (n – 1)
(B) n
(C) n + 1
(D) n + 2

(v) ପ୍ରଥମ n ସଂଖ୍ୟକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ଯମାନ କେତେ ?
(A) (n – 11)
(B) n
(C) n + 1
(D) n + 2

(vi) ‘m’ ମାଧମାନ ବିଶିଷ୍ଟ 10ଟି ଲବ୍‌ଧାଙ୍କ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକକୁ 2 ବଢ଼ାଇଲେ ନୂତନ ଲବ୍‌ଧାଙ୍କ 10ଟିର ମାଧ୍ଯମାନ କେତେ ହେବ ?
(A) (n – 11)
(B) n
(C) n + 1
(D) n + 2

(vii) ‘M’ ମାଧ୍ୟମାନ ବିଶିଷ୍ଟ n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କମାନଙ୍କ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକକୁ 4 ଗୁଣ କରିଦେଲେ ନୂତନ ଲବ୍‌ଧାଙ୍କମାନଙ୍କର ମାଧ୍ଯମାନ କେତେ ହେବ ?
(A) \(\frac{M)}{4}\)
(B) M
(C) 4M
(D) \(\frac{4}{M}\)

BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a)

(viii) ‘M’ ମାଧ୍ଯମାନ ବିଶିଷ୍ଟ n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କମାନଙ୍କ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକରୁ x ବିୟୋଗ କଲେ ନୂତନ ଲବ୍‌ଧାଙ୍କମାନଙ୍କର ମାଧ୍ଯମାନ କେତେ ହେବ ?
(A) M
(B) (M + x)
(C) Mx
(D) (M – x)

(ix) ‘M’ ମାଧ୍ଯମାନ ବିଶିଷ୍ଟ n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କମାନଙ୍କ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକକୁ 5 ଦ୍ଵାରା ଭାଗକଲେ ନୂତନ ଲବ୍‌ଧାଙ୍କମାନଙ୍କର ମାଧ୍ଯମାନ କେତେ ହେବ ?
(A) M
(B) \(\frac{M}{5}\)
(C) 5M
(D) M – 5

(x) ଯଦି à ସଂଖ୍ୟକ ବାଳକମାନଙ୍କର ମାଧ୍ଯମାନ ବୟସ 12 ବର୍ଷ ଓ b ସଂଖ୍ୟକ ବାଳିକାଙ୍କର ମାଧ୍ଯମାନ ବୟସ 10 ବର୍ଷ ହୁଏ, ତେବେ ଉପରୋକ୍ତ ସମସ୍ତ ବାଳକ ବାଳିକାଙ୍କର ମାଧ୍ଯମାନ ବୟସ କେତେ ବର୍ଷ ହେବ ?
(A) \(\frac{10a+12b}{a+b}\)
(B) \(\frac{12a+10b}{a+b}\)
(C) \(\frac{10a+12b}{10+12}\)
(D) \(\frac{12a+10b}{10+12}\)

(xi) 998.9, 999.1, 1000-3, 1000-6, 1000.1 ର ମାଧ୍ୟମାନ କେତେ?
(A) 998
(B) 999
(C) 1000
(D) 1001

(xii) 6,8, 5, 7, x ଏବଂ 4 ଲବ୍‌ଧାଙ୍କଗୁଡ଼ିକର ମାଧ୍ଯମାନ 7 ହେଲେ xର ମାନ କେତେ ହେବ ?
(A) 10
(B) 11
(C) 12
(D) 13

(xiii) E1, E2, E3, E4, E5, E6ଲବ୍‌ଧାଙ୍କଗୁଡ଼ିକର ମାଧ୍ଯମାନ M ହେଲେ 6Σi=1(x1 – M)ର ମାନ କେତେ ହେବ ?
(A) 0
(B) 6
(C) 36
(D) -6

(xiv) x, x + 2, x + 4, x + 6, x + 8ର ମାଧ୍ୟମାନ କେତେ ?
(A) x+2
(B) x + 4
(C) x+6
(D) x

(xv) 18ର ସମସ୍ତ୍ର ଗୁଣନୀୟକମାନଙ୍କର ମାଧ୍ୟମାନ କେତେ
(A) 5
(B) 6
(C) 6.5
(D) 7

ଉତ୍ତର:
(i) 69
(ii) 10½
(iii) \(\frac{n-1}{2}\)
(iv) n + 1
(v) n
(vi) m + 2
(vii) 4M
(viii) (M – x)
(ix) \(\frac{M}{5}\)
(x) \(\frac{12a+10b}{a+b}\)
(xi) 1000
(xi) 1000
(xii) 12
(xiii) 0
(xiv) x + 4
(xv) 6.5

BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a)

(ଖ – ବିଭାଗ )

Question 3.
ଦଶଥର ଖେଳି ଜଣେ କ୍ରିକେଟ୍ ଖେଳାଳୀ ସଂଗ୍ରହ କରିଥିବା ରଗୁଡ଼ିକ ହେଲା – 47, 41, 50, 39, 45, 48,
42, 32, 60 ଏବଂ 20 । ତାଙ୍କଦ୍ୱାରା ସଂଗୃହୀତ ରନ୍‌ର ମାଧ୍ଯମାନ ସଂକ୍ଷିପ୍ତ ପ୍ରଣାଳୀରେ (ଉପଯୁକ୍ତ ଆରମ୍ଭ ବିନ୍ଦୁ ନେଇ) ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର ଆରମ୍ଭ ବିନ୍ଦୁ 45 1 ( ∵ ସର୍ବନିମ୍ନ ଏବଂ ସର୍ବାଧ‌ିକ ରନ୍ ଯଥାକ୍ରମେ 20 ଏବଂ 60) ।
∴ ଲବ୍ଧାଙ୍କମାନଙ୍କର ବିଚ୍ୟୁତିମାନ 2, − 4, 5, 6, 0, 3, – 3, −13, 15, – 25
ବିଚ୍ୟୁତିମାନଙ୍କର ସମଷ୍ଟି = – 26
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -1
∴ ଦଶଥର ଖେଳି ସଂଗୃହୀତ ରନ୍‌ର ମାଧ୍ୟମାନ = 42.4

Question 4.
କିଲୋଗ୍ରାମ୍ ଓଜନରେ 30 ଜଣ ପିଲାଙ୍କର ଓଜନ ହେଲା 21, 30, 40, 25, 26, 22, 26, 31, 22, 36, 30, 25, 25, 33, 30, 25, 27, 27, 25, 31, 33, 22, 21, 36, 40, 31, 33, 30, 37, 36 | ଏହି ତଥ୍ୟାବଳୀକୁ ବାରମ୍ବାରତା ବଣ୍ଟନରେ ସଜ୍ଜିତ କରି ମାଧ୍ଯମାନ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଓଜନ କିଲୋଗ୍ରାମ୍ ମାପରେ ଥ‌ିବା ଲବ୍‌ଧାଙ୍କମାନଙ୍କୁ ବାରମ୍ବାରତା ବଣ୍ଟନ ସାରଣୀରେ ରଖିଲେ –
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -2
ଉକ୍ତ ସାରଣୀରୁ 2f = 30 ଏବଂ Efx = 876.. ମାଧ୍ୟମାନ = \(\frac{Σf_x}{Σf}\)

Question 5.
କିଛି ରାସାୟନିକ ପଦାର୍ଥର ଓଜନ 30 ଥର ନିଆଯାଇ ଫଳାଫଳକୁ ନିମ୍ନ ସାରଣୀରେ ସଜାଯାଇଛି । ମାଧ୍ଯମାନ ଓଜନ ନିର୍ଣ୍ଣୟ କର ।

ଓଜନ (ଗ୍ରାମ୍‌ରେ) 3.8 3.9 4.0 4.1 4.2 4.3 4.4 4.5 4.6
ବାରମ୍ବାରତା 1 1 6 6 7 5 2 1 1

ସମାଧାନ :

ଓଜନ (ଗ୍ରାମ୍‌ରେ) (x) ବାରମ୍ବାରତା (f) ଓଜନ × ବାରମ୍ବାରତା (fx)
3.8 1 3.8
3.9 1 3.9
4.0 6 24.0
4.1 6 24.6
4.2 7 29.4
4.3 5 21.5
4.4 2 8.8
4.5 1 4.5
4.6 1 4.6
Σf=30 Σfx=125.1

∴ ମାଧ୍ଯମାନ = \(\frac{Σf_x}{Σf}=\frac{125.1}{30}=4.17\)
∴ ମାଧ୍ୟମାନ ଓଜନ 4.17 ଗ୍ରାମ୍ ।

Question 6.
ଏକ ଶ୍ରେଣୀରେ 30 ଜଣ ଛାତ୍ରଙ୍କର ହାରାହାରି ବୟସ 12 ବର୍ଷ । ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ସହିତ ସେମାନଙ୍କର ହାରାହାରି ବୟସ 13 ବର୍ଷ ହେଲେ, ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ବୟସ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଏକ ଶ୍ରେଣୀରେ 30 ଜଣ ଛାତ୍ରଙ୍କର ହାରାହାରି ବୟସ 12 ବର୍ଷ ।
30 ଜଣ ଛାତ୍ରଙ୍କର ମୋଟ ବୟସ = 30 × 12 = 360 ବର୍ଷ ।
ଛାତ୍ରମାନଙ୍କ ସହ ତାଙ୍କର ଶ୍ରେଣୀଶିକ୍ଷକ ମିଶିବାରୁ ହାରାହାରି ବୟସ 13 ବର୍ଷ ହେଲା ।
∴ 31 ଜଣ ଅର୍ଥାତ୍ 30 ଜଣ ଛାତ୍ର ଓ ଜଣେ ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ମୋଟ ବୟସ = 31 × 13 = 403 ବର୍ଷ ।
ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ବୟସ = 403 – 360 = 43 ବର୍ଷ ।
∴ ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ବୟସ 43 ବର୍ଷ ।

Question 7.
x1, x2, x3 …… ପ୍ରଭୃତି n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କର ମାଧ୍ଯମାନ m । ଯଦି ପ୍ରତ୍ୟେକ ଲବ୍‌ଧାଙ୍କରେ (a + b) ଯୋଗ କରାଯାଏ ଦର୍ଶାଅ ଯେ, ନୂତନ ଲବ୍‌ଧାଙ୍କଗୁଡ଼ିକର ମାଧମାନ (m + a + b) ହେବ ।
ସମାଧାନ :
ଲବ୍‌ଧାଙ୍କଗୁଡ଼ିକ ହେଲେ x1, x2, x3 ……… xn
ଉକ୍ତ n-ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କମାନଙ୍କର ମାଧ୍ଯମାନ (m) = \(\frac{x_1+x_2+x_3+…..x_n}{n}\)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -3

BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a)

(ଗ – ବିଭାଗ )

Question 8.
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -4
ସମାଧାନ :

ଉଚ୍ଚତା (x) ବାରମ୍ବାରତା (f) ଫଭାଗର ମଧ୍ୟବିନ୍ଦୁ ମଧ୍ୟବିନ୍ଦୁ × ବାରମ୍ବାରତା (fy)
70-65 4 67.5 270.0
65-60 7 62.5 437.5
60-55 8 57.5 460.0
55-50 10 52.5 525.0
50-45 5 47.5 237.5
45-40 6 42.5 255.0
40-35 3 37.5 112.5
35-30 7 32.5 227.5
30-25 2 27.5 55.0
Σf = 52 Σfy = 2580.00

∴ ମାଧ୍ଯମାନ = \(\frac{Σfy}{Σf}=\frac{2580}{52}=49.6\)
ବିକଳ୍ପ ପ୍ରଣାଳୀ : (ସଂକ୍ଷିପ୍ତ ପ୍ରଣାଳୀ)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -5
ଆରମ୍ଭ ବିନ୍ଦୁ = 47.5, ସଂଭାଗ ବିସ୍ତାର (i) = 5
ମାଧ୍ୟମାନ = ଆରମ୍ଭ ବିନ୍ଦୁ + \(\frac{Σfy’}{Σf}\) × i = 47.5 + \(\frac{22×5}{52}\) (y’ = ବିଚ୍ୟୁତି) = 47.5 + \(\frac{110}{52}\) = 47.5+2.1 = 49.6
ମଧ୍ୟବିନ୍ଦୁ = \(\frac{\text { ସଂଭାଗର ନିମ୍ନସୀମା + ସଂଭାଗର ଉଚ୍ଚସୀମା }}{2}\)
ଅନ୍ତର୍ଭୁକ୍ତ ସଂଭାଗୀକରଣରେ ସଂଭାଗ ବିସ୍ତାର = ସଂଭାଗର ଉଚ୍ଚସୀମା – ସଂଭାଗର ନିମ୍ନସୀମା

Question 9.
ସଂକ୍ଷିପ୍ତ ପ୍ରଣାଳୀର ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ନିରୂପଣ କର ।

ସଂଭାଗ 84-90 90-96 96-102 102-108 108-114 114-120
ବାରମ୍ବାରତା 8 10 16 23 12 11

ସମାଧାନ :
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -6
∴ ମାଧ୍ଯମାନ = A + \(\frac{Σfy}{Σf}=100+\frac{244}{80}\) = 100 + 3.05 = 103.05

Question 10.
ନିମ୍ନ ଭାଗ-ବିଭକ୍ତ ବାରମ୍ବାରତା ବିତରଣ ସାରଣୀରେ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ସୋପାନ-ବିଦ୍ୟୁତ ପ୍ରଣାଳୀରେ ସ୍ଥିର କର ।

ସଂଭାଗ 0-4 4-8 8-12 12-16 16-20 20-24
ବାରମ୍ବାରତା 5 7 10 15 9 4

ସମାଧାନ :
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -7
∴ ମାଧ୍ୟମାନ = ଆରମ୍ଭ ବିନ୍ଦୁ + \(\frac{Σfy’}{Σf}\) × c = 12 + \(\frac{6}{50}\) × 2 = 12 + 0.24 = 12.24
ବିକଳ୍ପ ପ୍ରଣାଳୀ : ମାଧମାନ (M) = A + \(\frac{Σfy}{Σf}\) × i
ସୂତ୍ରର ପ୍ରୟୋଗ କରି ସମାଧାନ କରାଯାଇପାରିବ । ଯେଉଁଠାରେ i = ସଂଭାଗବିସ୍ତାର ହେବ ।

Question 11.
ନିମ୍ନ ସାରଣୀରେ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ଉଭୟ ସଂକ୍ଷିପ୍ତ ପ୍ରଣାଳୀ ଓ ସୋପାନ-ବିଦ୍ୟୁତ ପ୍ରଣାଳୀ ଅବକମୂଳରେ ସ୍ଥିର କର ।

ସଂଭାଗ (C.I.) 0-50 50-100 100-150 150-200 200-250 250-300
ବାରମ୍ବାରତା (f) 4 10 12 10 8 8

ସମାଧାନ :
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -8
∴ ମାଧ୍ଯମାନ = A + \(\frac{Σfy}{Σf}=150+\frac{300}{52}\) = 150 + 5.77 = 155.77
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -9
∴ ମାଧ୍ୟମାନ (M)= A + \(\frac{Σfy’}{Σf}\) × c = 150 + \(\frac{12}{52}\) × 25 = 150 + 5.77 = 155.77

Question 12.
ସୋପାନ ବିଚ୍ୟୁତି ପ୍ରଣାଳୀ ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ, ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ସ୍ଥିର କର ।

ସଂଭାଗ 20-30 30-40 40-50 50-60 60-70 70-80
ବାରମ୍ବାରତା 10 6 8 12 5 9

ସମାଧାନ :
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -10
ଏଠାରେ A = 55, i = 60 – 50 = 10
∴ ମାଧ୍ୟମାନ (M)= A + \(\frac{Σfy’}{Σf}\) × i = 55 + \(\frac{-27}{50}\) × 10 = 55 + (-5.4) = 49.6

Question 13.
(i) ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ 7.5 ହେଲେ ‘f” ର ନିରୂପଣ କର ।

ସଂଭାଗ 5 6 7 8 9 10 11 12
ବାରମ୍ବାରତା 20 17 f 10 8 6 7 6

(ii) ମୂଲ୍ୟ ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ 6 ହେଲେ ‘P’ ର ମୂଲ୍ୟ ନିରୂପଣ କର ।

ସଂଭାଗ 3 6 7 4 P+3 8
ବାରମ୍ବାରତା 5 2 3 2 4 6

ସମାଧାନ :
(i) ବଡ ତଥ୍ୟାବଳୀର ମାଧ୍ୟମାନ = 7.5
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -11
ମାଧ୍ୟମାନ (M) = \(\frac{Σfy}{Σf}\) ⇒ 7.5 = \(\frac{563+7f}{74+f}\)
⇒ 555 + 7.5f = 563 + 7f ⇒ 7.5f – 7f = 563 – 555
⇒ 0.5f = 8 ⇒ f = 8 ⇒ \(\frac{1}{2}\)f = 8 × 2 = 16

(ii) ବଡ ତଥ୍ୟାବଳୀର ମାଧ୍ୟମାନ = 6
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -12
ମାଧ୍ୟମାନ (M) = \(\frac{Σfy}{Σf}\) ⇒ 6 = \(\frac{116+4p}{22}\)
⇒ 4p + 116 = 132 ⇒ 4p = 16
⇒ p = \(\frac{16}{4}\) = 4

Question 14.
ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧମାନ 50 ଏବଂ ବାରମ୍ବାରତାଗୁଡ଼ିକର ସମଷ୍ଟି 120 ହେଲେ f1 ଓ f2 ନିର୍ଣ୍ଣୟ କର ।

ସଂଭାଗ 0-20 20-40 40-60 60-80 80-100
ବାରମ୍ବାରତା 17 f1 32 F2 19

ସମାଧାନ :
ବଡ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧମାନ = 50, ବାରମ୍ବାରତାଗୁଡ଼ିକର ସମଷ୍ଟି = 120
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -13
ପ୍ରଶ୍ନନୁସାରେ, 68 + f1 + f2 = 120
f1 + f2 = 52
Σfx = 3480 + 30f1 + 70f2 = 3480 + 30(f1 + f2) + 40f2
=3480 + 30 × 52 ÷ 40f2 = 3480+ 1560 + 40f2 = 5040 + 40f2
∴ ମାଧ୍ୟମାନ (m) = \(\frac{Σfx}{Σf}=\frac{5040+4f_2}{120}\)
⇒ 50= \(\frac{5040+4f_2}{120}\) ⇒ 40f2 = 6000 – 5040 ⇒ f2 = \(\frac{960}{40}\) = 24

ଆଗରୁ ପ୍ରମାଣିତ f1 + f2 =52 f1 = 52 – 24 = 28
∴ f1 = 28 ଏବଂ f2 = 24

BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a)

Question 15.
ସୋପାନ-ବିଚ୍ୟୁତି ପ୍ରଣାଳୀ ଅବଲମ୍ବନରେ ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ସ୍ଥିର କର ।

ସଂଭାଗ 10-19 20-29 30-39 40-49 50-59 60-69 70-79
ବାରମ୍ବାରତା 5 65 222 112 53 40 3

ସମାଧାନ :
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -14
∴ ମାଧ୍ୟମାନ = A + \(\frac{Σfy’}{Σf}\) × i = 44.5 + \(\frac{-225}{500}\) × 10 = 44.5 + \(\frac{-450}{100}\) = 44.5 – 4.5 = 40

Question 16.
x1, x2, x3 ……. ପ୍ରଭୃତି n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କର ମାଧ୍ଯମାନ M । ଯଦି \(\sum_{i=1}^n\left(x_i-5\right)=60\) ଏବଂ \(\sum_{i=1}^n\left(x_i-8\right)\) = 24 ହୁଏ ତେବେ ‘n’ ଓ M ସ୍ଥିର କର ।
ସମାଧାନ :
x1, x2, x3 ………. ପ୍ରଭୃତି n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କର ମାଧମାନ M ।
⇒ \(\frac{x_1+x_2+x_3+…..x_n}{n}=M\)
⇒ x1 + x2 + x3 ……. + xn = nM
\(\sum_{i=1}^n\left(x_i-5\right)=60\)
⇒ (x1 – 5) + (x2 – 5) + (x3 – 5) ……. + (xn – 5) = 60
⇒ (x1 + x2 + x3 ……. + xn) – 5n = 60
⇒ nM – 5n = 60 ………(i)
⇒ \(\sum_{i=1}^n\left(x_i-8\right)\) = 24 ⇒ nM – 8n = 24 ………(ii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ
BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) -15
‘n’ ର ମାନ ସମୀକରଣ (i)ରେ ପ୍ରୟୋଗ କଲେ nM – 5n = 60
⇒ 12M – 60 = 60 ⇒ 12M = 120
⇒ M = \(\frac{120}{12}\) = 10
∴ n = 12 ଓ M = 10

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(b)

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(b) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(b)

Question 1.
ନିମ୍ନ ଉକ୍ତିଗୁଡ଼ିକରେ ଠିକ୍ ଉକ୍ତି ପାଇଁ T ଓ ଭୁଲ ଉକ୍ତି ପାଇଁ F ଲେଖ ।
(i) ବୃତ୍ତର ଏକ ଉପସେଟ୍‌କୁ ଚାପ କହନ୍ତି ।
(ii) ଚାପର ଏକ ଅନ୍ତ୍ରମ୍ମ ଦିନ୍ଦୁ ସମ୍ରକ୍ର ଦୃଭର ଅନ୍ତ୍ରମ ବନ୍ଧୁ ନ୍ମ6ଦୃ |
(iii) ଗୋଟିଏ ବୃତ୍ତରେ P ଓ Q ଦୁଇଟି ଚାପର ସାଧାରଣ ପ୍ରାନ୍ତବିନ୍ଦୁ ହେଲେ ଚାପଦ୍ଵୟ ପରସ୍ପରର ପରିପୂରକ ଚାପ
(iv) ପ୍ରତ୍ୟେକ ଚାପର ପ୍ରାନ୍ତବିନ୍ଦୁକୁ କେନ୍ଦ୍ର ସହିତ ଯୋଗ କଲେ ଯେଉଁ କୋଣ ଉତ୍ପନ୍ନ ହୁଏ ତାହା ଉକ୍ତ ଚାପର କେନ୍ଦ୍ରସ୍ଥ କୋଣ ଅଟେ ।
(v) ଦୁଇଟି ଚାପର ଡିଗ୍ରୀ ପରିମାପର ସମଷ୍ଟି 360°ରୁ ଅଧ‌ିକ ହୋଇ ପାରିବ ନାହିଁ ।
(vi) ବୃତ୍ତ ଏକ ଉତ୍ତଳ ସେଟ୍ ନୁହେଁ ।
(vii) ଗୋଟିଏ ବୃତ୍ତରେ ଦୁଇଟି ଚାପର ଗୋଟିଏ ସାଧାରଣ ପ୍ରାନ୍ତ ବିନ୍ଦୁ ଥିଲେ ଚାପ ଦୁଇଟି ସନ୍ନିହିତ ଚାପ ହେବେ ।
(viii) ଦୁଇଟି ସର୍ବସମ ଜ୍ୟା ସହ ସମ୍ପୃକ୍ତ ଚାପଦ୍ଵୟ ସନ୍ନିହିତ ଚାପ ହେଲେ ଚାପଦ୍ଵୟର ସଂଯୋଗରେ ସର୍ବଦା ବୃହତ୍ ଚାପ ଗଠିତ ହେବ ।
(ix) ଦୁଇଟି ସର୍ବସମ ଜ୍ୟା ପରସ୍ପରକୁ ଲମ୍ବ ଭାବରେ ବୃତ୍ତର ଏକ ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ P ରେ ଛେଦ କରନ୍ତି । ବୃତ୍ତର କେନ୍ଦ୍ର O ଠାରୁ ସେମାନଙ୍କ ପ୍ରତି \(\overline{\mathrm{OQ}}\), \(\overline{\mathrm{OR}}\) ଲମ୍ବ ଗଠନ କରାଯାଇଛି । ତେବେ ଠ, Q, P ଓ R ଏକ ବର୍ଗଚିତ୍ରର ଶୀର୍ଷବିନ୍ଦୁ ହେବେ ।
(x) \(\overparen{B P C}\) ର ଡିଗ୍ରୀ ପରିମାପ 30° । A ବୃତ୍ତ ଉପରିସ୍ଥ ଏକ ବିନ୍ଦୁ ହେଲେ △ABC ରେ ∠Aର ପରିମାଣ ସର୍ବଦା 15° 6ଦ୍ଵଦା
(xi) ଗୋଟିଏ ଚାପ ଅସଂଖ୍ୟ ବିନ୍ଦୁର ସମାହାର ଅଟେ ।
(xii) ବୃତ୍ତାନ୍ତର୍ଲିଖ ରମ୍ବସ୍ ଏକ ବର୍ଗଚିତ୍ର ।
Solution:
(i) T
(ii) T
(iii) T
(iv) F
(v) F
(vi) T
(vii) T
(viii) F
(ix) F
(x) F
(xi) T
(xii) T

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(b)

Question 2.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(i) ଏକ ବୃହତ୍ ଚାପର ଡିଗ୍ରୀ ପରିମାପ …………….. ରୁ ବେଶୀ ।
(ii) ଗୋଟିଏ ସୁଷମ୍ ଷଡ଼ଭୁଜର ପ୍ରତ୍ୟେକ ବାହୁ ଏହାର ପରିବୃତ୍ତର କେନ୍ଦ୍ରଠାରେ ଉତ୍ପନ୍ନ କରୁଥିବା କେନ୍ଦ୍ରସ୍ଥ କୋଣର ପରିମାଣ ……………….. |
(iii)
(iv) ଗୋଟିଏ ବୃତ୍ତରେ ଦୁଇଟି ସର୍ବସମ ଜ୍ଯା \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{CD}}\) ପରସ୍ପରକୁ ବୃତ୍ତର ଏକ ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ Pରେ ଛେଦ କରନ୍ତି । O ବୃତ୍ତର କେନ୍ଦ୍ର ଏବଂ B ଓ C \(\overline{\mathrm{OP}}\)ର ଏକ ପାର୍ଶ୍ଵରେ ଥିଲେ \(\overparen{A D}\) ଓ ……………….. ଦୁହେଁ ସର୍ବସମ ।
(v) ଗୋଟିଏ ବୃତ୍ତରେ ଏକ ଜ୍ୟାର ଦୈର୍ଘ୍ୟ ବ୍ୟାସାର୍ଦ୍ଧ ସହ ସମାନ ହେଲେ ଉକ୍ତ ଜ୍ୟା ଦ୍ଵାରା ଛେଦିତ କ୍ଷୁଦ୍ର ଚାପର ତିଗ୍ରା ପରିମାପ ………….. |
(vi) \(\overline{\mathrm{AB}}\) ର ଏକ ପାଣ୍ଡରେ C ଓ D ଦୁଇଟି ଦିନ୍ଦୁ | m∠ACB = m∠ADB = 20° △ACD ର ପରିଦର୍ଭର 6କହ O 6ଦୃବେ m∠AOB ……………. |
(vii) m∠ABC = 90° ଚତୁର୍ଭୁକ △ABC ର ପରିଦଉଭେ AC ଏକ ………………… |
(viii) ABCD ଏକ ଦ୍ଵରାନ୍ତକଖର ଚତୁର୍ଭୁକ m∠BAD ………………. ଚାପର ଡିଗ୍ରୀ ପରିମାପର ଅର୍ଦ୍ଧେକ ।
(ix) ଏକ ଅର୍ଥବୃତ୍ତର ଡିଗ୍ରୀ ପରିମାପ ………….. |
(x) ଗୋଟିଏ ବୃତ୍ତରେ ଏକ ଚାପର ଡିଗ୍ରୀ ପରିମାପ 90° ହେଲେ, ସଂପୃକ୍ତ ଜ୍ୟା ଓ ବ୍ୟାସାର୍ଦ୍ଧର ଅନୁ ପାତ …………….. |
Solution:
(i) 180°
(ii) 60° \(\left(\frac{360^{\circ}}{6}=60^{\circ}\right)\)
(iii) 70° m∠D = 180° – 120° = 60°
m∠C = 180° – 50° = 130°
∴ m∠C – m∠D = 130° – 60° = 70°

(iv) \(\overparen{B C}\) \(\overparen{A C B}\) ≅ \(\overparen{C A D}\) ( ∵ AB ≅ CD (ଦତ୍ତ))
⇒ BC ≅ AD

(v) 60° ଜ୍ୟାର ପ୍ରାନ୍ତବିନ୍ଦୁଦ୍ଵୟ ଓ କେନ୍ଦ୍ରବିନ୍ଦୁ ଏକ ସମବାହୁ ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁ ହେବେ |
(vi) 40° O ବିଦୁଟି ବୃତ୍ତର କେନ୍ଦ୍ର ହେବ ।
(vii) ବ୍ୟାସ (ଅର୍ଦ୍ଧବୃତ୍ତଖଣ୍ଡସ୍ଥ କୋଣ ସମକୋଣ ।)
(viii) \(\overparen{B C D}\)
(ix) 90°
(x) √ 2 : 1 (ସଂପୃକ୍ତ ଜ୍ୟାଟି ବୃତ୍ତର ବ୍ୟାସ)

Question 3.
ଚିତ୍ରରେ △ABC ବୃତ୍ତାନ୍ତର୍ଲିଖ ଏବଂ ସୂକ୍ଷ୍ମକୋଣୀ । D, E, F ବୃତ୍ତ ଉପରିସ୍ଥ ତିନୋଟି ବିନ୍ଦୁ ହେଲେ ନିମ୍ନ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 1
(i) ∠B କେଉଁ ଚାପର ଅନ୍ତର୍ଲିଖ୍ ?
(ii) ∠B ଦ୍ବାରା କେଉଁ ଚାପ ଛେଦିତ ?
(iii) \(\overline{\mathrm{AB}}\) ଜ୍ୟା ଦ୍ବାରା ଛେଦିତ କ୍ଷୁଦ୍ରଚାପ ଓ ବୃହତ୍ ଚାପ କିଏ ?
(iv) ∠A ର ପରିମାଣ କେଉଁ କେନ୍ଦ୍ରସ୍ଥ କୋଣ ପରିମାଣର ଅର୍ଦ୍ଧେକ ?
(v) △ABC ରେ ଯଦି AB = BC ହୁଏ ତେବେ କେଉଁ ଚାପ ଦ୍ଵୟ ସର୍ବସମ ହେବେ ?
(vi) ଦୁଇଟି ସନ୍ନିହିତ ଚାପର ନାମ ଲେଖ ଯେପରିକି ସେମାନଙ୍କ ସଂଯୋଗରେ \(\overparen{B A D}\) ଗଠିତ ହେବ ।
(vii) \(\overparen{B F C}\) ଉପରେ ଏପରି ଏକ ବିନ୍ଦୁ P ନିଅ ଯେପରିକି m∠BPA = m∠C । ଏପରି କେତୋଟି ବିନ୍ଦୁ ଅଛି ? ADC ଉପରେ ଏପରି କୌଣସି ବିନ୍ଦୁ ଅଛି କି ? \(\overparen{B E A}\) ଉପରେ ଏପରି କୌଣସି ବିନ୍ଦୁ ଅଛି କି ?
Solution::
(i) ∠B, \(\overparen{E B F}\) ର୍ଥିଥଚା \(\overparen{A B C}\) ର ଅନ୍ତ୍ର କିଣତ |
(ii) ∠Bଦ୍ଵାରା \(\overparen{A D C}\) ଛେଦିତ ।
(iii) \(\overline{\mathrm{BC}}\) ଖ୍ୟା ଦ୍ଵାରା ଛେଦିତ ସୁଦୃତାପ \(\overparen{B F C}\) ଦ୍ଦଦ୍ର ତ୍ତାପ \(\overparen{B A C}\) |
(iv) m∠A = \(\frac { 1 }{ 2 }\) m∠BOC
(v) △ABC ରେ \(\overparen{A E B}\) ≅ \(\overparen{B F C}\)
(vi) \(\overparen{B E A}\) ∪ \(\overparen{A D}\) = \(\overparen{B A D}\), \(\overparen{B E}\) ∪ \(\overparen{E A D}\) = \(\overparen{B A D}\)
(vii) ଏପରି ଅସଂଖ୍ୟ ବିନ୍ଦୁ ଅଛି । \(\overparen{A D C}\) ଉପରେ ମଧ୍ୟ ଅସଂଖ୍ୟ ବିନ୍ଦୁ ଅଛି । \(\overparen{B E A}\) ଚାପ ଉପରେ ଏପରି ବିନ୍ଦୁ ରହିବା ସମ୍ଭବ ନୁହେଁ ।

Question 4.
ଚିତ୍ରରେ ABCD ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖ୍ ଚତୁର୍ଭୁଜ ଯାହର କର୍ଣ୍ଣଦ୍ଵୟ ବୃତ୍ତର କେନ୍ଦ୍ରଠାରେ ଛେଦ କରନ୍ତି ।
m\(\overparen{A E B}\) = 100° 6 ଦ୍ଵ6କ |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 2
(i) ଚତୁର୍ଭୁଜର ସମସ୍ତ କୋଣ ପରିମାଣ ନିର୍ଣ୍ଣୟ କର ।
(ii) \(\overparen{A H D}\) ଓ \(\overparen{B F C}\) ମଧ୍ୟରେ କି ସମ୍ପର୍କ ଦେଖୁଛ ?
(iii) ABCD କି ପ୍ରକାର ଚତୁର୍ଭୁଜ ?
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 3
m \(\overparen{A E B}\) : 100°
⇒ m∠ADB = m∠ACB = 50°
m∠COD = 100° (ତ୍ପତାପ 6କାଶ)
⇒ m∠CBD = m∠CAD = 50°
m∠BOC = 180° – 100° = 80°
⇒ m∠BAC = m∠BDC = 40°
⇒ m∠AOD = 80°
⇒ m∠ACD = m∠ABC = 40°

(i) m∠A = m∠BAC + m∠CAD = 40° + 50° = 90°
6ସ ଦ୍ଵିପରି m∠С = 90°, m∠B = 90° ଓ m∠D = 90° |
(ii) \(\overparen{A H D}\) ≅ △\(\overparen{B F C}\)
(ii) ABCD ଏକ ଆୟତଚିତ୍ର ।

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(b)

Question 5.
ଚିତ୍ରରେ \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{CD}}\) ଜ୍ୟା ଦ୍ଵୟ ପରସ୍ପରକୁ ବୃତ୍ତର ଏକ ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ P ଠାରେ ଛେଦ କରନ୍ତି । m∠PBD = 80°, m∠CAP = 45° 6ଦ୍ଵଲେ :
(i) △BPDର କୋଣ ପରିମାଣଗୁଡ଼ିକ ନିର୍ଣ୍ଣୟ କର ।
(ii) △APCର କୋଣ ପରିମାଣଗୁଡ଼ିକ ନିର୍ଣ୍ଣୟ କର ।
(iii) △APC ଓ △BPD ମଧ୍ୟରେ କି ସମ୍ପର୍କ ଦେଖୁଛ ?
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 4
Solution:
m∠PBD = 80°, m∠CAP = 45°

(i) \(\overparen{B C}\) ଉପରିସ୍ଥ m∠CDB = m∠CAP = 45°
∴ m∠BPD = 180° – 80° – 45° = 55°
∴ △BPDର m∠BPD = 55°
m∠PDB = 45°, m∠PBD = 80°
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 5
(ii)
△APCର m∠CAP = 45° (ଦଉ)
m∠APC = m∠BPD = 55° (ପ୍ରତାପ 6କାଣ)
∴ m∠ACP = m∠PBD = 80° (\(\overparen{A D}\) ଉପରିସ୍ଥ ପରିଧ୍ଵସ୍ଥ ଏକ କୋଣ)

(iii) △APC ~ △BPD (କୋ . କୋ . କୋ . ପାଦଣ୍ୟ)

Question 6.
△ABCରେ ∠Aର ସମଦ୍ବିଖଣ୍ଡକ ତ୍ରିଭୁଜର ପରିବୃତ୍ତକୁ D ବିନ୍ଦୁରେ ଛେଦ କଲେ ପ୍ରମାଣ କର ଯେ, △BDC ସମଦ୍ବିବାହୁ ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 6
ଦତ୍ତ : △ABCର ∠Aର ସମଦ୍ଵିଖଣ୍ଡକ ତ୍ରିଭୁଜର ପରିବୃତ୍ତକୁ D ବିନ୍ଦୁରେ ଛେଦକରେ ।
ପ୍ରାମାଣ୍ୟ : △BDC ସମଦ୍ବିବାହୁ ।
ପ୍ରମାଣ : ∠BAD ≅ ∠CAD (ଦଉ)
⇒ \(\overparen{B D}\) ≅ \(\overparen{D C}\) ⇒ \(\overline{\mathrm{BD}}\) ≅ \(\overline{\mathrm{DC}}\) (ଚାପ ସର୍ବସମ ହେତୁ ଜ୍ୟାଦ୍ଵୟ ସର୍ବସମ)
=> △BDC ସମଦିବାହୁ ।

Question 7.
ଚିତ୍ରରେ ଗୋଟିଏ ବୃତ୍ତର ଏକ ବହିଃସ୍ଥ ବିନ୍ଦୁ A ଠାରୁ \(\overrightarrow{\mathbf{AP}}\) ଓ \(\overrightarrow{\mathbf{AR}}\) ରଶ୍ମିଦ୍ଵୟ ବୃତ୍ତକୁ ଯଥାକ୍ରମେ P, Q ଏବଂ R, S ଠାରେ ଛେଦ କରନ୍ତି ଯେପରି A-P-Q ଏବଂ A-R-S |
(a) ପ୍ରମାଣ କର ଯେ △APR ~ △AQS
(b) ପ୍ରମାଣ କର ଯେ △APS ~ △ARQ
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 7
(C) ଯଦି \(\overline{\mathrm{PS}}\) ଓ \(\overline{\mathrm{QR}}\) ର ଛେନ୍ଦ୍ରବିନ୍ଦୁ T ହୁଏ, ତେଦେ
(i) ପ୍ରମାଣ କର ଯେ TP • TS = TR • TQ
(ii) ପ୍ତମାଣ କାର 6ପ m∠PTR = \(\frac { 1 }{ 2 }\) (m\(\overparen{Q S}\) + m\(\overparen{P R C}\))

(d) m∠PAR = 15° ଏବଂ m\(\overparen{Q X S}\) = 50° ହେଲେ m∠PTR ନିର୍ଣ୍ଣୟ କର ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 8
(a) ପ୍ରମାଣ : PRSQ ଏକ ଦ୍ଵରାନୁଲଖତ ଚତୁରୁଜ |
⇒ m∠RSQ + m∠RPQ = 180°
କିନ୍ନ m∠RPQ + m∠APR = 180°
⇒ m∠RSQ + m∠RPQ = m∠RPQ + m∠APR
m∠RSQ = m∠APR
△APR ଓ △AQS ମଧ୍ୟ6ର m∠RAP = m∠QAS
ଓ m∠RSQ = m∠APR |
⇒ △APR ~ △AQS (6କା-6କା ସାଦ୍ୱଣ)

(b) △APS ଓ △ARQ ମଧ୍ୟ6ର
m∠PAS = m∠RAQ (ମଧ୍ୟ6ର 6କା)
m∠ASP = m∠AQR (ଏକ ଚାପ ଉପରିମ ପରିଧମ 6କାଣ)
⇒ △APS ~ △ARQ

(c) (i) △ TPQ ଓ △TRS ମଧ୍ୟ6ର
m∠TPQ = m∠TRS (ଏକ ଚାପ ଉପରିମ ପରିଧମ 6କାଣ)
m∠PTQ = m∠RTS (ପ୍ତତାପ 6କାଣ)
⇒ △TPQ ~ △TRS (6କା-6କା ସାଦ୍ୱଣ)
⇒ \(\frac { TP }{ TR }\) = \(\frac { TQ }{ TS }\) ⇒ TP • TS = TR • TQ
=m∠SPQ + m∠PQR = \(\frac { 1 }{ 2 }\)m\(\overparen{Q S}\) + \(\frac { 1 }{ 2 }\)m\(\overparen{P R}\) = \(\frac { 1 }{ 2 }\) (m\(\overparen{Q S}\) + m\(\overparen{P R}\)) (9flêe)

(d) m∠PAR = 15°
m\(\overparen{Q X S}\) = 50° ⇒ m∠QPS = \(\frac { 50° }{ 2 }\) = 25°
△APS 6ର 6କାଣ m∠QPS = m∠PAR + m∠PSR
⇒ 25° = 15° +m∠PSR ⇒ m∠PSR = 25° – 15° = 10°
m∠QRS = m∠QPS = 25°
∴ m∠PTR = m∠QRS + m∠TSR = 25° + 10° = 35°

Question 8.
ଚିତ୍ରରେ ABC ଦଉର \(\overparen{A X B}\) ଓ \(\overparen{B Y C}\) ହୁକରି ଚାପର ଗିଗାମାପ ଯଥାକୃମେ 80° ଓ 140° |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 9
(i) m∠BAC କିଣ୍ଡଯ କର |
(ii) m\(\overparen{A B C}\) କିଣ୍ଡଯ କର |
(iii) m\(\overparen{A C B}\) କିଣ୍ଡଯ କର |
(iv) \(\overparen{A Z C}\) ଓ \(\overparen{B Y C}\) ମଧ୍ୟରେ କି ସମଳ ଅଛି ?
Solution:
m\(\overparen{A X B}\) + m\(\overparen{B Y C}\) + m\(\overparen{A Z C}\) = 360° ⇒ 80° + 140° + m\(\overparen{A Z C}\) = 360°
⇒ m\(\overparen{A Z C}\) = 360° – 80° – 140° = 140°
(i) M∠BAC = \(\frac { 1 }{ 2 }\) m\(\overparen{B Y C}\) = \(\frac { 1 }{ 2 }\) × 140° = 70°
(ii) \(\overparen{A B C}\) = 360° – 140° = 220°
(iii) \(\overparen{A C B}\) = M\(\overparen{A Z C}\) + m\(\overparen{B Y C}\) = 140° + 140° = 280°
(iv) \(\overparen{A Z C}\) ≅ \(\overparen{B Y C}\) (“: m∠AOC = m∠BOC)

Question 9.
ଏକ ବୃତ୍ତର କେନ୍ଦ୍ର O ଏବଂ \(\overline{\mathrm{AB}}\) ଏକ ବ୍ୟାସ । ବୃତ୍ତ ଉପରିସ୍ଥ P ଓ ଠୁ ବିଦୁ୍ୟଦ୍ୱୟ \(\overline{\mathrm{AB}}\) ର ଏକ ପାର୍ଶ୍ଵରେ ଅବସ୍ଥିତ । ଯଦି A ଓ P ପ୍ରାନ୍ତ ବିନ୍ଦୁ ବିଶିଷ୍ଟ ଚାପର ଡିଗ୍ରୀ ପରିମାପ 60° ଏବଂ B ଓ ( ପ୍ରାନ୍ତବିନ୍ଦୁ ବିଶିଷ୍ଟ ଚାପର ଡିଗ୍ରୀ ପରିମାପ 50° ହୁଏ ତେବେ–
(i) A ଓ Q ପ୍ରାନ୍ତବିନ୍ଦୁ ବିଶିଷ୍ଟ କ୍ଷୁଦ୍ରଚାପର ଡିଗ୍ରୀ
(ii) P ଓ B ପ୍ରାନ୍ତବିନ୍ଦୁ ବିଶିଷ୍ଟ ବୃହତ୍ ଚାପର ଡିଗ୍ରୀ
(iii) P ଓ Q ପ୍ରାନ୍ତବିନ୍ଦୁ ବିଶିଷ୍ଟ ବୃହତ୍ ଚାପର ଡିଗ୍ରୀ ପରିମାପ ନିର୍ଣ୍ଣୟ କର ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 10
Solution:
m∠AOP = 60°
m∠BOQ = 50°
⇒ m⇒POQ = 180° – (60° + 50°) = 70°
(i) m\(\overparen{A Q}\) = 60° + 70° = 130°
(ii) m\(\overparen{P B}\) = 70° + 50° = 120°
(iii) m\(\overparen{P Q}\) = 70°

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(b)

Question 10.
\(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{CD}}\) ଦୁଇଟି ସମାନ୍ତର ଜ୍ୟା । ପ୍ରମାଣ କର ଯେ,
(i) M\(\overparen{A X C}\) = m\(\overparen{B Y D}\), (ii) AC = BD |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 11
Solution:
ଦତ୍ତ : ABCD 96 \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{AB}}\) |
ପ୍ରାମାଣ୍ୟ :
(i) M\(\overparen{A X C}\) = m\(\overparen{B Y D}\)
(ii) AC = BD
ଅଙ୍କନ : \(\overline{\mathrm{BC}}\) ଅଙ୍କନ କର ।
ତ୍ପମାଣ : \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{AB}}\) (ଦଉ)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 12
⇒ m∠ABC = m∠BCD
⇒ \(\frac { 1 }{ 2 }\) m\(\overparen{A X C}\) = \(\frac { 1 }{ 2 }\) m\(\overparen{B Y D}\)
⇒ m\(\overparen{A X C}\) = m\(\overparen{B Y D}\)

(ii) \(\overparen{A X C}\) = \(\overparen{B Y D}\) ⇒ AC = BD (ଚାପ ସବସମ ହେତ୍ ଲ୍ୟା ସଦସ୍ୟ)

Question 11.
ABCD ଏକ ଦ୍ଵରୀନ୍ତ୍ର କିଖବ ତତ୍କଲକ |
(i) AC = BD ଏବଂ \(\overline{\mathbf{AB}}\)||\(\overline{\mathbf{CD}}\) 6ଦ୍ର6କ ପ୍ରମାଣ କର ସେ, AD = BC |
(ii) AD = BC 6ଦ୍ର6କ ପ୍ରମାଣ କର ସେ, AC = BD ଏବଂ \(\overline{\mathbf{AB}}\)||\(\overline{\mathbf{CD}}\) |
Solution:
(i) ଦର : ABCD ଦୃଭାନ୍ତ୍ରରଖତ ଚତୁରୁକରେ
AC = BD ଏବଂ \(\overline{\mathbf{AB}}\)||\(\overline{\mathbf{CD}}\) |
ପ୍ରାମଣ୍ୟ: AD = BC
ପ୍ରମାଣ : AC = BD (ଦତ୍ତ)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 13
⇒ \(\overparen{A D C}\) ≅ \(\overparen{B C D}\)
⇒ l \(\overparen{A X D}\) + l \(\overparen{D Y C}\) = l \(\overparen{D Y C}\) + l \(\overparen{B Z C}\)
⇒ l \(\overparen{A X D}\) = l \(\overparen{B Z C}\) ⇒ \(\overparen{A X D}\) ≅ \(\overparen{B Z C}\)
⇒ AD = BC

(ii) ଦଉ : ABCD ଏକ ତ୍ରଭାନ୍ତକଖତ ତଡୁରୁକ | AD = BC
ପ୍ତମାଣ୍ୟ: (i) AC = BD (ii) \(\overline{\mathbf{AB}}\)||\(\overline{\mathbf{CD}}\)
ପ୍ରମଣ : AD = BC ⇒ \(\overparen{A X D}\) ≅ \(\overparen{B Z C}\)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 14
⇒ l \(\overparen{A X D}\) = l \(\overparen{B Z C}\)
⇒ l \(\overparen{A X D}\) + l \(\overparen{D Y C}\) = l \(\overparen{D Y C}\) + l \(\overparen{B Z C}\)
⇒ l \(\overparen{A D C}\) = l \(\overparen{B C D}\) ⇒ \(\overparen{A D C}\) ≅ \(\overparen{B C D}\) ⇒ AC = BD (i)
ପୁକଣ୍ଠ ∵ \(\overparen{A X D}\) ≅ \(\overparen{B Z C}\)
⇒ m \(\overparen{A X D}\) + m \(\overparen{B Z C}\) ⇒ \(\frac { 1 }{ 2 }\) m \(\overparen{A X D}\) = \(\frac { 1 }{ 2 }\) m \(\overparen{B Z C}\)
⇒ m∠ABD = m∠BDC (ଏକାନ୍ତ୍ରର) ⇒ \(\overline{\mathbf{AB}}\)||\(\overline{\mathbf{CD}}\) …(ii)

Question 12.
(i) ଗୋଟିଏ ବୃତ୍ତରେ \(\overparen{A X B}\) ଏକ ଚାପ । ପ୍ରମାଣ କର ଯେ \(\overparen{A X B}\)ର ଅନ୍ତଃସ୍ଥ ଗୋଟିଏ ଏବଂ କେବଳ ଗୋଟିଏ ବିନ୍ଦୁ C ଅଛି ଯେପରି \(\overparen{A C}\) ଓ \(\overparen{B C}\) ଚାପଦ୍ଵୟ ସର୍ବସମ ହେବେ । (C ବିନ୍ଦୁକୁ \(\overparen{A X B}\) ର ମଧ୍ୟବିନ୍ଦୁ କୁହାଯାଏ ।)
(ii) ଚାପର ମଧ୍ୟବିନ୍ଦୁ ଧାରଣାକୁ ବ୍ୟବହାର କରି ପ୍ରମାଣ କର ଯେ, AXBରେ ଅସଂଖ୍ୟ ବିନ୍ଦୁ ଅଛି ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 15
Solution:
(i) ଦତ୍ତ : O ବୃତ୍ତର କେନ୍ଦ୍ର । \(\overparen{A X B}\) ଏକ ଚାପ ।
ପ୍ରାମାଣ୍ୟ : (i) \(\overparen{A X B}\) ର ଅନ୍ତଃସ୍ଥ ଗୋଟିଏ
ଏବଂ କେବଳ ଗୋଟିଏ ବିନ୍ଦୁ C ଅଛି,
ଯେପରି \(\overparen{A C}\) = \(\overparen{C B}\)ହେବ ।
(ii) \(\overparen{A X B}\) ରେ ଅସଂଖ୍ୟ ବିନ୍ଦୁ ରହିଅଛି ।
∠AOB ର ସମଦ୍ଵିଖଣ୍ଡକ \(\overrightarrow{\mathrm{OC}}\) ଅଙ୍କନ କର ।
ଅଙ୍କନ :
ପ୍ରମାଣ :
(i) \(\overparen{A X B}\) ଚାପ ଉପରେ ଅବସ୍ଥିତ ଆବଶ୍ୟକ C ବିନ୍ଦୁଟି ଅନନ୍ୟ ଅର୍ଥାତ୍ ଗୋଟିଏ ଓ କେବଳ ଗୋଟିଏ ବିନ୍ଦୁ ହେବ ଯାହା m∠AOB ର ସମଦ୍ବିଖଣ୍ଡକ ରଶ୍ମି \(\overrightarrow{\mathrm{OC}}\) ଉପରେ ଅବସ୍ଥିତ ହେବ । ପୁନଶ୍ଚ ଚାପର ସର୍ବସମତା ଅନୁସାରେ ଦୁଇଟି ଚାପର ଡିଗ୍ରୀ ପରିମାପ ସମାନ ହେଲେ ଚାପଦ୍ଵୟ ସର୍ବସମ ହେବେ ।
ଦର ଚିତ୍ର6ର m∠AOC = m∠BOC 6ଦ୍ରଦ \(\overparen{A C}\) = \(\overparen{C B}\) 6ଦ୍ରଦ | (ପ୍ରମାଣିତ)

(ii) ଦତ୍ତ ଚିତ୍ରରେ A ଓ B ବିନ୍ଦୁ ସମେତ ‘A’ ଠାରୁ ‘B’ ପର୍ଯ୍ୟନ୍ତ ବୃତ୍ତ ଉପରିସ୍ଥ ସମସ୍ତ ବିନ୍ଦୁ ମାନଙ୍କର ସେଟ୍‌କୁ ଏକ ଚାପ କୁହାଯାଏ । A ଓ B ଏହି ଚାପର ଦୁଇଟି ପ୍ରାନ୍ତବିନ୍ଦୁ ଅଟନ୍ତି । ପ୍ରାନ୍ତବିନ୍ଦୁ ଭିନ୍ନ ଚାପ ଉପରିସ୍ଥ ଅନ୍ୟ ସମସ୍ତ ବିନ୍ଦୁମାନଙ୍କୁ ଚାପର ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ କୁହାଯାଏ; ଯାହା ଅସଂଖ୍ୟ ବିନ୍ଦୁମାନଙ୍କର ସେଟ୍ ।(ପ୍ରମାଣିତ)

Question 13.
ଚିତ୍ରରେ AB ବୃତ୍ତର ଏକ ବ୍ୟାସ ଏବଂ O କେନ୍ଦ୍ର । OD ଯେକୌଣସି ଏକ ବ୍ୟାସାର୍ଦ୍ଧ | \(\overline{\mathbf{AC}}\)||\(\overline{\mathbf{OD}}\) ହେଲେ, ପ୍ରମାଣ କର ଯେ, \(\overparen{B X D}\) ଓ \(\overparen{D Y C}\) ସର୍ବସମ ଅର୍ଥାତ୍ D, \(\overparen{B D C}\)ର ମଧ୍ୟବିନ୍ଦୁ । (ସୂଚନା : \(\overline{\mathbf{OC}}\) ଅଙ୍କନ କରି ଦର୍ଶାଅ ଯେ, m∠BOD = m∠DOC)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 16
Solution:
ଦତ୍ତ : ବୃତ୍ତର କେନ୍ଦ୍ର O | \(\overline{\mathbf{AB}}\) ବୃତ୍ତର ବ୍ୟାସ । \(\overline{\mathbf{AC}}\)||\(\overline{\mathbf{OD}}\) |
ପ୍ରାମାଣ୍ୟ : \(\overparen{B X D}\) ≅ \(\overparen{D Y C}\) ଅର୍ଥାତ୍ D, \(\overparen{B D C}\) ର ମଧ୍ୟବିନ୍ଦୁ ।
ଅଙ୍କନ : \(\overline{\mathbf{CO}}\) ଅଙ୍କନ କର ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 17
ପ୍ରମାଣ : △AOC ରେ OA = OC ⇒ m∠OAC = m∠OCA
କିନ୍ତୁ m∠OAC – m∠BOD (ଅନୁରୁପ)
∴ m∠OAC = m∠BOD (i)
ପୁନଶ୍ଚ m∠OCA=m∠COD …(ii)
(i) ଓ (ii) ରୁ m∠BOD = m∠COD ⇒ \(\overparen{B X D}\) ≅ \(\overparen{D Y C}\)
⇒ D, \(\overparen{B D C}\) ର ମଧ୍ୟବିନ୍ଦୁ ।

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(b)

Question 14.
ଚିତ୍ରରେ \(\overline{\mathbf{CD}}\) ଜ୍ୟା \(\overline{\mathbf{AB}}\) ବ୍ୟାସ ସହ ସମାନ୍ତର ଏବଂ CD = OB |
ପ୍ରମାଣ୍ୟ କର ଯେ m∠BDC = 2m∠OBD |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 18
Solution:
ଦତ୍ତ : ବୃତ୍ତର \(\overline{\mathbf{AB}}\) ବ୍ୟାସ । \(\overline{\mathbf{CD}}\) ଜ୍ୟା । CD = OB \(\overline{\mathbf{CD}}\) ର ଦୈର୍ଘ୍ୟ, ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ସହ ସମାନ ।
ପ୍ରାମାଣ୍ୟ : m∠BDC = 2m∠OBD
ଅଙ୍କନ : \(\overline{\mathbf{OC}}\) ଏବଂ \(\overline{\mathbf{OD}}\) ଅଙ୍କନ କର ।
ସ୍ତମାଣ : CD = OB = OC = OD ∴ △OCD ସମବାହୁ ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 32
⇒ m∠OCD = 60°
ପୁନଶ୍ଚ, COBD ଏକ ରମଣ (∵ CD = OB, \(\overline{\mathbf{CD}}\)||\(\overline{\mathbf{OB}}\) ଏବଂ OB = OC = CD)
∴ m∠OBD =m∠OCD = 60° ଏବଂ m∠BDC = 120°
∴ m∠BDC = 2m∠OBD (ପ୍ରମାଣିତ)

Question 15.
ABCD ବୃତ୍ତାନ୍ତର୍ଲିଖୂତ ଚତୁର୍ଭୁଜର \(\overline{\mathbf{AC}}\) ଓ \(\overline{\mathbf{BD}}\) କର୍ଣ୍ଣଦ୍ୱୟ ପରସ୍ପରକୁ P ଠାରେ ଛେଦ କରନ୍ତି । O ବୃତ୍ତର କେନ୍ଦ୍ର ଏବଂ B ଓ C, \(\overleftrightarrow{O P}\) ର ବିପରୀତ ପାର୍ଶ୍ଵରେ ଅବସ୍ଥିତ । ଯଦି AC = BD ହୁଏ, ତେବେ ପ୍ରମାଣ କର ଯେ,
(i) AB = CD, (ii) PA = PD 1° (iii) \(\overline{\mathbf{BC}}\) || \(\overline{\mathbf{AD}}\) |
Solution:
ଦତ୍ତ : ABCD ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖ୍ ଚତୁର୍ଭୁଜ । \(\overline{\mathbf{AC}}\) ଓ \(\overline{\mathbf{BD}}\) କର୍ଣ୍ଣଦ୍ଵୟ ପରସ୍ପରକୁ P ଠାରେ ଛେଦ କରନ୍ତି ।
AC = BD |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 20
ପ୍ରାମାଣ୍ୟ : (i) AB = CD (ii) PA = PD (iii) \(\overline{\mathbf{BC}}\) || \(\overline{\mathbf{AD}}\)
ପ୍ତମାଣ: (i) AC = BD ⇒ \(\overparen{A D C}\) ≅ \(\overparen{B A D}\)
⇒ \(\overparen{A Y B}\) ≅ \(\overparen{C Z D}\)
(∵ \(\overparen{A X D}\) ଭରଯ \(\overparen{A D C}\) ଓ \(\overparen{B A D}\) ର ପାଧାରଣ ଚାପ )
⇒ AB = CD ….(i)

(ii) △ABD ଏବଂ △ADC ଦଯ6ର \(\overline{\mathbf{AD}}\) ପାଧାରଣ, AB = CD [(i) ରେ ପ୍ତମାଣିତ]
ଏବଂ AC = BD (ଦର)
∴ m∠ADB = m∠CAD ⇒ m∠ADP = m∠PAD
⇒ PA = PD …(ii)

(iii) m∠DAC = m∠DBC (ଏକ ଦ୍ଵରଖଣ୍ଡମ 6କାଣ)
କିନ୍ତୁ m∠DAC = m∠ADB
∴ m∠ADB = m∠PBC
କିନ୍ତୁ ଏମା6ନ ଏକାନ୍ତ୍ରର |
∴ \(\overline{\mathbf{BC}}\) || \(\overline{\mathbf{AD}}\)

Question 16.
(i) ପ୍ରମାଣ କର ଯେ ପ୍ରମାଣ କରଯେ ର ଥନ୍ତ୍ର କିଣିତ 6କାଶ ଏକ ପୁର6କାଣ |
(ii) ପ୍ରମାଣ କର ଯେ ଏକ ବୃହତ୍ ଚାପର ଅନ୍ତର୍ଲିଖ କୋଣ ଏକ ସୂକ୍ଷ୍ମକୋଣ ।
(ସୂଚନା : \(\overparen{A P B}\) ଏକ କ୍ଷୁଦ୍ର ଚାପ ଓ \(\overparen{A Q B}\) ଏକ ବୃହତ୍ ଚାପ ହେଉ ।
\(\overline{\mathbf{AD}}\) ଦ୍ୟାସ ଅକନ କର | m∠APD = 90° m∠APB)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 21
Solution:
ଏବଂ m∠AQB < 90°
ଅଙ୍କନ : \(\overline{\mathbf{AR}}\) ବ୍ୟାସ ଅଙ୍କନ କରି । \(\overline{\mathbf{PR}}\) ଅଙ୍କନ କର ।
ପ୍ରମାଣ : (i) \(\overline{\mathbf{AR}}\) ବ୍ୟାସ । m∠APR ଅର୍ଥବୃତ୍ତଖଣ୍ଡସ୍ଥ କୋଣ । ∴ m∠APR = 90°
କିନ୍ତ୍ର R, ∠APB ର ଅନ୍ତ୍ ମ ହୋଇଥିବାରୁ m∠APR + m∠RPB = m∠APB
m∠APR = 90° ହେତୁ m∠APB > 90° ….. (i)
ଅର୍ଥାତ୍ ∠APB ଏକ ସ୍ଥୂଳକୋଣ ।

(iii) \(\overparen{A P B}\) କ୍ଷୁଦ୍ର ଚାପର ବିପରୀତ ଚାପ \(\overparen{A Q B}\) ଏକ ବୃହତ୍ ଚାପ ।
m∠APB+m∠AQB = 180° (·.· AQBP ଏକ ବୃହତ୍ ଚାପ ହେଉ )
(i) ରେ ପ୍ରମାଣିତ m∠APB > 90°
∴ m∠AQB < 90° ଅର୍ଥାତ୍ m∠AQB ଏକ ସୂକ୍ଷ୍ମକୋଣ । (ପ୍ରମାଣିତ)

Question 17.
(i) △ABCର ପରିବୃତ୍ତର କେନ୍ଦ୍ର ( ତ୍ରିଭୁଜଟିର ଏକ ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ ହେଲେ ପ୍ରମାଣ କର ଯେ,
m∠BAC + m∠OBC = 90° |
(ii) △ABCର ପରିବୃତ୍ତର କେନ୍ଦ୍ର O ତ୍ରିଭୁଜଟିର ଏକ ବହିଃସ୍ଥ ବିନ୍ଦୁ । O ଏବଂ A, \(\overline{\mathbf{BC}}\) ର ବିପରୀତ ପାଣମ 6ଦୃକେ, ଦେଲେ ପ୍ରମାଣ କର ଯେ, m∠BAC – m∠OBC = 90° |
Solution:
(i) ଦତ୍ତ : △ABCର ପରିବୃତ୍ତର କେନ୍ଦ୍ର ଠ, ତ୍ରିଭୁଜର ଏକ ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ ।
ପ୍ତାମାଣ୍ୟ: m∠BAC + m∠OBC = 90°
ଅଙ୍କନ : \(\overrightarrow{\mathrm{BO}}\) ବୃତ୍ତର ପରିଧ‌ିକୁ P ବିନ୍ଦୁରେ ଛେଦ କରୁ । \(\overline{\mathbf{PA}}\) ଅଙ୍କନ କର ।
ପ୍ରମାଣ : ABC ବୃତ୍ତର ବ୍ୟାସ \(\overline{\mathbf{BP}}\) |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 22
⇒ m∠BAP = 90° (ଅଦି ଦ୍ଵରଖଣ୍ଡମ କୋଣ)
⇒m∠BAC+m∠CAP = 90°
ମାତ୍ର m∠CAP = m∠PBC (ଏକ ବୃତ୍ତଖଣ୍ଡସ୍ଥ କୋଣ)
⇒ m∠PBC=m∠OBC
⇒ m∠BAC+m∠OBC = 90°

(ii) ଦତ୍ତ : △ABCର ପରିବୃତ୍ତର କେନ୍ଦ୍ର ଠ ତ୍ରିଭୁଜଟିର ଏକ ବହିଃସ୍ଥ ବିନ୍ଦୁ ।
ପ୍ତାମାଣ୍ୟ: m∠BAC – m∠OBC = 90°
ଅଙ୍କନ : \(\overrightarrow{\mathrm{BO}}\) ବୃତ୍ତର P ବିନ୍ଦୁରେ ଛେଦକରୁ । \(\overline{\mathbf{PA}}\) ଅଙ୍କନ କର ।
ପ୍ରମାଣ: ABC ଦ୍ଦଭର ବ୍ୟାସ \(\overline{\mathbf{BP}}\) |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 23
⇒ m∠BAP = 90° (ଅଦି ଦ୍ଵରଖଣ୍ଡମ କୋଣ)
⇒ m∠BAC – m∠CAP = 90°
⇒ m∠BAC – m∠CBP = 90°
(∵ m∠CAP – m∠CBP ଏକ ବୃତ୍ତଖଣ୍ଡସ୍ଥ କୋଣ)
⇒ m∠BAC – m∠OBC = 90°

Question 18.
ପ୍ରମାଣ କର ଯେ ଏକ ଟ୍ରାପିଜିୟମ୍‌ର ଅସମାନ୍ତର ବାହୁଦ୍ୱୟ ସର୍ବସମ ହେଲେ ଟ୍ରାପିଜିୟମ୍ ବୃତ୍ତାନ୍ତର୍ଲିଖୁତ ହେବ ।
Solution:
ଦତ୍ତ : ABCD ଏକ ଟ୍ରାପିଜିୟମ୍ । \(\overline{\mathrm{AD}}\) || \(\overline{\mathrm{BC}}\) ଏବଂ \(\overline{\mathrm{AB}}\) ≅ \(\overline{\mathrm{CD}}\) |
ପ୍ରାମାଣ୍ୟ : ABCD ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖ୍ ଚତୁର୍ଭୁଜ ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 24
ଅଙ୍କନ : \(\overline{\mathrm{DM}}\) || \(\overline{\mathrm{AB}}\) ଅଙ୍କନ କର ।.
\(\overline{\mathrm{DM}}\) , \(\overline{\mathrm{BC}}\) କୁ M ବିନ୍ଦୁରେ ଛେଦ କରୁ ।
ପ୍ରମାଣ : ADBM ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର ।
(: \(\overline{\mathrm{AD}}\) || \(\overline{\mathrm{BM}}\) ଏବଂ \(\overline{\mathrm{DM}}\) || \(\overline{\mathrm{AB}}\) )
⇒ \(\overline{\mathrm{AB}}\) ≅ \(\overline{\mathrm{DM}}\)
କିନ୍ତୁ ଦକ \(\overline{\mathrm{AB}}\) ≅ \(\overline{\mathrm{DC}}\)
∴ \(\overline{\mathrm{DM}}\) ≅ \(\overline{\mathrm{DC}}\) ⇒ m∠DMC = m∠DCM
କିନ୍ତୁ \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{DM}}\), \(\overline{\mathrm{BC}}\) ଛେଦକ । ⇒ m∠ABM = m∠DMC (ଅନୁରୂପ)
m∠ABM = m∠DCM ⇒ m∠ABC = m∠DCB
\(\overline{\mathrm{AD}}\) || \(\overline{\mathrm{BC}}\) ହେତ୍ର m∠DAB + m∠ABC = 180°
⇒ m∠DAB + m∠DCB = 180° (∵ m∠ABC = m∠DCB)
⇒ ABCD ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖ୍ ଚତୁର୍ଭୁଜ । (ପ୍ରମାଣିତ)

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(b)

Question 19.
ଦୁଇଟି ବୃତ୍ତ ପରସ୍ପରକୁ P ଓ () ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । P ବିନ୍ଦୁ ମଧ୍ୟଦେଇ ଏକ ସରଳରେଖା ବୃତ୍ତଦ୍ଵୟକୁ K ଓ L ବିନ୍ଦୁରେ ଛେଦ କରେ । ସେହିପରି Q ମଧ୍ୟଦେଇ ଏକ ସରଳରେଖା ବୃତ୍ତଦ୍ଵୟକୁ M ଓ N ବିନ୍ଦୁରେ ଛେଦ କରେ । K ଓ M \(\overline{\mathrm{PQ}}\) ର ଏକ ପାର୍ଶ୍ବରେ ଥିଲେ ପ୍ରମାଣ କର ଯେ, \(\overline{\mathrm{KM}}\) || \(\overline{\mathrm{LN}}\) |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 25
Solution:
ଦତ୍ତ : S1 ଓ S2 ବୃତ୍ତଦ୍ଵୟ ପରସ୍ପରକୁ P ଓ Q ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି ।
P ଓ Q ବିନ୍ଦୁ ମଧ୍ୟଦେଇ ଅଙ୍କିତ ସରଳରେଖା ବୃତ୍ତଦ୍ଵୟକୁ ଯଥାକ୍ରମେ
K, L ଏବଂ M, N ବିନ୍ଦୁରେ ଛେଦ କରୁଛନ୍ତି ।
ପ୍ରାମାଣ୍ୟ : \(\overline{\mathrm{KM}}\) || \(\overline{\mathrm{LN}}\)
ଅଙ୍କନ : \(\overline{\mathrm{PQ}}\) ଅଙ୍କନ କର ।
ପ୍ରମାଣ : m∠KMQ = m∠QPL (∵ ବୃତ୍ତାନ୍ତର୍ଲିଖୂତ ଚତୁର୍ଭୁଜର ବହିଃସ୍ଥ କୋଣର ପରିମାଣ, ଏହାର ବିପରୀତ ଅନ୍ତଃସ୍ଥ କୋଣର ପରିମାଣ ସହ ସମାନ ।)
କିନ୍ତୁ m∠QPL + m∠QNL = 180° (ଦୃଭାନ୍ତ୍ରକଖତ ଚତୁରୁକର ବିପର।ତ କୋଣ)
∴ m∠KMQ + m∠QNL = 180°; ମାତ୍ର ଏହି କୋଣଦ୍ଵୟ ଏକ ପାର୍ଶ୍ୱସ୍ଥ ଅନ୍ତଃସ୍ଥ କୋଣ ।
⇒ \(\overline{\mathrm{KM}}\) || \(\overline{\mathrm{LN}}\) (ପ୍ରମାଣିତ)

Question 20.
ABCD ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖ୍ ଚତୁର୍ଭୁଜରେ ∠B ଓ ∠Dର ସମତ୍ତିଖଣ୍ଡକ ଦ୍ଵୟ ପରସ୍ପରକୁ E ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । \(\stackrel{\longleftrightarrow}{\mathbf{D} E}\) ବୃତ୍ତକୁ F ବିନ୍ଦୁରେ ଛେଦ କଲେ ପ୍ରମାଣ କର ଯେ, \(\overline{\mathrm{BE}}\) ⊥ \(\overline{\mathrm{BF}}\) |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 26
Solution:
ଦତ୍ତ : ABCD ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖତ ଚତୁର୍ଭୁଜ । ∠B ଓ ∠D ର
ସମଦ୍ୱିଖଣ୍ଡକଦ୍ୱୟ ପରସ୍ପରକୁ E ବିନ୍ଦୁରେ ଛେଦ କରୁଛନ୍ତି ।
\(\overrightarrow{\mathrm{DE}}\) ବୃତ୍ତକୁ F ବିନ୍ଦୁରେ ଛେଦ କରୁଛି ।
ପ୍ରାମାଣ୍ୟ : \(\overline{\mathrm{BE}}\) ⊥ \(\overline{\mathrm{BF}}\)
ପ୍ରମାଣ : m∠ADC + m∠ABC = 180° (ABCD ବୃତ୍ତାନ୍ତର୍ଲିଖ ଚତୁର୍ଭୁଜ)
\(\frac { 1 }{ 2 }\) m∠ADC + \(\frac { 1 }{ 2 }\) m∠ABC = 90°
⇒ m∠CDF + m∠EBC = 90°
କିନ୍ତୁ m∠CDF = m∠CBF (ଏକ ବୃତ୍ତଖଣ୍ଡସ୍ଥ କୋଣ)
m∠CBF + m∠EBC = 90° ⇒ m∠ERF = 90°
⇒ \(\overline{\mathrm{BE}}\) ⊥ \(\overline{\mathrm{BF}}\) (ପ୍ରମାଣିତ)

Question 21.
△ABCର କୋଣମାନଙ୍କର ସମଦ୍ବିଖଣ୍ଡକମାନେ ତ୍ରିଭୁଜର ପରିବୃତ୍ତକୁ X, Y ଓ Z ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । ପ୍ରମାଣ କର ଯେ, △XYZର କୋଣମାନଙ୍କର ପରିମାଣ ଯଥାକ୍ରମେ 90° – \(\frac { 1 }{ 2 }\) m∠A, 90° – \(\frac { 1 }{ 2 }\) m∠B ଓ 90° – \(\frac { 1 }{ 2 }\) m∠C |
Solution:
ଦତ୍ତ : △ABC ଦୁଲାନ୍ତ୍ରଖତ ∠A, ∠B ଓ ∠C ର ସମଦ୍ଵିଖଣ୍ଡକ ବୃତ୍ତକୁ ଯଥାକ୍ରମେ X, Y ଏବଂ Z ବିନ୍ଦୁରେ ଛେଦ କରେ ।
ପ୍ରାମାଣ୍ୟ : m∠X = 90° \(\frac { 1 }{ 2 }\)m∠A, \(\frac { 1 }{ 2 }\) m∠B ଏଦ m∠Z = 90° – \(\frac { 1 }{ 2 }\) m∠C
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 27
ପ୍ରମାଣ : \(\overparen{A Z}\) ର ଦିପରାତ ଗାପାନୁଇଖତ m∠AXZ = m∠ACZ
ଏବଂ \(\overparen{A Y}\) ଚାପର ବିପରୀତ ଚାପାନ୍ତର୍ଲିଖ କୋଣ m∠AXY = m∠ABY
∴ m∠AXZ + m∠AXY = m∠ACZ = m∠ABY
⇒ m∠X = \(\frac { m∠C }{ 2 }\) + \(\frac { m∠B }{ 2 }\)
⇒ m∠X = 90° – \(\frac { m∠A }{ 2 }\) (∵ \(\frac { m∠A }{ 2 }\) + \(\frac { m∠B }{ 2 }\) + \(\frac { m∠C }{ 2 }\) = 90°)
ସେହିପରି ପ୍ରମାଣ କରାଯାଇ ପାରେ ଯେ,
m∠Y = 90° – \(\frac { m∠B }{ 2 }\) ଏବଂ m∠Z = 90° – \(\frac { m∠C }{ 2 }\) ହେବ ।

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(b)

Question 22.
△ABC ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖ ସମବାହୁ ତ୍ରିଭୁଜ । \(\overline{\mathbf{BC}}\) ଜ୍ୟା ସହ ସମ୍ପୃକ୍ତ କ୍ଷୁଦ୍ର ଚାପ ଉପରେ P ଏକ ବିନ୍ଦୁ । ପ୍ରମାଣ କର ଯେ PA = PB + PC । (ସୂଚନା : \(\overrightarrow{\mathbf{B P}}\) ଉପରେ D ନିଅ ଯେପରି PC = PD ହେବ । △BCD ଓ △ACP ର ତୁଳନା କର ।)
Solution:
ଦତ୍ତ : △ABC ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖ ସମବାହୁ ତ୍ରିଭୁଜ । \(\overline{\mathbf{BC}}\) ଜ୍ୟା ସହ ସଂପୃକ୍ତ କ୍ଷୁଦ୍ରଚାପ ଉପରେ P ଏକ ବିନ୍ଦୁ ।
ପ୍ରାମାଣ୍ୟ : PA = PB + PC
ଅଙ୍କନ : \(\overrightarrow{\mathbf{B P}}\) ଉପରେ D ଏକ ବିନ୍ଦୁ ନିଅ ଯେପରିକି PC = PD ହେବ । \(\overline{\mathbf{CD}}\) ଅଙ୍କନ କର ।
ପ୍ରମାଣ : △ABC ଏକ ସମବାହୁ ତ୍ରିଭୁଜ ।
m∠BAC = m∠CPD (ବୃତ୍ତାନ୍ତର୍ଲିଖ୍ ଚତୁର୍ଭୁଜର ବହିଃସ୍ଥ କୋଣ)
∴ m∠CPD = 60°
ସୁନଶ୍ଚ, △PCD ରେ PC = PD |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 28
∴ △PCD ଏକ ସମବାହୁ ତ୍ରିଭୁଜ । ⇒ PC = CD = PD
ବର୍ତ୍ତମାନ m ∠ACB = m∠PCD = 60°
⇒ m∠ACB+m∠BCP=m∠PCD+m∠BCP
⇒ m∠ACP=m∠BCD
△APC ଓ △BCD ଦୟରେ AC = BC, PC = CD
ଏକ m∠ACP=m∠BCD
∴ △ACP ≅ △BCD
⇒ AP = BD ⇒ AP = BP + PD ⇒ AP = BP + PC

Question 23.
△ABCରେ ∠Aର ସମଦ୍ବିଖଣ୍ଡକ △ABCର ପରିବୃତ୍ତକୁ P ବିନ୍ଦୁରେ ଛେଦ କରେ । P ବିନ୍ଦୁରୁ \(\overrightarrow{\mathbf{AB}}\) ଓ \(\overline{\mathbf{AC}}\) ପ୍ରତି ଅଙ୍କିତ ଲମ୍ବ ଦ୍ବୟର ପାଦବିନ୍ଦୁ ଯଥାକ୍ରମେ Q ଏବଂ R । ପ୍ରମାଣ କର ଯେ, AQ = AR = \(\frac { AB+AC }{ 2 }\) | (ସ୍ମତନା : ଦଶାଥ ଯେ △PBQ ≅ △PCR ⇒ BQ = CR )
Solution:
ଦତ୍ତ : △ABC ର ∠A ର ସମଦ୍ବିଖଣ୍ଡକ ତ୍ରିଭୁଜର ପରିବୃତ୍ତକୁ P ବିନ୍ଦୁରେ ଛେଦ କରୁଛି । P ବିନ୍ଦୁରୁ \(\overrightarrow{\mathbf{AB}}\) ଓ \(\overrightarrow{\mathbf{AC}}\) ପ୍ରତି ଅଙ୍କିତ ଲମ୍ବର ପାଦବିନ୍ଦୁ ଯଥାକ୍ରମେ Q ଏବଂ R । (ଏଠାରେ △ABCର AC > AB)
ପ୍ରାମାଣ୍ୟ : AQ = \(\frac { AB+AC }{ 2 }\) = AR
ଅଙ୍କନ : \(\overline{\mathbf{PB}}\) ଓ \(\overline{\mathbf{PC}}\) ଅଙ୍କନ କର ।
ପ୍ରମାଣ : ∠A ର ସମଦ୍ବିଖଣ୍ଡକ ପରିବୃତ୍ତକୁ P ବିନ୍ଦୁରେ ଛେଦକରେ ।
⇒ \(\overparen{B P}\) = \(\overparen{P C}\) ⇒ BP = PC
△BPQ ଏକ △CPR ଦଯରେ
BP = PC, m∠BQP = m∠CRP (= 90°)
ଏବଂ PQ = PR
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 29
(∵ କୋଣର ବାହୁମାନଙ୍କଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ବିନ୍ଦୁମାନ, କୋଣ ସମଦ୍ବିଖଣ୍ଡକ ଉପରେ ଅବସ୍ଥାନ କରିବେ ।)
∴ △BPQ ≅ △CPR ⇒ BQ = CR
ପୁନଶ୍ଚ, △AQP ଓ △APR ଦ୍ବୟରେ
PQ = PR, \(\overline{\mathbf{AP}}\) ସାଧାରଣ ଏବଂ M∠AQP = m∠ARP
∴ △AQP ⇒ △APR ⇒ AQ = AR

ଚଇଂଲାନ 2AQ = AQ + AQ = AQ + AR = AB + BQ + AC – CR
= AB + AC (∵ BQ = CR)
∴ AQ = \(\frac { AB + AC }{ 2 }\) ⇒ AR = \(\frac { AB + AC }{ 2 }\)
⇒ AQ = \(\frac { AB + AC }{ 2 }\) = AR

Question 24.
△ABCରେ ∠Aର ସମଦ୍ବିଖଣ୍ଡକ △ABCର ପରିବୃତ୍ତକୁ P ବିନ୍ଦୁରେ ଛେଦ କରେ । \(\overline{\mathbf{AP}}\) ଓ \(\overline{\mathbf{BC}}\)ର ଛେଦ ବିନ୍ଦୁ D ହେଲେ ପ୍ରମାଣ କର ଯେ △ABD ଓ △APC ସଦୃଶ ଅଟନ୍ତି । ସୁତରାଂ ଦର୍ଶାଅ ଯେ, AB • AC = BD • DC + AD2 |
(ପୁଚନା : △ABD ଓ △APC ପଦଣ ⇒ AB.AC = AD.AP, AD2 = AD (AP – PD))
Solution:
ଦତ୍ତ : △ABC ର ∠A ର ସମଦ୍ବିଖଣ୍ଡକ, ଏହାର ପରିବୃତ୍ତକୁ P ବିନ୍ଦୁରେ ଛେଦ କରେ । \(\overline{\mathbf{BC}}\) ଓ \(\overline{\mathbf{AP}}\) ର ଛେଦବିନ୍ଦୁ D |
ପ୍ରାମାଣ୍ୟ : (i) △ABD ~ △APC
(ii) AB AC = BD · DC + AD2
ପ୍ରମାଣ : △ABD ଓ △APC ଦ୍ବୟରେ
m∠ABD = m∠APC (ଏକ ବୃତ୍ତଖଣ୍ଡସ୍ଥ କୋଣ)
m∠BAD=m∠PAC ଅବଶିଷ m∠ADB = m∠ACP
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 30
∴ △ABD ~ △APC
⇒ \(\frac { AB }{ AP }\) = \(\frac { AD }{ AC }\) ⇒ AB . AC = AD . AP
⇒ AB . AC = AD (AD + DP)
= AD2 + AD . DP …..(i)
ପୁନଣ୍ଡ △ABD ~ △PDC
(∵m∠BAD = m∠DCP, m∠ADB = m∠PDC)
⇒ \(\frac { BD }{ DP }\) = \(\frac { AD }{ DC }\) ⇒ BD . DC = AD . DP
(i) ରେ ପ୍ତ6ଯାଗ କଲେ AB . AC = AD2 + BD . DC

Question 25.
(ଟଲେମୀଙ୍କ ଉପପାଦ୍ୟ) ABCD ଏକ ବୃତ୍ତାନ୍ତଲିଖତ ଚତୁର୍ଭୁଜ ହେଲେ ପ୍ରମାଣ କର ଯେ,AC · BD = AB · CD + BC · AD | ଗୁଣଫଳ, ଚତୁର୍ଭୁଜର ସମ୍ମୁଖୀନ ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟର ଗୁଣଫଳର ସମଷ୍ଟି ସଙ୍ଗେ ସମାନ ।)
(ସୂଚନା : ମନେକର m∠ADB > m∠BDC | E, AC ଉପରେ ଏପରି ଏକ ବିନ୍ଦୁ ହେଉ ଯେପରି m∠BDC = m∠ADE | ବର୍ତ୍ତମାନ △ADE ଏବଂ △BDC ସଦୃଶ ⇒ \(\frac { AE }{ BC }\) = \(\frac { AD }{ BD }\) ପୁନଶ୍ଚ △ADB ଏବଂ △EDC ସଦୃଶ ⇒ \(\frac { CD }{ BD }\) = \(\frac { EC }{ AB }\) | )
Solution:
ଦଭ : ABCD ଏକ ବୃତ୍ତାନ୍ତଲିଖତ ଚତୁର୍ଭୁଜ ହେଲେ |
ପ୍ରାମାଣ୍ୟ : AC . BD = AB . CD + BC . AD
ଅକନ : ମନେକର m∠ADB > m∠BDC |
\(\overline{\mathbf{AC}}\) ଉପରିସ୍ଥ E ଏପରି ଏକ ବିନ୍ଦୁ ନିଅ ।
ଯେପରିକି m∠ADE = m∠BDC ହେବ ।
ପ୍ରମାଣ : ବର୍ତ୍ତମାନ △ADE ଏବଂ △BDC ଦ୍ଵୟରେ
m∠ADE = m∠BDC ଏବଂ m∠DAE = m∠DBC
ଥଗଣିପୁ m∠AED = m∠BCD
BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 Img 31
∴ △ADE ~ △BDC
⇒\(\frac { AE }{ BC }\) = \(\frac { AD }{ BD }\) ⇒ AE . BD = AD . BC
ପୁନ୍ଦଣ, △ADB ଏବଂ △EDC ଦ୍ଵପ୍ରେଭେ
m∠ABD + m∠ECD = m∠ADB + m∠EDC)
(∵m∠ADE = m∠BDC ⇒ m∠ADE + m∠EDB = m∠BDC + m∠EDB)
∴ △ADB ~ △EDC
⇒\(\frac { BD }{ CD }\) = \(\frac { AB }{ EC }\) ⇒ EC . BD = AB . CD
(i) ଓ (ii) ରୁ AE . BD + EC. BD = AD. BC + AB. CD
⇒ BD (AE + EC) = AB. CD + BC. AD
⇒ BD. AC = AB. CD + BC. AD

BSE Odisha 8th Class English Solutions Test-1

Odisha State Board BSE Odisha 8th Class English Solutions Test-1 Textbook Exercise Questions and Answers.

BSE Odisha Class 8 English Solutions Test-1

Test -1

1. Your teacher gives you a dictation of five 3/4 lettered words. Write them.    [05]
(ତୁମ ଶିକ୍ଷକ ତୁମକୁ ୫ଟି ତିନି କିମ୍ବା ଚାରି ଅକ୍ଷରିଆ ଇଂରାଜୀ ଶବ୍ଦ ଡାକିବେ । ତୁମେ ଶୁଣି ଲେଖିବ ।)
huge, pant, drag, old, do.

2. Your teacher will read aloud the following lines. Listen to him/her and fill in the gaps.   [07]
(ତୁମ ଶିକ୍ଷକ ଏହି ଧାଡ଼ିଗୁଡ଼ିକ ପାଟି କରି ପଢ଼ିବେ । ତୁମେ ତାହା ସାବଧାନତା ସହ ଶୁଣି ଦ୍ବିତୀୟ ଥର ପଢ଼ିବା ପରେ ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କରିବ ।) (Questions with Answers)
“They _____________ into the king’s ___________. The ___________ man said to the king’s officer, Sir, I was ___________ to the town in my __________. This __________ wanted to __________ to the town market. He is ____________. So, I brought him to the ____________ on my horse. _________ he says that the horse is __________. Sir, ___________ help me to ___________ my horse from ____________.

Answer:
“They entered into the king’s court. The first man said to the king’s officer, Sir, I was rushing to the town in my horse. This man wanted to take my horse to the town market. He is a thief. So, I brought him to the court on my horse. Now he says that the horse is not mine. Sir, please help me to get my horse from the thief.

BSE Odisha 8th Class English Solutions Test-1

3. There is some relationship between spelling and pronunciation. Generally, there is, ie, ea, oo, ee, or ou, in the spelling of a word; this signals a long sound. And Odia speakers of English have problems with long sounds. They have a tendency to pronounce long sounds as short sounds. Given below are some words, underline which of them have long sounds.    [07]
(ସାଧାରଣତଃ ବନାନ ଏବଂ ଉଚ୍ଚାରଣ ମଧ୍ୟରେ କିଛିଟା ସମ୍ବନ୍ଧ ରହିଥାଏ । ସାଧାରଣତଃ ଗୋଟିଏ ଇଂରାଜୀ ଶବ୍ଦ ବନାନରେ je, ea, o୦, ce କିମ୍ବା ou ଦୀର୍ଘ ଭାବରେ ଉଚ୍ଚାରିତ ହୋଇଥା’ନ୍ତି ଏବଂ ଇଂରାଜୀ କହୁଥ‌ିବା ଓଡ଼ିଆ ବକ୍ତାମାନଙ୍କର ସମସ୍ୟା ହୋଇଥାଏ । ସେମାନେ ସାଧାରଣତଃ ଦୀର୍ଘ ଉଚ୍ଚାରଣକୁ ସୂକ୍ଷ୍ମ ଭାବରେ ଉଚ୍ଚାରଣ କରିଥା’ନ୍ତି । ତଳେ କେତେକ ଶବ୍ଦ ଦିଆଯାଇଛି । ସେଗୁଡ଼ିକରେ ଦୀର୍ଘ ଉଚ୍ଚାରିତ ଶବ୍ଦଗୁଡ଼ିକ ତଳେ ଗାର ଦିଅ ।
(Questions with Answers)

agree, market, reach, please, cover, thought, punish, village, need. speed, under, thief, steal, peace, deep, honey, seed, fields, spring, bean.

4. Write the following Odia names in English. (Teacher will give four names of persons in Odia.) (Questions with Answers)        [08]
____________________  _______________________
____________________  _______________________
____________________  _______________________
____________________  _______________________
Answer:
ମଧୁ — Madhu
ରମେଶ — Ramesh
ସଦାନନ୍ଦ — Sadananda
ବୀରେନ୍ଦ୍ର — Birendra

5. Write the following names of places in English. (Teacher will give four names of places in Odia.)     [08]
____________________  _______________________
____________________  _______________________
____________________  _______________________
____________________  _______________________
Answer:
କଟକ — Cuttack
ଭୁବନେଶ୍ୱର — Bhubaneswar
ରାଉରକେଲା — Rourkela
ସୁନ୍ଦରଗଡ଼ — Sundargarha

BSE Odisha 8th Class English Solutions Test-1

6. Match the words under ‘A’ with the words under ‘B’ —(who lives where).    [06]
Match the words under ‘A’ with the words under ‘B’
Answer:
Match the words under ‘A’ with the words under ‘B’. Answer

7. Read the following text and answer the questions. (ନିମ୍ନଲିଖ ଅନୁଚ୍ଛେଦଗୁଡ଼ିକ ପାଠ କରି ପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

1. Akbar was very fond of jewellery. He had hundreds of rings, rings in which diamonds and many other gems had been set. But all of the rings he had, he liked one of them the most. It was a large ring with a number of pearls and diamonds set in it. The ring was a present to Akbar from the Queen.
2. At Akbar’s palace, there were eight servants who looked after the Emperor’s clothes and jewellery. Every day one of these eight servants used to help Akbar get ready to go to the court. None other than these eight servants could enter the Emperor’s room.
3. One day the Emperor was getting ready to go to the court. He wanted to wear his favourite ring that day. He asked one of his servants to bring it. But the servants came back saying that he could not find the ring. Akbar ordered to search for the ring, but it could not be found.
4. Akbar was very angry. He felt that one of his servants had stolen the ring. He sent for Birbal. When Birbal came, he told him what had happened and asked him to find out the thief.

BSE Odisha 8th Class English Solutions Test-1

(a). Answer the following questions each in one complete sentence. [05 ]
(ଏହି ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଗୋଟିଏ ଗୋଟିଏ ପୂର୍ଣ୍ଣ ବାକ୍ୟରେ ପ୍ରକାଶ କର ।)

Question 1.
What was Akbar fond of?
Answer:
Akbar was fond of jewellery.

Question 2.
Who presented the ring to Akbar?
Answer:
The Queen presented the ring to Akbar.

Question 3.
Why was Akbar angry?
Answer:
Akbar was angry when he felt that one of his servants had stolen his most favourite ring.

Question 4.
Who did Akbar tell what had happened?
Answer:
Akbar told Birbal what had happened.

Question 5.
How many servants looked after Akbar’s clothes and jewellery?
Answer:
Eight servants looked after Akbar’s clothes and jewellery.

BSE Odisha 8th Class English Solutions Test-1

(b). From the text, write five sentences about Akbar.   [10]
(ଏହି ପାଠରୁ ପାଞ୍ଚୋଟି ବାକ୍ୟ ଆକବରଙ୍କ ସମ୍ପର୍କରେ ଲେଖ ।)
_______________________________________________
_______________________________________________
_______________________________________________
_______________________________________________
Answer:
Akbar was a great Emperor of India in the Mughal period. He was a good administrator. He was liked by all classes of people in India. He married a Hindu girl and made her Queen. He was fond of jewellery.

(c). Given below are some sentences. As per the text, the sentences are not in order. Order them by putting serial numbers in brackets provided against each sentence. (Questions with Answers)    [09]
(ନିମ୍ନରେ କେତେକ ବାକ୍ୟ ରହିଛି । ବହି ପାଠ ଅନୁଯାୟୀ ଏହା ଠିକ୍ କ୍ରମରେ ନାହିଁ । ସେମାନଙ୍କୁ ଠିକ୍ କ୍ରମରେ ଲେଖିବା ପାଇଁ ଡାହାଣ ପାଖରେ ଥ‌ିବା ଖାଲି ଘରେ କ୍ରମିକ ନମ୍ବରଗୁଡ଼ିକୁ ଲେଖ ।)

Akbar was very angry. [  ]
Akbar was fond of jewellery. [  ]
The queen presented the ring to Akbar. [  ]
The ring was not to be found. [  ]
Eight servants were in charge of the jewellery. [  ]
Akbar sent for Birbal. [  ]
Akbar ordered to search for the ring. [  ]

Answer:
Akbar was very angry. [ 5 ]
Akbar was fond of jewellery. [ 1 ]
The queen presented the ring to Akbar. [ 3 ]
The ring was not to be found. [ 4 ]
Eight servants were in charge of the jewellery. [ 2 ]
Akbar sent for Birbal. [ 6 ]
Akbar ordered to search for the ring. [ 7 ]

BSE Odisha 8th Class English Solutions Test-1

(d). See the use of the following four phrases in the text. The paragraph number is given against each phrase. Try to understand the meaning and use of the phrase from the context. Next, read the paragraph given and fill in the gaps with the right phrases. [8]
(ନିମ୍ନଲିଖ୍ ୪ଟି ଖଣ୍ଡବାକ୍ୟର ବ୍ୟବହାର ଲକ୍ଷ୍ୟ କର । ପ୍ରତ୍ୟେକ ଖଣ୍ଡବାକ୍ୟର ଡାହାଣରେ ଅନୁଚ୍ଛେଦର ସୂଚନା ଦିଆଯାଇଛି । ସେଗୁଡ଼ିକର ବ୍ୟବହାର ଏବଂ ଅର୍ଥ ତୁମ ବିଷୟ ମାଧ୍ୟମରେ ଜାଣିବାକୁ ଚେଷ୍ଟା କର । ଏହାପରେ ପ୍ରଦତ୍ତ ଅନୁଚ୍ଛେଦଟିକୁ ପଢ଼ି ଠିକ୍ ଭାବରେ ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକୁ ଯୋଗ କର ।)
(Question with Answer)

Find out (5), looked after (2), sent for (5), fond of (1)

Abdul had a pet baby donkey. He was very __________ the baby donkey. He _____________ the donkey very well. One day the baby donkey went somewhere. He ___________ his faithful servant Ali. He asked Ali to _____________ the baby donkey.
Answer:
Abdul had a pet baby donkey. He was very fond of the baby donkey. He looked after the donkey very well. One day the baby donkey went somewhere. He sent for his faithful servant Ali. He asked Ali to find out the baby donkey.

8. Read the following text and do the tasks that.
(ନିମ୍ନ ବିଷୟଟିକୁ ପାଠ କର ଏବଂ ପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

1. Long long ago, on the bank of the river Nagabali there was a small village named Hatibadi, and at the one end of this village was the chatasali, or village school, run by Ghana Rath, where many children from villages nearby came to study Ghana Ratha taught all the subjects himself, including Mathematics, Literature and Social Sciences.
(ବହୁଦିନ ତଳେ ନାଗାବଳୀ ନଦୀ କୂଳରେ ହାତୀବାଡ଼ି ବୋଲି ଏକ କ୍ଷୁଦ୍ର ଗ୍ରାମ ଥିଲା ଏବଂ ଏହି ଗ୍ରାମର ଶେଷମୁଣ୍ଡରେ ଗୋଟିଏ ଚାଟଶାଳୀ ବା ଗାଁ ସ୍କୁଲ ଘନରଥ ନାମକ ଏକ ବ୍ୟକ୍ତିଙ୍କଦ୍ୱାରା ଚାଲୁଥୁଲା, ଯେଉଁଠି ଆଖାପାଖ ଗାଁର ବହୁ ପିଲା ପଢ଼ିବାକୁ ଆସୁଥିଲେ । ଘନରଥ ନିଜେ ଗଣିତ, ସାହିତ୍ୟ ଓ ସାମାଜିକ ଶିକ୍ଷାସହ ସବୁ ବିଷୟ ଶିକ୍ଷାଦାନ କରୁଥିଲେ ।)

2. By the side of the Chatasali ran a narrow road that led to the river and on this road, early every morning, you could see a boy named Hatia riding a donkey and leading another by a rope. He was the son of a washerman. But as his parents were dead, he supported himself by washing the dirty clothes in the village. Every day he took a donkey load of clothes to the river, where he washed and dried them. When his work was finished, he returned home by the same road, together with his two donkeys. One was named Bhadra and the other Madri.
(ଚାଟଶାଳୀର ପାଖଦେଇ ନଈ ଆଡ଼କୁ ଗୋଟିଏ ଅଣଓସାରିଆ ରାସ୍ତା ଯାଉଥିଲା ଏବଂ ଏହି ରାସ୍ତା ଉପରେ ତୁମେ ଦେଖିବ ପ୍ରତିଦିନ ବଢ଼ିଭୋରରୁ ହଟିଆ ନାମରେ ଜଣେ ଯୁବକ ଗୋଟିଏ ଗଧ ଉପରେ ଚଢ଼ି ଏବଂ ଆଉ ଗୋଟାକୁ ଦଉଡ଼ିରେ ବାନ୍ଧି ନେଇ ଯାଉଥବ । ସେ ଗୋଟିଏ ଧୋବାର ପୁଅ । ତା’ର ବାପ ମା’ ମରିଯାଇଥିବାରୁ ସେ ନିଜେ ସବୁ ମଇଳା ଲୁଗାପଟା ଗାଁ ଲୋକଙ୍କର ନେଇ ନଈକୁ ଯାଏ । ପ୍ରତ୍ୟେକ ଦିନ ଏକ ଗଧ ବୋଝେଇ ଗାଡ଼ିରେ ମଇଳା ଲୁଗା ନଦୀକୁ ନେଉଥିଲା । ଯେଉଁଠାରେ ସେ ତାକୁ ଧୋଇସାରି ଶୁଖାଇ ଦେଉଥିଲା ଏବଂ ସେଇ ବାଟଦେଇ ପୁଣି ଘରକୁ ଫେରୁଥିଲା ଦି ଗଧଙ୍କୁ ଧରି । ଗୋଟିକର ନାମ ଥିଲା ଭଦ୍ର ଏବଂ ଅନ୍ୟଟିର ନାମ ମାଦ୍ରି ।)

BSE Odisha 8th Class English Solutions Test-1

(a). Answer the following questions. [07]
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

Question (i)
What was the name of the river?
Answer:
The name of the river was Nagabali.

Question (ii)
What was the name of the village?
Answer:
The name of the village was Hatibadi.

Question (iii)
What was the name of the teacher?
Answer:
The name of the teacher was Ghana Ratha.

Question (iv)
What subjects did Ghana Ratha teach?
Answer:
Ghana Ratha taught all the subjects including Mathematics. Literature and Social Science.

Question (v)
What was the name of the boy?
Answer:
The name of the boy was Hatia.

Question (vi)
How many donkeys did Hatia have?
Answer:
Hatia had two donkeys.

BSE Odisha 8th Class English Solutions Test-1

Question (vii)
What were their names?
Answer:
The names of his two donkeys were Bhadra and Madri.

(b). Write four sentences about Hatia. [10]
(ହଟିଆ ବିଷୟରେ ୪ଟି ବାକ୍ୟ ଲେଖ ।)
(Question with Answer)
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
Answer:
1. Hatia was the son of a washerman.
2. Every day he was going to the rider riding a donkey and leading another by a rope for washing clothes.
3. His parents were dead.
4. Every day he took a donkey load of clothes to the river and there he washed and dried them.

BSE Odisha 8th Class English Solutions Test-1

(c). Rewrite paragraph 1 of the text replacing the following words/ phrases at the right places. [10]
(ପ୍ରଥମ ଅନୁଚ୍ଛେଦକୁ ଆଉ ଥରେ ଲେଖ ଯେପରିକି ନିମ୍ନରେ ପ୍ରଦତ୍ତ ଇଂରାଜୀ ଶବ୍ଦଗୁଡ଼ିକ ତାଙ୍କ ମଧ୍ୟରେ ରହିବେ ।)

Mahanadi, town, Cuttack, college, principal, Satpathy, Odia, Sanskrit and English.

Answer:
Long long ago on the bank of the river Mahanadi. There was a small town named Cuttack and at the one end of this small town was a college or Mahavidyalaya run by Mr Satpathy. Where many children from the town nearby came to study. Mr Satapathy taught all the subjects himself including Odia, Sanskrit and English.

CHSE Odisha Class 11 Math Notes Chapter 11 Straight Lines

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 11 Straight Lines will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 11 Straight Lines

Distance formula:
Distance between two points A (x1, y1) and A (x2, y2) = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Section Formula:
If C(x, y) divides the join of A (x1, y1) and A (x2, y2) in the ratio m: n internally then, x = \(\frac{m x_2+n x_1}{m+n}\), y = \(\frac{m y_2+n y_1}{m+n}\)

Note:

  • If the division is external then, x = \(\frac{m x_2-n x_1}{m-n}\), y = \(\frac{m y_2-n y_1}{m-n}\)
  • If C(x, y) is the midpoint then x = \(\frac{x_1+x_2}{2}\), y = \(\frac{y_1+y_2}{2}\)

Area of triangle formula:
The area of triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is given by  = \(\frac{1}{2}\)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Different points related to a triangle:
(a) Centroid of the triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is = G \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

(b) In centre of a triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is = I \(\left(\frac{a x_1+b x_2+c x_3}{3}, \frac{a y_1+b y_2+c y_3}{3}\right)\)

Slope Of A Line:
(a) Angle of inclination: the angle θ made by a line with positive x-axis is the angle of inclination.
(b) Slope of a line: Slope of a line is the tangent of angle of inclination. i,.e m = tan θ.
(c) Slope of a line joining A(x1, y1), and B(x2, y2) = \(\frac{y_2-y_1}{x_2-x_1}\)

Note:

(i) Slope of x-axis = 0
Slope of any line parallel to x-axis = 0

(ii) Slope of y-axis  = ∞
Slope of any line parallel to y-axis = ∞

CHSE Odisha Class 11 Math Notes Chapter 11 Straight Lines

Angle Between two Lines:
Angle Φ between two lines with slope m1 and m2 is given by tan Φ = \(\pm \frac{\left(m_1-m_2\right)}{1+m_1 m_2}\)

Note:

  • To find the acute angle between two lines use the formula. tan Φ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
  • Two lines are parallel if m1 = m2
  • Two lines are perpendicular if m1m2 = (-1).

Collinearity Of Three Points:
Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear if
(i) Sum of distances between two pairs of points = Distance between the 3rd pair.
Or, (ii) Area of Δ ABC = 0
Or, (iii) Let B(x2, y2) divides the join of AC in ratio k: 1
∴ \(x_2=\frac{k x_3+x_1}{k+1}, y_2=\frac{k y_3+y_1}{k+1}\)
The value of k obtained from two cases are equal.
Or, (iv) Slope of AB = Slope of AC.

Equation of a straight line:
Lines parallel to co-ordinate axes:
(i) Equation of any line parallel to x-axis is, y = k
⇒ Equation of x-axis is, y = 0

(ii) Equation of any line parallel to y-axis is, x = k
⇒ Equation of y-axis is, x = 0

Lines Not Parallel To Any Axes:
(i) Slope intercept form:
Equation of a line with slope ‘m’ and y-intercepts ‘c’ is: y = mx + c

(ii) Point slope form:
Equation of a line with slope ‘m’ and passing through a point A(x1, y1) is: y – y1 = m(x – x1)

(iii) Two point form:
Equation of the line passing through A(x1, y1) and B(x2, y2) is : \(\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\)

(iv) Intercept form:
Equation of a line with x-intercept ‘a’ and y-intercept ‘b’ is \(\frac{x}{a}+\frac{y}{b}=1\)

(v) Normal form:
Equation of a line whose distance form origin is P and the perpendicular drawn form origin to the line makes an angle α with positive direction of x-axis is: x cos α + y sin α = P

(vi) Parameteric form or symmetric form:
Equation of the line passing through A(x1, y1) and making an angle θ with positive direction of x-axis is: \(\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}\) = r
Or, x = x1 + r cos θ, y = y1 + r sin θ
where r = The directed distance between points P(x, y) and A(x1, y1)

(vii) General form:
General equation of a straight line is Ax + By + C = 0

CHSE Odisha Class 11 Math Notes Chapter 11 Straight Lines

Note:

  • Slope of this line = –\(\frac{\mathrm{A}}{\mathrm{B}}\)
  • x-intercept = –\(\frac{\mathrm{C}}{\mathrm{A}}\)
  • y-intercept = – \(\frac{\mathrm{C}}{\mathrm{B}}\)
  • Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) perpendicular if a1a2 + b1b2 = 0 and coincident if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Condition of concurrency of three lines:
Three lines a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 are concurrent if \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) = 0

Family Of Lines:

(i) Equation of lines parallel to the line ax + by + c = 0 is given by: ax + by + λ = 0
(ii) Equation of lines perpendicular to the line ax + by + c = 0 is given by bx – ay + λ = 0
(iii) Equation of lines passing through the point of intersection of two lines.
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is given by: (a1x + b1y + c1) + λ(a2x + b2y + c2)

Distance of a point from a line:
The perpendicular distance of A(x1, y1) from the line ax + by + c  = 0 is: d = \(\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right|\)

Distance between two parallel lines:
ax + by + c1 = 0 and  ax + by + c2 = 0 is d = \(\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|\)

Position of a point with respect to a line:
A point A(x1, y1) lies
(i) above the line ax + by + c = 0 if \(\frac{a x_1+b y_1+c}{b}\) > 0
(ii) below the line ax + by + c = 0 if \(\frac{a x_1+b y_1+c}{b}\) < 0

Equation of bisectors of angle between two intersecting lines:
(i) Equation of angle bisector of two lines. a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is given by \(\frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_2 x+b_2 y+c_2}{\sqrt{a_2^2+b_2^2}}\)

Note:

Out of two bisector take one and find the angle between that bisector and one line. If the angle is less than 45° then that bisector is the bisector of acute angle, otherwise, the other bisector is the bisector of acute angle.

(ii) Bisector of angle containing a given point (h, k):

Step – 1: Check the sign of a1h + b1k + c1  and a2h + b2k + c2

  • If they have same sign then the bisector of angle containing (h, k) is: \(\frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_2 x+b_2 y+c_2}{\sqrt{a_2^2+b_2^2}}\)
  • If they have opposite sign then the bisector of angle containing (h, k) is: \(\frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_2 x+b_2 y+c_2}{\sqrt{a_2^2+b_2^2}}\)

CHSE Odisha Class 11 Math Notes Chapter 11 Straight Lines

Change Of Axes (Shifting Of Origin):

(i) Translation of coordinate axes.
Let O'(h, k) is the origin of system S’ with respect to origin O(0, 0) of the system S. S’ is the translation of S. If (x, y) and (x’, y’) are the coordinate of a point P in the system S and S’ respectively then
x’ = x – h and y’ = y – k Or, x = x’ + h, y = y’ + k

(ii) Rotation of axes:
Let S’ is a rotation of S, α is the measure of rotation
If (x, y) and (x’, y’) are the coordinate of a point P with respect to S and S’ then x = x’ cos α – y’ sin α and y = x’ sin α + y’ cos α

(iii) Translation as well as a rotation:
If S’ is a combination of translation followed by a rotation then x = h + x’ cos α – y’ sin α, y = k + x’ sin α + y’ cos α

CHSE Odisha Class 11 Math Notes Chapter 6 Complex Numbers and Quadratic Equations

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 6 Complex Numbers and Quadratic Equations will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 6 Complex Numbers and Quadratic Equations

Unit imaginary number ‘i’.
The unit imaginary number i = √-1
i2 = -1
i3 = -i
i4 = 1
In general (i)4n = 1, (i)4n+1 = i, (i)4n+2 = -1, and (i)4n+3 = -i.
⇒ If a and b are positive real numbers then
√-a × √-b = -√ab
√a × √b = √ab

Complex Number
General form: = z = a +ib

  • a = Real part of (z)  = Re (z)
  • b = Imaginary part of (z) = Im(z)
  • a + i0 is purely real and 0 + ib is purely imaginary .
  • a + ib = c + id iff a = c and b = d

Complex Algebra
(a) Addition of complex numbers
If z1 = a + ib and z2 = c + id then z1 + z2 = (a + c) + i(b + d)

Properties:

  • Addition is commutative: z1 + z2 = z2 + z1
  • Addition is associative: (z1 + z2) + z3 = z1 + (z2 + z3)
  • 0 + i0 is the additive identity.
  • -z is the additive inverse of z.

(b) Subtraction of complex numbers:
z1 = a + ib and z2 = c + id then z1 – z2 = (a – c) + i(b – d)

(c) Multiplication of complex numbers:
z1 = a + ib and z2 = c + id then z1z2 = (ac – bd) + i(bc + ad)

Properties:

  • Multiplication is commutative: z1z2 = z2z1
  • Multiplication is associative: z1(z2z3) = z1z2(z3)
  • 1 = 1 + i0 is the multiplicative identity.
  • If z = a + ib then the inverse of z.
    z-1 = \(\frac{1}{a+i b}=\frac{a-i b}{(a+i b)(a-i b)}\)
    = \(\frac{a-i b}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{i b}{a^2+b^2}\)
  • Multiplication is distributive over addition. z1(z2 + z3) = z1z2 + z1z3

Conjugate and modulus of a complex number:
If  z = a + ib the conjugate of z is \(\bar{Z}\) = a – ib.
⇒ We get conjugate by replacing i by (-i) Modulus of z = a + ib is denoted by |z| and |z| = \(\sqrt{a^2+b^2}\)

CHSE Odisha Class 11 Math Notes Chapter 6 Complex Numbers and Quadratic Equations

Properties Of Conjugate:
(i) \((\overline{\bar{z}})\) = z
(ii) z + \(\bar{z}\) = 2 Re (z)
(iii) z – \(\bar{z}\) = 2i m̂ (z)
(iv) z – \(\bar{z}\) ⇔ z is purely real
(v) Conjugate of real number is itself.
(vi) z + \(\bar{z}\) = 0 ⇒ z is purely imaginary.
(vii) z. \(\bar{z}\) = [Re(z)]2 + [m̂(z)]2
= a2 + b2
= |z|2
(viii) \(\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\)
(ix) \(\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}\)
(x) \(\overline{z_1z_2}=\overline{z_1}\overline{z_2}\)
(xi) \(\left(\overline{\frac{z_1}{z_2}}\right)=\frac{\overline{z_1}}{\overline{z_2}}\)

Properties of modulus:
(1) Order relations are not defined for complex numbers. i,e,. z1 > z2 or z1 < z2 has no meaning but |z1| < |z2| or |z1| > |z2| is meaningful because |z1| and |z2| are real numbers.
(2) |z|  = 0 ⇔ z = 0
(3) |z| = |\(\bar{z}\)| = |-z|
(4) |z| ≤ Re (z) ≤ |z| and -|z| ≤ m̂ (z) ≤ |z|
(5) |z1z2| = |z1| |z2|
(6) \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\)
(7) |z1 ± z2|2 = |z1|2 + |z2|2 ± 2 Re (z1\(\bar{z}_2\))
(8) |z1 + z2|2 = |z1 – z2|2 = 2(|z1|2 + |z2|2)
(9) |z1 + z2|2 ≤ |z1| + |z2|

Square Root Of Complex Number:
Let z = a + ib
Let √z = x + iy
CHSE Odisha Class 11 Math Notes Chapter 6 Complex Numbers and Quadratic Equations
If b > 0 then x and y are taken as same sign.
If b < 0 then x and y are of opposite sign.

Representation of a complex number:
We represent a complex number in different forms like
(i) Geometrical form
(ii) Vector form
(iii) Polar form
(iv) Eulerian form or Exponential form

(i) Geometrical form:
Geometrically z = x + iy = (x, y) represents a point in a coordinate plane known as Argand plane or Gaussian plane.

(ii) Vector form:
In vector form a complex number z = x + iy is the vector \(\overrightarrow{\mathrm{OP}}\) where p(x, y) is the point in the cartesian plane.

(iii) Polar form:
A complex number z = x + iy  in polar form can be written as z = r(cos θ + i sin θ) where r = \(\sqrt{x^2+y^2}\) = |z| and θ is called the argument and -π < θ ≤ π. Technique to write z = x + iy in polar form.
Step – 1: Find r = |z| = \(\sqrt{x^2+y^2}\)
Step – 2: Find α = tan-1 \(\left|\frac{y}{x}\right|\)
Step – 3:
θ = α for x > 0, y > 0
θ = π – α for x > 0, y > 0
θ = -π + α for x > 0, y > 0
θ = -α for x > 0, y > 0
Step – 4: Write z = r(cos θ + i sin θ)

(iv) Eulerian form or Exponential form z = r e, because e = cos θ + i sin θ where θ is the argument and r is the modulus if z.

Note:
(1) |z1 z2 z3 ….. zn| = |z1||z2| …. |zn|
(2) arg (z1z2 …. Zn) = arg (z1) + arg (z2) + ….. + arg (zn)
(3) arg \(\left(\frac{z_1}{z_2}\right)\) = arg (z1) – arg (z2)
(4) arg \((\bar{z})\) = -arg (z)

Cube Roots Of Unity:
Cube roots of unity are 1, ω, ω2 where ω = \(\frac{-1 \pm i \sqrt{3}}{2}\)

Properties of Cube roots of unity:
(i) Cube roots of unity lie on unit circle |z| = 1
(ii) 1 + ω + ω2 = 0
(iii) Cube roots of -1 are -1, -ω, -ω2
(iv) 1 + ωn + ω2n \(=\left\{\begin{array}{l}
0 \text { if } n \text { is not a multiple of } 3 \\
3 \text { if } n \text { is a multiple of } 3
\end{array}\right.\)
(v) z3 + 1 = (z + 1) (z + ω) (z + ω2)
(vi) -ω and -ω2 are roots of z2 – z + 1  = 0.

De-moivre’s theorem:
(a) (De-moivre’s theorem for integral index)
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)

(b) (De-moivre’s theorem for rational index)
cos (nθ) + i sin (nθ) is one of the values of (cos θ + i sin θ)n

(c) nth roots of unity
nth roots of unity are 1, α, α2, α3 …..αn-1. where α = ei\(\frac{2 \pi}{n}\) = cos \(\frac{2 \pi}{n}\) + i sin \(\frac{2 \pi}{n}\)

Properties:

  • 1 + α + α2 ….. + αn-1 = 0
  • 1 + αp + α2p + ….. + α(n-1)p \(= \begin{cases}0 & \text { if } p \text { is not a multiple of } n \\ n & \text { if } p \text { is a multiple of } n\end{cases}\)
  • 1. α. α2 ….. αn-1 = (-1)n-1
  • zn – 1 = (z – 1) (z – α) (z – α2) …..(z – αn-1)

CHSE Odisha Class 11 Math Notes Chapter 6 Complex Numbers and Quadratic Equations

Quadratic Equations:
The general form: ax2 + bx + c = 0  …(i)
Solutions of quadratic equation(1) are
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
D = b2 – 4ac is called the discrimination of a quadratic equation.
D > 0 ⇒ The equation has real and distinct roots.
D = 0 ⇒ The equation has real and equal roots.
D < 0 ⇒ The equation has complex roots.

Note:
In a quadratic equation with real coefficients, the complex roots occur in conjugate pairs.

CHSE Odisha Class 11 Math Notes Chapter 9 Binomial Theorem

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 9 Binomial Theorem will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 9 Binomial Theorem

Binomial Theorem For Positive Integral Index:
For any a,b ∈ R, and n ∈ N
(a + b)n = nC0 an + nC1 an-1b + ….. nCn bn

Note:

(a) (a + b)n = an + nan-1 b + \(\frac{n(n-1)}{2 !}\) an-2b2 ….. + bn
(b) (1 + x)n = nC0 + nC1 x + nC2 x2 + ….. + nCn xn
(c) (a – b)n = nC0 annC1 an-1 b + nC2 an-2b2 ….. + (-1)n bn
(d) (1 – x)n = nC0nC1 x + nC2 x2 ….. + (-1)n xn

Some conclusions from the Binomial theorem:

  • There are (n + 1) terms in the expansion of (a + b)n
  • We can write (a + b)n = \(\sum_{r=0}^n{ }^n \mathrm{C}_r a^{n-r} b^r\) and (a – b)n = \(\sum_{r=0}^n(-1)^r{ }^n \mathrm{C}_r a^{n-r} b^r\)
  • The sum of powers of a and b in each term = n
  • As nCr = nCn-r (The coefficient of terms equidistant from the beginning and the end are equal).
  • (r + 1)th term (General term)
    = tr+1 = nCr an-rbr
  • (a + b)n + (a – b)n = 2[nC0an + nC2 an-2b2 + ….]
  • (a + b)n – (a – b)n = 2[nC1 an-1b + nC3 an-3b3 + ….]
  • (middle terms):
    ⇒ If n is even then the middle term = \(t_{\left(\frac{n+2}{2}\right)}=t_{\left(\frac{n}{2}+1\right)}\)
    ⇒ If n is odd there are two middle terms. They are = \(t_{\left(\frac{n+1}{2}\right)} \text { and } t_{\left(\frac{n+3}{2}\right)}\)
  • tr+1 from the end in the expansion of (a + b)n = tr+1 from the beginning in the expansion of (b + a)n.

CHSE Odisha Class 11 Math Notes Chapter 9 Binomial Theorem

Binomial Theorem For Any Rational Index:
If n ∈ Q and x ∈ R such that |x| < 1 then (1 + x)n = 1 + nx + \(\frac{n(n-1)}{2 !} x^2\) + \(\frac{n(n-1)(n-2)}{3 !} x^3+\ldots .\)

Note:

(1) (1 + x)-1 = 1 – x + x2 – x3 + …..
(2) (1 – x)-1 = 1 + x + x2 + …..
(3) (1 + x)-2 = 1 – 2x + 3x2 – 4x3 + …..
(4) (1 – x)-2 = 1 + 2x + 3x2 + 4x3 + …..

CHSE Odisha Class 11 Math Notes Chapter 8 Permutations And Combinations

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 8 Permutations And Combinations will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 8 Permutations And Combinations

Fundamental Principle Of Counting:
(a) Fundamental principle of Multiplication:
If we choose an element from set A with m element and then one element from set B  with n elements, then are total number of ways we can make a choice is exactly mn.

OR

If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of ways in which both the events can occur in succession in mn ways.

(b) Fundamental Principle of addition: If there are two events such that they can be performed independently in m and n different ways respectively, then either of two events can be performed in (m + n) ways.

Note:
(a) Use the multiplication principle if by doing one part of the job, the job remains incomplete.
(b) Use the addition principle if by doing one part of the job, the job is completed.

Factorial Notation:
If n ∈ N then the factorial of n, denoted by n! or ∠n is defined as
n! = n (n – 1). (n – 2) … 3.2.1.

Note:
0! = 1

Properties of Factorial:
(1) Factorial of negative integers is not defined
(2) n! = n(n – 1)!
= n(n – 1) (n – 2)!
= n(n – 1) (n – 2) (n – 3)!
(3) \(\frac{n !}{r !}\) = n(n – 1) (n – 2) ….. (r + 1)
(4) Exponent of a prime number p in n! denoted by
\(\mathrm{E}_p(n !)=\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\ldots \ldots\)

CHSE Odisha Class 11 Math Notes Chapter 8 Permutations And Combinations

Permutation:
Each of the arrangements which can be made by taking some or all objects or things at a time is called a permutation.

(a) Permutation of n different objects:

  • Number of permutations of n different objects have taken all at a time = \({ }^n \mathrm{P}_n\) = n!.
  • Number of permutations of n different objects taken none at a time = \({ }^n \mathrm{P}_0\) = 1
  • Number of permutations of n different objects taken r at a time = \({ }^n \mathrm{P}_r\) = P(n, r) = \(\frac{n !}{(n-r) !}\)

(b) Permutation ofnon-distinct objects:
(1) Number of permutations of n objects taken all at a time of which p objects are of same kind and others are distinct = \(\frac{n !}{p !}\)
(2) Number of permutations of n objects taken all at a time of which p objects are of one kind, q objects are of a second kind and other are distinct = \(\frac{n !}{p ! q !}\)
(3) Number of permutations of n objects taken all at a time in which p1 objects are of one kind, p2 are of second kind, p3 are 3rd kind ….. and
pn are of nth kind and other are distinct. = \(\frac{n !}{p_{1} ! p_{2} ! \cdots p_{n} !}\)

(c) Restricted permutations:

  • Permutation of distinct objects with repetition: The number of permutations of n different things taken r at a time when each thing may be repeated any number of times = nr
  • Number of permutations of n different things taken r at a time when a particular thing is to be always included in each arrangement = r. n-1Pr-1.
  • Number of permutations of n different things, taken r at the time when p particular are to be always included in each arrangement = P(r – (p – 1) n-pPr-p.
  • Number of permutations of n different things taken r at a time, when a particular thing is never taken in each arrangement = n-1Pr.
  • Number of permutations of n different things taken r at a time, when p particular things never taken in each arrangement = n-pPr.

(d) Circular permutation:
(1) When we do an arrangement of objects along a closed curve we call it the circular permutation.
(2) Number of circular permutations of n distinct objects taken all at a time = (n – 1)!, where clockwise and anti-clockwise orders are taken as different, as arrangements round a table.
(3) Number of circular permutations of n distinct objects taken all at a time, where clockwise and anti-clockwise orders make no difference as beads or flowers in a necklace or garland.
= \(\frac{(n-1) !}{2}\)
(4) Number of circular permutations of n different things taken r at a time where clockwise and anti-clockwise orders are different = \(\frac{\left({ }^n \mathrm{P}_r\right)}{r}\)
(5) Number of circular permutations of n different things taken r at a time where clockwise and anti-clockwise orders make no difference = \(\frac{\left({ }^n \mathrm{P}_r\right)}{2 r}\)

(e) Some more restricted permutations:

  • Number of permutations of n different things taken all at a time, when m specified things come together = m!(n – m + 1)!.
  • Number of permutations of n different things taken all at a time when m specified things never come together = n!  – m!(n – m + 1)!.

Combinations:
Each of the different selections made by taking some or all objects at a time irrespective of any order is called a combination.

(a) Difference between permutation and combination:

  • A combination is a selection but a permutation is not a selection but an arrangement.
  • In combination the order of appearance of objects is immaterial, whereas in a permutation the ordering is essential.
  • Practically to find permutations of n different objects taken r at a time, we first select objects then we arrange them.
  • One combination corresponds to many permutations.

(b) Combinations of n different things taken r at a time:
The number of combinations of n different things have taken r at a time ncr = C(n, r) = \(\left(\begin{array}{l}
n \\
r
\end{array}\right)=\frac{n !}{r !(n-r) !}\)

(c) Properties of ncr :
(1) ncr = nC0 = 1, nC1 = n
(2) nCr = nCn-r
(3) nCr + nCr-1 = n+1Cr (Euler’s formula)
(4) nCx = nCy ⇒ x = y or x + y = n
(5) n. n-1Cr-1 = (n – r + 1) nCr-1
(6) nCr = \(\frac{n}{r}{ }^{n-1} \mathrm{C}_{r-1}\)
(7) \(\frac{{ }^n \mathrm{C}_r}{{ }^n \mathrm{C}_{r-1}}=\frac{n-r+1}{r}\)
(8) If n is even then the greatest value of nCr is nCn/2.
⇒ If n is odd then the greatest value of nCr is \({ }^n \mathrm{C}_{\left(\frac{n+1}{2}\right)} \text { or }{ }^n \mathrm{C}_{\left(\frac{n-1}{2}\right)}\)

(d) Number of combinations of n different things taken r at a time, when k particular things always occur = n-kCr-k

(e) The number of combinations of n different things, taken r at a time where k particular things never occur = n-kCr

(f) The total number of combinations of n different things taken one or more at a time (or the number of ways of n different things selecting at least one of them) = nC1 + nC2 + nC3 + ….. + nCn = 2n -1

(g) The number of combinations of n identical things taken r at a time = 1.

(h) Number of ways of selecting r things out of n alike things where r = 0, 1, 2, 3 ….. n is (n+ 1).

(i) Division into groups:

  • The number of ways in which (m + n) different things can be divided into two groups which contain m and n things respectively = \(\frac{(m+n) !}{m ! n !}\) for m ≠ n.
  • If m-n then the groups are of equal size. Thus, division can be done in two ways as:
    ⇒ If order of groups is not important: In this case the number of ways = \(\frac{(2 n) !}{2 !(n !)^2}\)
    ⇒ If order of groups is important: In this case the number of ways = \(\frac{(2 n) !}{(n !)^2}\)

CHSE Odisha Class 11 Math Notes Chapter 8 Permutations And Combinations

(j) Arrangement in groups:

  1. The number of ways in which n different things can be arranged into r different groups = n+r-1Pn or n! n-1Cr-1
  2. The number of ways in which n different things can be distributed into r different groups = rnrC1(r – 1)n + rC2(r – 2)n ….. + (-1)r-1 . rCr-1. (Blank groups are not allowed)
  3. The number of ways in which n identical things can be distributed into r different groups where blank groups are allowed
    = (n+r-1)C(r-1)
    = (n+r-1)Cn
  4. Number of ways in which n identical things can be distributed into r different groups where blank groups are not allowed (each group receives at least one item) = n-1Cr-1

(k) Number of divisors:
CHSE Odisha Class 11 Math Notes Chapter 8 Permutations And Combinations

BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(b)

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

BSE Odisha Class 10 Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(b)

Question 1.
ନିମ୍ନଲିଖ ଉକ୍ତି ମଧ୍ୟରୁ କେଉଁଟି ଠିକ୍ ଦର୍ଶାଅ।
(i) ଘଟଣାଟି ϕ ହେଲେ ଏହାର ସମ୍ଭାବ୍ୟତା ଶୂନ ।
(ii) ଘଟଣା E = S, ଯେଉଁଠାରେ S (Sample Space) ତେବେ P(E) < 1।
(iii) ଗୋଟିଏ ମୁଦ୍ରାକୁ ଥରେ ଟସ୍ କଲେ Sample Spaceର ଉପାଦାନ ସଂଖ୍ୟା 4 ଅଟେ।
(iv) ‘Probability’ ଶବ୍ଦରୁ ଗୋଟିଏ ଅକ୍ଷର ‘i’ ବାଛିବାର ସମ୍ଭାବ୍ୟତା \(\frac{2}{11}\)।
(v) E1 ଓ E2 (E1 E2 ⊂ S) ପରସ୍ପର ବର୍ହିଭୁକ୍ତ ଘଟଣା ଦ୍ଵୟର ସମ୍ଭାବ୍ୟତାର ଯୋଗଫଳ 1 ।
(vi) ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଏକ ସଙ୍ଗେ ଦୁଇ ଥର ଗଡ଼ାଇଲେ ଲବ୍‌ଧ ସାମ୍ପଲ ସେସ୍‌ର ଉପାଦାନ ସଂଖ୍ୟା 36 ।
(vi) ଗୋଟିଏ ମୁଦ୍ରାକୁ 3 ଥର ଟସ୍ କଲେ ଲବ୍‌ଧ ସାମ୍ପଲ ସ୍ପେସ୍‌ରେ ବିଦ୍ୟମାନ ଉପାଦାନମାନଙ୍କ ସଂଖ୍ୟା
32 = 9।
(viii) ଗୋଟିଏ sample spaceର E1 ଏବଂ E2 ଦ୍ଵୟ ବହିର୍ଭୁକ୍ତ ଘଟଣା ହେଲେ
P(E1 ∪ E2) = P (E1) + P(E2)।
(ix) ଥରେ ମୁଦ୍ରାକୁ ଟସ୍ କଲେ E1 = {H} ଘଟଣାଟିର ପରିପୂରକ ଘଟଣାଟି E2 = {H, T}।
ଉ –
ଠିକରକ୍ତି: (i), (iv), (vi) ଓ (viii)

BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(b)

Question 2.
ଏକ ପରୀକ୍ଷଣରେ E1, E2, E3 ଏବଂ E4 ଚାରିଗୋଟି ବହିର୍ଭୁକ୍ତ ଘଟଣା । ଏଠାରେ (E1 ∪ E2 ∪ E3 ∪ E4) ନିଶ୍ଚିତ ରୂପେ ଘଟୁଥିବା ଘଟଣା । ଦତ୍ତ ଘଟଣାଗୁଡ଼ିକ ସମ ସମ୍ଭାବ୍ୟତା ବିଶିଷ୍ଟ ହେଲେ ପ୍ରତ୍ୟେକର ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
ସମାଧାନ:
ପରୀକ୍ଷଣରେ E1, 2, E3, E4 ଚାରୋଟି ବର୍ହିଭୁକ୍ତ ଘଟଣା ଅର୍ଥାତ୍ E1 ∩ E2 ∩ E3 ∩ E4 = ϕ
(E1 ∪ E2 ∪ E3 ∪ E4) ଏକ ନିଶ୍ଚିତ ଘଟଣା ହେତୁ ଏହାର ସମ୍ଭାବ୍ୟତା 1।
P (E1 ∪ E2 ∪ E3 ∪ E4) = P(E1) + P(E2) + P(E3) + P(E4)
⇒ 1 = P (E1) + P(E2) + P(E3) + P(E4)
⇒ P(E1) = P (E2) = P (E3) = P (E4) = \(\frac{1}{4}\)
କାରଣ ଘଟଣାଗୁଡିକ ସମ ସମ୍ଭାବ୍ୟତାବିଶିଷ୍ଟ ।

Question 3.
ଗୋଟିଏ ଲୁଡୁଗୋଟି ଥରେ ଗଡ଼ାଇ ଦିଆଗଲା । ତେବେ ନିମ୍ନଲିଖ୍ ଘଟଣାମାନଙ୍କ ସମ୍ଭାବ୍ୟତା ସ୍ଥିର କର ।
(i) ଫଳ ≤ 3
(ii) ଫଳ < 3
(iii) ଫଳ ≤ 4
(iv) ଫଳ < 6 (v) ଫଳ ≤ 6 (vi) ଫଳ > 6
ସମାଧାନ:
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଇ ଦିଆଗଲା ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(b) - 1
(i) ଫଳ ≤ 3 ଏକ ଘଟଣା = E1 ∴ E1 = {1, 2, 3} ଏବଂ |E1| = 3
ଦତ୍ତ ପ୍ରଶ୍ନରେ ସାମ୍ପଲ ସେଟ୍ |S| = = 6 ଓ | E1| = 3
∴ ଫଳ ≤ 3ର ସମ୍ଭାବ୍ୟତା P(E1) = \(\frac{\left|E_1\right|}{|S|}=\frac{3}{6}=\frac{1}{2}\)

(ii) ଫଳ < 3 ଏକ ଘଟଣା E2 ∴ E2 = {1, 2} ⇒ |E2| = 2
∴ ଫଳ < 3ର ସମ୍ଭାବ୍ୟତା P(E2) = \(\frac{\left|E_2\right|}{|S|}=\frac{2}{6}=\frac{1}{3}\)

(iii) ଫଳ ≤ 4 ଏକ ଘଟଣା E3 ∴ E3 = {1, 2, 3, 4} ⇒ |E3| = 4
∴ ଫଳ ≤ 4ର ସମ୍ଭାବ୍ୟତା P(E3) = \(\frac{\left|E_3\right|}{|S|}=\frac{2}{6}=\frac{1}{3}\)

(iv) ଫଳ < 6 ଏକ ଘଟଣା E4 ∴ E4 = {1, 2, 3, 4, 5} ⇒ |E4| = 5
∴ ଫଳ < 6 ର ସମ୍ଭାବ୍ୟତା P(E4) = \(\frac{\left|E_4\right|}{|S|}=\frac{5}{6}\)

(v) ଫଳ ≤ 6 ଏକ ଘଟଣା E5
∴ E5 = {1, 2, 3, 4, 5, 6} ⇒ |E5| = 6
∴ ଫଳ ≤ 6 ର ସମ୍ଭାବ୍ୟତା P(E5) = \(\frac{\left|E_5\right|}{|S|}=\frac{6}{6}=1\)
ବି.ଦ୍ର. : ଫଳ ≤ 6 ଘଟଣାଟି ଏକ ନିଶ୍ଚିତ ଘଟଣା ହେତୁ ସମ୍ଭାବ୍ୟତା l ହେବ ।

(vi) ଫଳ > 6 ଏକ ଘଟଣା E6
∴ E = ϕ ⇒ |E6| = 0
∴ ଫଳ > 6ରେ ସମ୍ଭାବ୍ୟତା P(E6) = \(\frac{0}{6}=0\)
ବି.ଦ୍ର. : ଫଳ > 6 ଏକ ଅନିଶ୍ଚିତ ଘଟଣା ହେତୁ ସମ୍ଭାବ୍ୟତା 0 ହେବ ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(b)

Question 4.
ଗୋଟିଏ ଜାର୍‌ରେ 5 ଗୋଟି ନାଲି, 6 ଗୋଟି ସବୁଜ ଏବଂ 4 ଗୋଟି ନୀଳ ମାର୍ବଲ ରହିଛି । ଜାରୁରୁ ଯଦୃଚ୍ଛା ଗୋଟିଏ ସବୁଜ ମାର୍ବଲ୍ ବାହାର କରିବାର ସମ୍ଭାବ୍ୟତା ସ୍ଥିର କର ।
ସମାଧାନ:
ଗୋଟିଏ ଜାର୍‌ରେ 5ଟି ନାଲି, ଟି ସବୁଜ ଓ 4 ଗୋଟି ନୀଳ ମାର୍ବଲ ଅଛି ।
ସବୁଜ ମାର୍ବଲ ସଂଖ୍ୟା = 6
ସମୁଦାୟ ମାର୍ବଲ ସଂଖ୍ୟା = 5 + 6 + 4 = 15
ଗୋଟିଏ ସବୁଜ ମାର୍ବଲ ବାହାର କରିବାର ସମ୍ଭାବ୍ୟତା
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(b) - 2

Question 5.
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଢ଼ାଗଲା । ଯଦି E ଘଟଣାଟି ‘ଫଳ ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟା’’କୁ ସୂଚାଏ ତେବେ E ଘଟଣାଟି ଘଟିବାର ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
ସମାଧାନ:
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଗଲା ।
ଏହାର Sample space, S = {1, 2, 3, 4, 5, 6} ⇒ |S|= 6
‘‘ଫଳ ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟା’’ ଏକ ଘଟଣା = E ∴ E = {2, 4, 6} = |E|= 3
‘‘ଫଳ ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟା’’ର ସମ୍ଭାବ୍ୟତା P(E) = \(\frac{|E|}{|S|}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 6.
ଗୋଟିଏ ଲୁଡୁ ଗୋଟିକୁ ଥରେ ଗଡ଼ାଇଲେ ‘‘ଫଳ ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା’’କୁ ସୂଚାଉଥବା ଘଟଣାଟି ଘଟିବାର ସମ୍ଭାବ୍ୟତା କେତେ ?
ସମାଧାନ:
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଗଲା ।
ଏଠାରେ Sample space S = {1, 2, 3, 4, 5, 6} = |S| = 6
“ଫଳ ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା” ଏକ ଘଟଣା = E, |E| = 3
∴ ଏହାର ସମ୍ଭାବ୍ୟତା P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 7.
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଗଲା । ଯଦି ‘‘ଫଳ ≤ 5’’କୁ ସୂଚାଉ ଥ‌ିବା ଘଟଣା E ହୁଏ, ତେବେ ଉକ୍ତ ଘଟଣାଟି ଘଟିବାର ସମ୍ଭାବ୍ୟତା କେତେ?
ସମାଧାନ:
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଗଲା ।
ଏଠାରେ Sample space, S = {1, 2, 3, 4, 5, 6} ⇒ |S| = 6
‘‘ଫଳ ≤ 5 ଏକ ଘଟଣା’’ = E
∴ |E| = {1, 2, 3, 4, 5} ⇒ |E|= 5
∴ ଏହାର ସମ୍ଭାବ୍ୟତା P(E) = \(\frac{|E|}{|S|}=\frac{5}{6}\)

Question 8.
ଗୋଟିଏ ମୁଦ୍ରାକୁ 2 ଥର ଟସ୍ କରାଗଲେ ନିମ୍ନଲିଖ୍ ଘଟଣାଗୁଡ଼ିକୁ ସ୍ଥିର କରି ସେମାନଙ୍କ ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
(i) ଅତି କମ୍‌ରେ ଗୋଟିଏ H;
(ii) ଫଳରେ କେବଳ T ରହିବା;
(i) ଫଳରେ ଅତି ବେଶିରେ ଗୋଟିଏ H ରହିବା ଓ
(iv) ଫଳରେ H ନ ରହିବା
ଗୋଟିଏ ମୁଦ୍ରାକୁ 2 ଥର ଟସ୍ କଲେ Sample space S = {HH, HT, TH, TT} ଏବଂ | S | = 4
(i) ମନେକର ଅତି କମ୍‌ରେ ଗୋଟିଏ H ଆସିବାର ଏକ ଘଟଣା =
∴ E1 = {HH, HT, TH} ⇒ |E1| = 3
ଅତି କମ୍‌ରେ ଗୋଟିଏ H ଆସିବାର ସମ୍ଭାବ୍ୟତା P(E) = \(\frac{\left|E_1\right|}{|S|}=\frac{3}{4}\)

(ii) ଫଳରେ କେବଳ T ଆସିବା ଏକ ଘଟଣା = E2
∴ E2 = {TT} ⇒ |E2| = 1
∴ P(E2) = \(\frac{\left|E_2\right|}{|S|}=\frac{1}{4}\)

(iii) ଫଳରେ ଅତିବେଶିରେ ଗୋଟିଏ H ରହିବା ଏକ ଘଟଣା = E
∴ E3 = {HH, TH, TT} ⇒ |E3| = 3
∴ P(E3) = \(\frac{\left|E_3\right|}{|S|}=\frac{3}{4}\)

(iv) ଫଳରେ H ନରହିବା ଏକ ଘଟଣା E4
∴ E4 = {TT} ⇒ |E4| = 1
∴ P(E4) = \(\frac{\left|E_4\right|}{|S|}=\frac{1}{4}\)

BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(b)

Question 9.
ଗୋଟିଏ ମୁଦ୍ରାକୁ 3 ଥର ଟସ୍ କରାଗଲା । ସାମ୍ପଲ ସ୍ପେସ୍ଟ ଲେଖ ଓ ନିମ୍ନଲିଖ୍ ଘଟଣାମାନଙ୍କ ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
(i) ଫଳରେ କେବଳ T ରହିବା
(ii) ଫଳରେ ଅତି କମ୍ରେ ଦୁଇଟି H ଥ‌ିବା
(iii) ଫଳରେ ଅତି ବେଶିରେ ଦୁଇଟି T ରହିବା
(iv) ଫଳରେ କେବଳ H କିମ୍ବା କେବଳ T ଥ‌ିବା ଓ
(v) କୌଣସି ଫଳରେ T ନ ଥ‌ିବା
ସମାଧାନ :
ଗୋଟିଏ ମୁଦ୍ରାକୁ 3 ଥର ଟସ୍ କରାଗଲା ।
ଏଠାରେ Sample space S = {HHH, HTH, HHT, HTT, TTT, TTH, THT, THH) ଏବଂ | S|=8
(i) ଫଳରେ କେବଳ ‘T ରହିବା ଏକ ଘଟଣା = E1
∴ E1 = {TTT} ଏଠାରେ |E1|= 1
∴ P(E1) = \(\frac{\left|E_1\right|}{|S|}=\frac{1}{8}\)

(ii) ଫଳରେ ଅତି କମ୍ରେ ଦୁଇଟି H ଥିବା ଏକ ଘଟଣା = F
∴ F = {HTH, HHT, THH, HHH} ⇒ |F|= 1
∴ P(F) = \(\frac{|F|}{|S|}=\frac{4}{8}=\frac{1}{2}\)

(iii) ଫଳରେ ଅତି ବେଶିରେ ଦୁଇଟି T ଥିବା ଏକ ଘଟଣା = A
∴ A = {HHH, HHT, HTH, HTT,THH, THT, TTH} = |A| = 7
P(F) = \(\frac{|A|}{|S|}=\frac{7}{8}\)

(iv) ଫଳରେ କେବଳ H ଥ‌ିବା ଏକ ଘଟଣା E = {HHH}, ⇒ |E| = 1
P(E) = \(\frac{|E|}{|S|}=\frac{1}{8}\)
ସେହିପରି F କେବଳ T ଥିବା ଏକ ଘଟଣା । P(F) =
ଫଳରେ କେବଳ H ଥ‌ିବା କିମ୍ବା କେବଳ T ଥ‌ିବା ଘଟଣାଟି E ∪ F
P(E ∪ F) = P(E) + P(F) = \(\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}\)
(∵ E ଓ F ଘଟଣାଦ୍ଵୟ ପରସ୍ପର ବହିଃର୍ଭୁକ୍ତ)

(v) କୌଣସି ଫଳରେ T ନଥିବା ଏକ ଘଟଣା = E
∴ E = {HHH} = |E|=1
∴ P(E) = \(\frac{|E|}{|S|}=\frac{1}{8}\)

Question 10.
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଦୁଇଥର ଗଡ଼ାଇ ଦିଆଯିବାରେ ନିମ୍ନଲିଖତ ଫଳ ଲବ୍‌ଧ ହେବାର ସମ୍ଭାବ୍ୟତା ସ୍ଥିର କର ।
(i) ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ = 6,
(ii) ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ = 4,
(iii) ସଂଖ୍ୟା ଦୁଇଟିରୁ ପ୍ରତ୍ୟେକଟି ଗୋଟିଏ ଗୋଟିଏ ବର୍ଗ ସଂଖ୍ୟା,
(iv) ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ 2 10,
(v) ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ < 6 ଓ
(vi) ପ୍ରଥମ ସଂଖ୍ୟାଟି ଅଯୁଗ୍ମ ଓ ଦ୍ବିତୀୟଟି 61
ସମାଧାନ :
(i) ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ 2 ଥର ଗଡ଼ାଇଲେ Sample space ସଂଖ୍ୟା |S| = 6² = 36 ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ 6 ଆସିବା ଏକ ଘଟଣା = E ∴ E = {15, 51, 24, 42, 33} |E| = 5 ∴ P(E) = \(\frac{|E|}{|S|}=\frac{5}{36}\)

(ii) ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ = 4 ଏକ ଘଟଣା T, ∴ T = {13, 31, 22} |T| = 5 ∴ P(T) = \(\frac{|T|}{|S|}=\frac{3}{36}=\frac{1}{12}\)

(iii) ଦୁଇଟି ସଂଖ୍ୟାରୁ ପ୍ରତ୍ୟେକଟି ଗୋଟିଏ ଗୋଟିଏ ବର୍ଗ ସଂଖ୍ୟା ଏକ ଘଟଣା = F ∴ F = {11, 44} |F| = 2 ∴ P(F) = \(\frac{|F|}{|S|}=\frac{2}{36}=\frac{1}{18}\) (iv) ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ > 10 ଏକ ଘଟଣା = E
∴ E = {46, 64, 55, 56, 65, 66} = |E| = 6
∴ P(E) = \(\frac{|E|}{|S|}=\frac{6}{36}=\frac{1}{6}\)

(v) ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ < 6 ଏକ ଘଟଣା = |E |
∴ E= (11, 12, 13, 14, 22, 23, 32, 41, 31, 21} |E| = 10
P(E) = \(\frac{|E|}{|S|}=\frac{10}{36}=\frac{5}{18}\)

(vi) ପ୍ରଥମ ସଂଖ୍ୟାଟି ଅଯୁଗ୍ମ ଓ 2ୟ ସଂଖ୍ୟାଟି 6 ଏକ ଘଟଣା ।
∴ E = {16, 36, 56) = |E| = 3
∴ P(E) = \(\frac{|E|}{|S|}=\frac{3}{36}=\frac{1}{12}\)

BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(b)

Question 11.
ଏକ ପରୀକ୍ଷଣରେ ପରସ୍ପର ବର୍ହିଭୁକ୍ତ ଦୁଇଟି ଘଟଣା E1 ଓ E2 ଏପରିକି P(E1) = 2P(E2) ଓ P(E1) + P(E2) = 0.9 । ତେବେ E1 ∪ E2 ଘଟଣା ତଥା E1, ଘଟଣାର ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
ସମାଧାନ :
ଏକ ପରୀକ୍ଷଣରେ ପରସ୍ପର ବର୍ହିଭୁକ୍ତ ଦୁଇଟି ଘଟଣା E1 ଓ E2
P(E1) = 2P (E2) P(E1) + P(E2) = 0.9
∴ P(E1) + P(E2) = 0.9 2P(E2) +P(E2) = 0.9
= 3P(E2) = 0.9 = P(E2) = 0.9 = 0.3 P(E1) = 2P (E2) = 2 × 0.3 = 0.6
∴ P(E1 ∪ E2) = P(E1) + P(E2) = 0.6 + 0.3 = 0.9.
P(E) = 0.6

Question 12.
ଯଦି E, ଓ E, ଏପରି ଦୁଇଟି ଘଟଣା ଯେଉଁଠାରେ P(E1) = \(\frac{5}{8}\), P(E2) = \(\frac{2}{8}\) ଓ P(E1 ∩ E2) = \(\frac{1}{8}\) ତେବେ ନିମ୍ନଲିଖତଗୁଡ଼ିକ ସ୍ଥିର କର ।
(i) P(E1 ∪ E2)
(ii) P(E1’)
(iii) P(E2’)
(iv) P(E’1 ∪ E’2)
ସମାଧାନ :
E, ଓ E, ଏପରି ଦୁଇଟି ଘଟଣା ଯେଉଁଠାରେ P(E1) = \(\frac{5}{8}\), P(E2) = \(\frac{2}{8}\)
P(E1 ∩ E2) = \(\frac{1}{8}\)
(i) P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2) = \(\frac{5}{8}+\frac{2}{8}-\frac{1}{8}=\frac{6}{8}=\frac{3}{4}\)

(ii) P(E1’) = 1 – P(E1) = 1 – \(\frac{5}{8}=\frac{3}{4}\)

(iii) P(E2’) = 1 – P(E2) = 1 – \(\frac{2}{8}=\frac{6}{8}=\frac{3}{4}\)

(iv) P(E’1 ∪ E’2) = P(E1 ∩ E2)’ = 1 – P(E1 ∩ E2) = 1 – \(\frac{1}{8}=\frac{7}{4}\)

Question 13.
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଇଲେ ‘ଫଳ 5 କିମ୍ବା ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା’ ଆସିବାର ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
ସମାଧାନ :
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଇଲେ ସାମ୍ପଲ୍ ସ୍ପେସ୍
S = {1, 2, 3, 4, 5, 6} = |S| = 6
ମନେକର ଫଳ 5 ଏକ ଘଟଣା = E1 ଏବଂ ଫଳ ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା ଘଟଣା = E2
E1 = {5} |E1| = 1
E2 = {1, 3, 5} = |E2|=3
E1 ∩ E2 = {5} = |E1 ∩ E2| = 1
ଫଳ ‘5’ କିମ୍ବା ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା ଘଟଣା E1 ∪ E2
P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2)
⇒ P(E1 ∪ E2)
= \(\frac{\left|E_1\right|}{|S|}+\frac{\left|E_2\right|}{|S|}-\frac{\left|E_1∩E_2\right|}{|S|}\)
= \(\frac{1}{6}+\frac{3}{6}-\frac{1}{6}-\frac{3}{6}=\frac{1}{2}\)

BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(b)

Question 14.
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଇବାରୁ ‘‘ଫଳ ଅଯୁଗ୍ମ କିମ୍ବା ଫଳ ≥ 3’ ଘଟଣାଟିର ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
ସମାଧାନ :
ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଇଲେ Sample space S = {1, 2, 3, 4, 5, 6} = | S | = 6
ମନେକର ‘‘ଫଳ ଅଯୁଗ୍ମ ଏବଂ ଫଳ ≥ 3’’ ଏକ ଘଟଣା = E2
∴ E = {1, 3, 5} = |E1| = 3 ଏବଂ E2 = {3, 4, 5, 6} = |E2| = 4
∴ (E1 ∩ E2) = {3, 5} = (E1 ∩ E2) = 2
ଫଳ ଅଯୁଗ୍ମ କିମ୍ବା ଫଳ ≥ 3 = E1 ∪ E2
∴ P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2)
= \(\frac{\left|E_1\right|}{|S|}+\frac{\left|E_2\right|}{|S|}-\frac{\left|E_1∩E_2\right|}{|S|}\)
= \(\frac{3}{6}+\frac{4}{6}-\frac{2}{6}=\frac{5}{2}\)

CHSE Odisha Class 12 Alternative English Grammar Prepositions

Odisha State Board CHSE Odisha Class 12 Alternative English Solutions Grammar Prepositions Exercise Questions and Answers.

CHSE Odisha Class 12 Alternative English Grammar Prepositions

A. Preposition
Definition:
A preposition is a world that is put before a noun or pronoun in order to show more relation between them.

Example:
1. The lamp is on the table
Here, the preposition ‘on’ shows the relation of the noun (the lamp) with another noun (the table)
2. She fell into the well.
Here, the preposition ‘into’ shows the relation of the pronoun (she) with the noun (well).
3. Akash gives respect to me.
Here, the preposition ‘to’ shows the relation of the noun (Akash) with the pronoun (me).
4. She is hot under me.
Here, the preposition ‘under’ shows the relation of the pronoun (she) with another pronoun (me).

CHSE Odisha Class 12 Alternative English Solutions Prepositions

Kinds Of Prepositions

Prepositions can be classified as the following five types: such as:

  1.  Simple prepositions: as
    At, after, by, to, in, up, with, over, etc.
  2. Compound prepositions: as
    about, across, against, before, beside, into, until, within, etc.
  3. Participle prepositions: as
    Accepting, during, etc.
  4. Phrasal prepositions: as
    Along with, by virtue of instead of on account of by way of on behalf of according to, etc.
  5. Double prepositions: as
    from among, from beneath, from under, out of, etc.

Usage Of Prepositions

Prepositions are generally used before a noun phrase. But sometimes, they are used at the end of a sentence.
Example:
A. Before the noun phrase:

  • Don’t sit on that broken chair.
  • That by writes with his left hand.

B. At the end of a sentence:

  • Used in a wh-question sentence.
    (i) What is that animal like?
    (ii) What are you looking for?
  • Used in a sentence with a relative clause.
    (i) Is this the book you were looking for?
    (ii) I know the may you are talking about

PREPOSITION OF POSITION

Use of ‘AT’
‘At’ is used in
(a) Public places:
Examples:
1. There are no buses at the bus stop now.
2. She will be at the cinema now.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

(b Addresses:
Examples:
1. Mr. Patra is living at 24, Mahatma Gandhi Road, BBSR.
2. You can find him at plot no, 5, Basanti Colony, Rourkela.

(c) A point in space:
Examples:
1. The watchman is standing at the gate.
2. The child is sitting at his desk.
3. He is at home today.
4. The temple is at the end of the village.

(d) With events:
Examples:
1. I met the girl at my sister’s wedding.
2. They are busy at the meeting.
3. There was a big crowd at the match.
4. They were at the party at that time.

(e) A place on the journey:
Examples:
1. We got down at Puri
2. This train doesn’t stop at Ballikuda.
3. You have changed the bus at Bhubaneswar.

Use Of ‘In’

‘In’ is used with:
(a) Large areas:
Examples:
1. Jagannath temple is in Odisha.
2. He lives in a town.

(b) Enclosed places:
Examples:
1. They bathed in a pond.
2. My mother is in the kitchen.
3. Keep your money in the box.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

(c) Roads or starts:
Examples:
1. My uncle lives in Kedar Gouri Road.

(d) Places of work:
Examples:
1. My mother works in a hospital.
2. Mahesh works in a factory.
3. I work in an office.
4. He works in a hotel.
5. My brother works in a bank.

When it refers to an open place, ‘on’ is used instead of ‘in’.
Examples:
1. My father works on a firm.
2. Siba works on the railway.
3. Ajit works on a tea plantation.

In the case of indefinite office, ‘in’ is used, as
Examples:
1. Kunal works in a bank.
2. Puspa works in a library.

In the case of a definite office, ‘at’ is used.
Examples:
1. He works at the main branch of the Bank of India, Bhubaneswar.
2. Pinki works at the University library.

Use Of ‘On’

‘On’ is used with:
(a) Surface:
Examples:
1. The books is on the table.
2. The clock is on the walk.
3. The children are playing on the beach.
4. A man is standing on the roof of the house.
5. There is a beautiful picture of Nehru at page 5.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

(b) A-line:
Examples:
1. We line on a small river that flows into Chilika.
2. This house stands on the main road.

But ‘in’ and ‘at’ with buildings:
1. I left my umbrella in school.
2. My youngest daughter is at school and the eldest one is at college.
3. I spent the whole evening at the public library.
4. There is a big hall in the library.

‘In’, ‘on’, ‘all’ with streets or roads:
Examples:
1. Kuni got a job in Netajee Road.
2. My house is in Netajee Road.
3. You will find meat 65, Netajee Road.

Position And Movement:

Some prepositions relate to both position and movements. Others relate to either position or movement.
1. Use of ‘to’ and ‘at’:
Examples:
1. The boy went to the blackboard.
2. He stood at the blackboard.
3. He threw the ball at me.
At is used for both position and movement, but ‘to’ is used for only movement.

2. Use of ‘above’ and ‘over’
Examples:
1. The clock is over/above the door.
2. There is a temple above my house.
3. A kite is flying over our house (Not above)
Hence, over is used for both position and movement, but above is used only for the position.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

3. Use of ‘below’ and ‘under’:
Examples:
1. I stood at the door under/below the clock.
2. The children are playing under the tree. (Not below)

4. Use of ‘between’and‘morning’:
Examples:
1. Divide the cake between Kuni and Muni.
2. There is a beautiful garden between the house and the road.
3. The teacher is standing among the pupils.
4. Divide the sweets between the four children.

5. Use of ‘In front oF and behind:
Examples:
1. The car is in front of the bus.
2. The bus is behind the car.

6. Use of ‘into’:
Examples:
1. We jumped into the pond.
2. The teacher walked into the classroom.
3. Pour the milk into the cup.

7. Use of ‘out oF and ‘outside’:
Examples:
1. He is out of town.
2. When the class finished the children rushed out of the classroom
3. He asked me to sit on a bench outside the closed door.
4. Some people are shouting outside the office.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

8. Use of ‘onto’:
Examples:
The cat jumped onto the table.

9. Use of ‘up’ and ‘down’:
Example:
Mantu is going up the stairs, but Puspa is coming down the stairs.

10. Use of ‘along’, ‘across’ and ‘through’:
Examples:
1. There are trees all along the road.
2. The car is running along the road.
3. There is a bridge across the stream.
4. He swam across the river.
5. A road goes through the village.
6. The train went through the tunnel.

11. Use of ‘round’:
Example:
The earth moves round the sun.

12. Use of ‘against’:
Example:
He placed the ladder against the wall.

13. Use of ‘near’:
Example:
The supermarket is very near the library.

14. Use of ‘try’:
Example:
Come and sit by me.

15. Use of ‘beside’:
Example:
Sit beside your sister.

Exercise For Practice:

1. Fill in the blanks choosing ‘in’, ‘on’, or ‘at’.
1. I will meet you __________ the bus stand.
2. We live __________ a bit building.
3. My village is __________ a river bank.
4. My family lives __________ 24, Gandhi Road.
5. You will find a very tall building __________ the main road.
6. Sunita lives __________ Park Street.
7. There is a beautiful picture __________ this page.
8. You will find him __________ the garden now.
9. He is not __________ home now.
10. Someone has drawn a picture the door.
11. He will be in two years more __________ school before the goes to any college.
12. The temple is __________ the top of the hill.
13. She is a teacher __________ a village school.
14. We walked for a long time __________ the beach.
15. Do not walk __________ the middle of the road.
16. Many laborers are working __________ my firm.
17. I last saw you __________ this theatre.
18. Sushanta left his bag __________ the college.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

2. Fill in the blanks by choosing the appropriate prepositions:
1. He walked __________ my room to see me. (into/to)
2. A river flows __________ the hill. (under / below)
3. We will need about to get __________ the river. (across / through)
4. A fan is moving __________ my head (over / above)
5. He threw the book __________ the table. (to/into)
6. I felt __________ in my pocket and took out a coin (into / in)
7. The cat jumped __________ the mouse (on/into)
8. I thought someone was standing __________ me (in front of / behind)
9. We drove __________ the road for fifty miles (on/along)
10. Water flows __________the pipe (along / through)
11. The car turned __________ the comer of the road. (on / round)
12. He falls __________ the horse (down/ off)
13. All the five brothers were quarreling __________ themselves. (between/among)
14. The bird flew __________ the nest (from/out of)
15. The squirrel hid __________ a bush. (blow / under)
16. I came __________ the stairs to welcome my friend. (below / down)

Preposition of Time:

When the duration of time of a happening is expressed we use the preposition life for, during, from, to, till, until by, etc. These are called prepositions of time.
1. Use of ‘At’
‘At’ is used with:
(a) Clock time:
Examples:
1. The train left at 7.30.
2. He reached at half past nine.

(b) Exact moment of day or night:
Example:
1. It is hot at noon.
at sunrise, at sunset, at midnight, at dawn, at dusk, at midday, etc.

(c) Lunch time:
Example:
1. He will be here at lunchtime, at breakfast, at dinner, etc.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

(d) Festival, short holidays:
Example:
1. He goes home at Dusserah every year, at Christmas, at Id, at Holi, etc.

(e) Age:
Example:
1. He was admitted to our school at the age of five.

(f) Others:
Example:
1. The villain was killed at the end. at the moment etc.

Use of ‘On’
‘On’ is used with:
(a) Day of the week:
Example:
1. Offices and schools are closed on Sundays.

(b) Dates:
Examples:
1. Mahatma Gandhi was bom on 2nd October 1869.
2. My friend usually visits us on my birthday.

(c) Day + part of day:
Example:
1. I will go home on Sunday morning in the night of the 25th, on a came after-noon, etc.

(d) Special days:
Example:
1. My father gave me gifts on my birthday on New Year’s Day on Republic Day etc.

Use of ‘In’
‘In’ us used with:
(a) Week/month/season/year/century:
Examples:
1. He was absent from college in the first week.
2. The First World War began in 1914.
3. We are now in the 21 st century.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

(b) The time something takes to be completed:
Examples:
1. The chief guest finished his speech in five minutes.
2. The clerk typed the report in half an hour.

(c) Part of day/night:
Example:
1. I get up in the morning every day.
2. The doctor is free in the afternoon.

Use of ‘on time’ and ‘in time’:
Examples:
1. We reached the station in time to buy the tickets.
2. The train reached on time.

Use of at night and in the night:
Examples:
1. I cannot sleep at night.
2. I woke up on night.

Use of During:

(a) We use during with a period of time in which something took place for a comparatively short period of time.
Examples:
1. During the summer vacation, I was ill for ten days.
2. I had slept only for two hours during the last night.

(b) We use during with an event:
Examples:
1. He broke his leg during the match.
2. Somebody shouted during the meeting.
3. The woman wept during her speech.

Use of ‘For’:

We use for to say how long something continued.
Examples:
1. We lived there for thirty years.
2. My father works in the factory for eight hours.
3. She is staying only for a week.
4. Pintu was at college for three years.
5. I have been waiting for two hours.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

Use of ‘From… To’:

When it implies ‘for’, we use from… to.
Example:
1. My father works in the factory from 10 am to 6 pm. every day:

Use of Since And From:

We use (a) since and (b) from with the starting point of a period of time.
(a) Examples:
1. I have not seen you since June last.
2. The baby has been crying since morning.
(b) Examples:
1. Books will be on sale from tomorrow.
2. I hope they will be friends from now.
3. She was deaf from birth.
4. I worked from 7 o’clock.

Use of Until, Till, Upto / By, Before:

We use until, till, upto, by, and before with terminal points of a period of time. Examples:
1. You can keep the book until /till Friday.
2. We didn’t get up until/till morning.
3. They worked up to 3 pm.
N.B: Both until and till are used in positive and negative sentences, but up to is not used in negative sentences.

We can’t write:
They didn’t work upto 3 pm

Use of By:

It is used for ‘on’ or ‘before’ but not for later than Examples:
1. You should return home by the evening.
(= not later than the evening)
2. Can you send the book by Friday?
(= on or before Friday)

CHSE Odisha Class 12 Alternative English Solutions Prepositions

Use of Before:

It is used for earlier than:
Example:
1. You must see me before Friday.

Prepositions are also used for:
1. Cause:
Examples:
1. The patient died of cholera.
2. Smita was lined for the offense.
3. The man died from a wound.
4. She helped me out of kindness.
5. The student is sick with a cold.

2. Means:
Examples:
1. My uncle came by bus.
2. Pinkicameinatrain.
3. He broke the lock with a hammer.
4. The village was destroyed by fire.

3. Accompaniment:
Examples:
1. The child is playing with his friends.
2. Mr. Patra came without his wife.

4. Support and Opposition:
Examples:
1. I stand for my brother.
2. She is against me.
3. We are with you all the time.

5. Having, possession:
Example:
1. She is a girl with long hair.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

6. Concession:
Example:
1. We went out in spite of the heavy rain.

7. State/condition:
Change of State
Examples:
1. The road is under repair.
2. Translate the passage into English.

8. Aim / Goal / Purpose / Target Example:
1. I bought a pen for Raju.

9. Origin/Source:
Examples:
1. This road runs from Bhubaneswar to Puri.
2. I borrowed a hundred rupees from my friend.

10. Separation:
Examples:
1. The rich man was robbed of his money.
2. The cat jumped off the chair.

11. Relation/possession:
Examples:
1. He is a friend of mine.
2. What is the name of your village.

12. Material / Ingredient:
Examples:
1. The chair is made of wood.
2. Flour is made from wheat.

13. Part/portion:
Example:
1. This is the door of this house.

14. Manner:
Examples:
1. She is like her mother.
2. He treated me with kindness.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

15. Subject matter:
Examples:
1. She is reading a book in English grammar.
2. I told him about my institution.

Exercise For Practice

1. Use in, on or at where necessary.

1. We go to the cinema usually __________ Sundays.
2. Rajat will return __________ the evening.
3. Where were you last Saturday?
4. The train arrived __________ 7 o ’ clock __________ the morning.
5. Tikun was born __________ the month of September.
6. The sun sets __________ 5 pm __________December.
7. My sister’s wedding will take place __________ 15th May.
8. He is busy __________ the moment.
9. Can you come __________ this evening?
10. My mother goes to the temple __________ Monday morning.
11. I will finish my breakfast __________ two minutes.
12. The bus left __________ time.
13. I heard a loud noise __________ night.
14. We reached station __________ time to catch the train.
15. Our college will close __________ two weeks time.
16. What are you doing __________ tomorrow morning.
17. I was born __________ 1985.
18. We visit Puri __________every summer.
19. Don’t move out __________ noon. It is very hot.
20. I will see you __________ next Friday.

2. Choose the correct prepositions and fill in the blanks.

1. I have known Smita __________ two years. (for/since).
2. They have been at college __________ 2005. (from/since)
3. You must finish the work __________ tomorrow. (by/until)
4. We visited many places __________ the holidays. (between/during)
5. They kept working __________ evening? (until / by)
6. My friend will be here any day __________ Monday and Wednesday. (by/between)
7. He will wait for me __________ 5 o’clock. (until/ by)
8. What have you been doing __________ morning? (from/since)

CHSE Odisha Class 12 Alternative English Solutions Prepositions

Preposition With Verbs, Adjectives & Nouns
Verb + Preposition

  • accuse of – The police accused him of stealing a radio.
  • add to – If you add 3 to 4, you get 7.
  • admit to – Ajit is admitted to +2 Arts of our college.
  • agree about – We agree about most things.
  • agree to – I don’t agree to his suggestion.
  • agree on – I agree on a date to start our journey.
  • agree with – I agree with my friends to go on a picnic.
  • aim at – The hunter aimed at the bird.
  • apologize to/for – I apologize to you for not replying to your letter.
  • appeal to, for – I apologize to you for not replying to your letter
  • apply to for – The poor man applied to the bank for a loan.
  • approve of – A approve of your decision.
  • argue for – The workers argued for a rise in their pay.
  • arrive at – We arrived at the station on time.
  • arrive in – We arrived in Bhubaneswar to attend to the meeting.
  • ask for – The workers asked for a higher wage.
  • attend on – The queen had a good doctor attending to her.
  • attend to – Attend to what your teacher is saying.
  • avail of – You should avail yourself of the opportunity.
  • believe in – I believe in God.
  • belong to – This cost belongs to my father.
  • beware of – Beware of the dog.
  • blame for – Ramesh blamed his teacher for his failure.
  • blame on – He blamed his failure on his teacher.
  • boast of – A wise man never boasts of his knowledge.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

  • care about – I care a lot about my reputation.
  • care for – She cared for his father all through his long illness.
  • charge with – The police charged the man with murder.
  • compare to/with – The poet compares her face to/with the moon.
  • complain about – The old man never complains about/of/ against his
  • of/against – suffering.
  • compete with /against – Ranjit is competing with Ajit for the first position.
  • confide to – I confide my troubles to my friend.
  • confide in – I confide in his honesty.
  • congratulate on – We congratulate him on his success.
  • consent on – The parents consented to their daughter’s marriage.
  • consist of – The committee consists often of members.
  • consist in – Happiness consists in contentment.
  • cure of – The doctor can cure you of your illness.
  • consult with – The head minister consulted with other teachers about the development of the school.
  • deal in – The shop deals in dress materials only.
  • deal with – Itisveryhardtodealwithhim.
  • deprive of – A prisoner is deprived of his rights of freedom in jail.
  • die from – The poor man died from hunger / a wound.
  • die off – Many people died of Malaria.
  • differs from – French differs from English.
  • differ with on – I am sorry to differ with you on that question.
  • dream about/of – Soldiers often dream of/ about their homes.
  • escape from – The thief escaped from the jail.
  • exempt from – Textbooks are exempted from tax.
  • gaze at – We gazed at the stars.
  • hope for – Let us hope for the best.
  • impose on – We should not impose more opinions on others.
  • inform about/of Inform the police about/of the theft.
  • insist on/upon – The children insisted on/upon visiting the park.
  • Interfere in – It is not good to interfere in other ‘s private life.
  • interfere with – Don’t interfere with this machine.
  • knock at/on – The visitor knocked at/on the door.
  • listen to – I often listen to good music.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

  • made of – The chair is made of wood.
  • made from – Cheese is made from milk.
  • mock at – They mocked my shyness.
  • object to – I object to your smoking in this room.
  • part with – Little children don’t want to part with their toys.
  • pick up – The mother picked up the baby.
  • pick out – She was picked out from thousands of applicants for the job.
  • point to – Both the hour hand and minute hand pointed to twelve.
  • point at – The hunter pointed a gun at the bird.
  • point at – Point out the spelling mistakes.
  • protect out – An umbrella protects us from the sun and the rain.
  • preside of – The Prime Minister presides at the meetings of the cabinet.
  • preside over – The City Council is presided over by the Mayor.
  • protest against – People protested against the government’s new food policy.
  • remind about – Please remind me about my promise.
  • remind of – The photo reminds me of many school days.
  • recover from – I have recovered from my illness.
  • repent of – The man repented of his misdeeds.
  • restrain from – Restrain the child from mischief
  • result from – The loss resulted from his negligence.
  • result in – Fifty percent of road accidents result in heat injuries.
  • retire from – My other retired from service last year.
  • shout at – The traffic police shouted at the car driver.
  • succeed in – He succeeded in his attempt.
  • suffer from – The child is suffering from fever.
  • supply with – The cyclone-affected people were supplied with relief materials.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

  • suspect of – The whole class suspended Biju for stealing the wall clock.
  • sympathize with – We all sympathized with the unfortunate man.
  • wait for – I am waiting for my friend.
  • worry about – He is worried about his daughter’s marriage.
  • worry over – The mother is worried about the health of her sick child.
  • wonder at – The village boy wondered at the tall buildings of the city.

Adjective + Preposition

  • absent from – I was absent from college yesterday as I had a headache.
  • accustomed to – I am not accustomed to such work.
  • afraid of – Who is not afraid of snakes?
  • angry with – My other was angry with me for my rude behavior.
  • angry about – He was angry about his own foolishness.
  • anxious about – I am always anxious about my father’s health.
  • anxious for – I am anxious for my examination results.
  • ashamed of – Kunal is ashamed of his bad conduct.
  • astonished at – All were astonished at his unexpected behavior.
  • aware of – I was quite aware of this before I joined this party.
  • blind to – Mothers are usually blind to the faults of their children.
  • busy with – Pravakar is now busy with his friends.
  • capable of – This child is capable of handling computers.
  • careful about/with – The rich man is very careful about I money.
  • careful of – Be careful of the ice on the road.
  • careful with – You should be careful with that man, he is dangerous.
  • confident of – Our soldiers are confident of winning the war.
  • conscious of – The boy is very conscious of his manners when he is among the elders.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

  • different from – Your habits are different from mine.
  • delighted with – The teacher was delighted with my success.
  • disappointed with- I am disappointed with your examination rescue.
  • disappointed at – I was disappointed at not finding her at home.
  • eligible for – Only graduates are eligible for this job.
  • envious of – We should not be envious of others’ happiness.
  • equal to – A pound is roughly equal to 500 grains.
  • essential to – Vitamins are essential to good health.
  • essential for – Land is essential for food.
  • familiar with – I am familiar with this city.
  • familiar to – This city is familiar to me.
  • famous for – Alexandar was famous for his bravery.
  • fit for – He is fit for the job.
  • fond of – Children are fond of sweets I playing.
  • good at /bad at hopeless at – The boy is very good at / bad at mathematics.
  • free with – He is free with his money.
  • free of – They are happy to give their services free of change.
  • free from – He is now free from disease.
  • grateful to for – The blind man was gratefûlto the child for his help.
  • glad of – I am glad of your success.
  • good / kind / nice of- I was kind of you to help me.
  • guilty of – He was found guilty of murder.
  • ignorant of – They were ignorant of any events outside.
  • ill with – The baby is ill with influenza.
  • innocent of – The young man pleaded that he was innocent of the charges made against him.
  • interested in – Some students are not interested in mathematics.
  • jealous of – My neighbors are jealous of me for my fortune.
  • keen on – I am keen on buying a house.
  • proud of – We are proud of our brave soldiers.
  • pleased with – Are you pleased with your new car?
  • responsible for – Who is responsible for this tembleness?
  • responsible to – The cabinet is directly responsible to Parliament.
  • satisfied with / pleased with – The boy is satisfied with his performance.
  • similar to – Your face is similar to your mother’s.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

  • sorry for / about – We are sorry for/about all our mistakes.
  • successful in – The Indian soldiers were successful in defeating the Pakistanis.
  • superior to – Man is superior to animals.
  • sure of – Are you sure of his honesty?
  • surprised/shocked at – We were surprised at the news.
  • suspicious of – I am suspicious of the man’s intentions.
  • tired of – He was tired to walk. He wanted to rest.
  • useful for – Soap is useful for removing dirt.
  • useful to – This book is very useful to me.
  • weak in – I am weak in mathematics.
  • worthy of – You are worthy of all my praise.
  • worried about – The government is very worried about the defènce of the country.

Noun + Preposition

  • admiration for – I have a lot of admiration for my teachers.
  • affection for – He felt a great affection for the old man.
  • agreement with – I have an agreement with my friend to exchange books between us.
  • belief in – She has lost her belief in God.
  • a consequence of – You have lost your job as a consequence of your laziness and rudeness.
  • love for – A mother’s love for her children is genuine.
  • love of – A soldier’s love of his country can’t be questioned.
  • objection to – I have an objection to your going out at night.
  • proof of – Gifts of nature are proof of God’s love for man.
  • reaction to – My reaction to his proposal was favorable.
  • demand for – There is a lot of demand for sugar during festival times.
  • difficulty in – I face difficulty in teaching the child.
  • difficulty with – IamhavingdifficultywithmyneigJbour.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

  • discussion on about of – We had a discussion on the discipline of the college.
  • effect on – The medicine has no effect on him.
  • influence on – Teachers have a great influence on their students.
  • example of – Lal Bahadur was an example of patriotism and honesty.
  • knowledge of – Our teacher has a good knowledge of English.
  • reason for – You must give reasons for beating that boy.
  • relief from – A heavy shower gave us great relief from the unbearable heat.
  • remedy for – There is no remedy for cancer.
  • reply to – I will reply to this letter tomorrow.
  • restriction on – There is a restriction on people entering this building.
  • solution to/for – Can you suggest a solution to/for this problem?
  • sympathy for – I feel sympathy for this helpless child.
  • need for – There is no need for you to come here again.
  • congratulation – Congratulation on your excellent performance.
  • a rise / an increase in – There has been a rise / an increase in the price of sugar recently.
  • a fall/decrease in – There has been a fall/decrease in the supply of wheat to our state.
  • cause of – What is the cause of Malaria?

Exercise For Practice.

1. Fill in the blanks with the appropriate prepositions.

1. Most people believe __________ God.
2. You can’t depend __________ a selfish person.
3. We laughed __________ his silly jokes.
4. Our class consists __________ forty students.
5. The baby is suffering __________ fever.
6. Every year many people die __________ hunger.
7. Children should care __________ their old parents.
8. Do not throw stones __________ others.
9. I agree __________ you on this proposal.
10. The people accused the old man __________ murder.
11. This shop deals __________ food items.
12. The two brothers diffèr __________ each other in many ways.
13. I succeeded __________ solving the problems.
14. Children should not be deprived __________ the joy of playing.
15. The students are going __________ a picnic.
16. Our house is built __________ bricks.
17. The property was divided __________ four parts.
18. I am looking __________ a furnished house.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

2. Put in the correct prepositions.

1. I am not afraid __________ snakes.
2. You should be sorry __________ your misbehavior.
3. We are all angry __________ his carelessness.
4. Latamangeshkar is famous __________ her singing.
5. Ranjit is good __________ mathematics.
6. Smoking is bad __________ health.
7. I am aware __________ your difficulties.
8. The young dancer was satisfied __________ her performance.
9. The child was keen __________ having chocolate.
1o. The oldman is worried __________ his poor health.
11. It was very clever __________ you to avoid that boy.
12. My problems are similar __________ yours.
13. I am not used __________ bad language.
14. You are late __________ college.
15. I am weak __________ English.
16. He is blind __________ one eye.
17. He is busy __________ his office work.
18. I can’t rely __________ you in this matter.

3. Fill in the blanks with appropriate prepositions.

1. There has been an increase __________ the price of sugar.
2. He developed an interest __________ music in his childhood.
3. I have a lot of differences __________ you.
4. What is the difference __________ a car and a bus?
5. I have got an invitation __________ the dinner party
6. My method __________ working is different from yours.
7. She is an expert __________ dogs.
8. What is the cause __________ your worry?
9. We made a request __________ more money.
10. An accident caused damage __________ the car.
11. I don’t have sympathy __________ people like you.
12. My advice has no effect __________ him.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

Fixed Prepositions

What are fixed prepositions?
The prepositions ofthe prepositional phrases which are fixed expressions in order to express a definite meaning in a sentence are called fixed prepositions. Their forms remain unchanged.
Examples:
1. I was attached by a dog.
2. The news was sent by telegram.
3. I go to school by bicycle.

Here, in the above sentences, the prepositional phrases; by a dog, by telegram, by bicycle remain unchanged. Because the sentence nos. can’t be written as:-
The news was sent by telegram.
Similarly, the third sentence can’t be written as:
1. I go to school by bicycle.
2. Rather we can write like this I go to school on my / bicycle.
3. We can only write, by bicycle.

Examples of Fixed Expressions

Use of ‘At’:

‘At’ as a fixed preposition is used when it denotes the following meanings.
1. at ease: comfortably
Ex:-1 am sitting at ease in this armchair.
2. at heart: deep inside
Ex:- At heart, he is very kind.
3. at length: in great detail or for a long time
Ex:- The matter was discussed at length in the meeting.
4. at war / at peace: in a state of war/peace
Ex:- The two countries are not at war now, they are at peace.
5. at rest: not doing anything active, not worrying about anything.
Ex:- you should see him when he is at rest.
6. at first: at the beginning
Ex:- I did not like him at first, but not I do.
7. at last: It has happened after he has been waiting for it for a long time.
Ex:- At last, he found a girl for his son to merry.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

8. at play: playing
Ex:- There was a noise of children at play.
9. at short notice: to be done very soon without very much warning.
Ex:- He had to live his previous quarters at short notice.
10. at work : Ex:- He had to live his previous quarters at short notice, busy doing a particular activity.
Ex:- He had been at work on a book.
11. at sight: as soon as one saw.
Ex:- The police were ordered to shoot at sight.
12. at hand: near
Ex:- Is there a doctor at hand?
13. at least: the number or amount mentioned is a minimum, and the actual number or quantity is much greater.
Ex:- Lend me at least 50 rupees.
14. at once: immediately
Ex:- He received a telegram and left for home at once.
15. at present: at the time of speaking
Ex:- He is not doing anything at present.
16. at the end: Towards the last part of something.
Ex:- There was a test at the end of the lesson.
17. at times: occasionally
Ex:- My friend visits me at times.
18. at best: taking the most hopeful view.
Ex:- I can give you 500 rupees at best.
19. at all times: doing something always.
Ex: – He remains busy in his business at all times.
20. at full speed: the highest speed, but no increase of further speed is possible. Ex:- He drove his car at full speed.
21. at a profit: at least some income is possible.
Ex:- Today, he is at a profit in his business.
22. at a loss: no income is possible.
Ex:- I am at a loss in my business this year.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

Use of ‘By’:

1. by oneself – alone, without help from anybody.
Ex:- One can’t play tennis by oneself
2. by name – using the name of somebody.
Ex:- The teacher knows all his students by name.
3. by accident/chance – meet someone or something accidentally.
Ex:- I only found it by accident/chance.
4. by heart: remember something entirely.
Ex:- Learn this poem by heart.
5. by mistake: as the result of carelessness and forgetfulness.
Ex:- I took your bag instead of mine by mistake.
6. by sight: know by appearance only, not as an acquaintance.
Ex:- I only know her by sight.
7. by surprise: unexpectedly
Ex:- The Cuttack was made by surprise.
8. by letter/telephone: to intimate news.
Ex:- The news was sent to him by letter.
9. by means of: doing something by its help.
Ex:- He stood first by means of hard labor.
10. by no means: no way of doing something.
Ex:- It is not possible to do by no means.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

11. by din’t of: doing something through it.
Ex:- He passed the examination by didn’t of hard labor.
12. by all means: certainly
Ex:- A graduate can complete all the tests by all means.
13. by hook or by crook: by fair means or foul
Ex:- The cunning people gain their ends by hook or by crook.
14. by and by gradually
Ex:- This medicine will show its effect by and by.
15. by way of: in the shape of
Ex:- He passed a silly remark on his beloved by way of a joke.
16. by force of: by virtue of
Ex:- The Indian team wins the hockey match by force of their superior game.

Use of ‘In’:

1. in brief: in a few words
Ex:- Tell me the whole story in brief.
you can’t make it happens quicker, it will happen when the time is
2. in due course: right font.
Ex:- Every home will have a computer in due course.
3. in general: in most cases, usually
Ex:- Many think that in general men are more hard-working than women.
4. in pieces: broken
Ex:- The mirror is in pieces.
5. in tears: is crying
Ex:- She was in tears when she heard about her sister’s death.
6. in case: fit so happens
Ex:- In case I forget, please remind me about my promise.
7. In fact: really, actually
Ex:- I do not like him, in fact, I hate him.
8. in public / in private: You say or do something when a group of people you do not know are present.
Ex:- He repeated in public what he said in private.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

9. in order: the condition on which everything is carefully arranged.
Ex:- Please keep your things in order.
Everything in the room is in order.
10. in common: having the same interests, characteristics features to make friends with.
Ex:- Both India and Pakistan have many things in common.
11. inflame: formally
Ex:- In the Koraput district some schools are rarely open, they exist in the name on.
12. in turn: In succession, one after the other.
Ex:- The patients saw the doctor in turn.

Other examples:
in-hospital/bed/prison/jail in all
in the sun/rain in the air
in trouble/danger/debt/difficulty in theory/practice in fun

Use of ‘Out of’:

1. out of print: not in print
Ex:- This book is out of print.
2. out of work: has no job
Ex:- He has been out of work for 10 years.
3. out of fashion: not fashionable
Ex:- Tight jeans are out of fashion now.
4. out of danger: not in danger
Ex:- The patient is now out of danger.
5. out of reach: Can’t be reached / not within reach.
Ex:- The picture on the wall is out of reach.
6. out of place: not suitable for a particular situation or occasion.
Ex:- I felt completely out of piece among all those smart rich people.
out of the question: not allowed, not possible
Ex:- You can’t go to the wedding in that old shirt, it is quite out of the question.
8. out of stock: no stock
Ex:- We haven’t any more wheat, we’re out of stock.
9. out of turn: before or after the permitted time.
Ex:- You must not speak out of turn.
10. out of doors: not inside a building but in the open air.
Ex:- Farmers spend most of their time out of doors.

CHSE Odisha Class 12 Alternative English Solutions Prepositions

More examples:
1. on the radio: relayed through radio
Ex:- I heard this news on the radio.
2. on television: news telecast on TV.
Ex:- You should have watched the 7.30 p.m. news on television today.
3. on fire: something is burning
Ex:- The house is on fire.
4. on sale: about to be sold
Ex:- These old books are on sale.
5. on duty: in the working hour
Ex:- The policeman is on duty.
6. on time: the reference of the perfect moment.
Ex:- The train arrived on time.

Exercise For Practice

1. Fill in the blanks with the correct prepositions.
1. The whole house was _________ fire (in/on/under)
2. I have got this book _________ loan from the library (by/on / under)
3. The patient is _________ treatment in hospital (under/ in, an)
4. I sold my old bike _________ a profit (at /on/with)
5. Mary is a Christian _________ birth. (in/with / by)
6. She has made this toy _________ hand (with/by/ through)
7. Minu’s father is a doctor _________ profession (in/by/ at)
8. We could not solve the problem _________ first, but we were able to do _________ it the end. (in/at/before)
9. The porter earns fifty rupees a day _________ an average. (on/by/with)
10. Father goes to his office _________ his car (with/in / by)
11. The policeman is _________ duty now (with/on / by)
12. The man sitting _________ my right is an engineer (to/on/ towards)

CHSE Odisha Class 12 Alternative English Solutions Prepositions

13. Please keep you things _________ order. (in/on/at).
14. Don’t disturb him, he is _________ work. (in/at/on)
15. I have a lot of work _________ on hand. (at/out/of)
16. Rice is _________ stock in the market. (with/out / of)
17. The thief is _________ prison now (in/inside/within)
18. My eldest daughter is _________ university now (in/at/inside)
19. Please send the letter_________ post (in/by/with)
20. You must keep your knowledge _________ date (out of! upto)
21. The crirninalis _________ arrest. (in/under at)
22. He is worthy _________ my praise. (to/of/for)
23. He is cure _________ his success. (fon in/of)
24. He is weak _________ mathematics (in/by/to)