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Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 11 Straight Lines Ex 11(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(a)

Question 1.
Find the distance between the following pairs of points.
(i) (3, 4), (-2, 1);
Solution:
Distance between points (3, 4) and (-2, 1) is
\(\sqrt{(3+2)^2+(4-1)^2}=\sqrt{25+9}=\sqrt{34}\)

(ii) (-1, 0), (5, 3)
Solution:
The distance between the points (-1, 0) and (5, 3) is
\(\sqrt{(-1-5)^2+(0-3)^2}\)
= \(\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}\)

Question 2.
If the distance between points (3, a) and (6, 1) is 5, find the value of a.
Solution:
Distance between the points. (3, a) and (6, 1) is
\(\sqrt{(3-6)^2+(a-1)^2}=\sqrt{9+(a-1)^2}\)
∴ \(\sqrt{9+(a-1)^2}=5\)
or, 9 + (a – 1)² = 25
or, (a – 1)² = 16
or, a – 1 = ± 4
a = 1 ± 4 = 5 or, – 3

Question 3.
Find the coordinate of the points which divides the line segment joining the points A (4, 6), B (-3, 1) in the ratio 2: 3 internally. Find also the coordinates of the point which divides \(\overline{\mathbf{A B}}\) in the same ratio externally.
Solution:

Question 4.
Find the coordinates of the mid-point of the following pairs of points.
(i) (-7, 3), (8, -4);
Solution:
Mid-point of the line segment joining the points (-7, 3) and (8, -4) are \(\left(\frac{-7+8}{2}, \frac{3-4}{2}\right)=\left(\frac{1}{2},-\frac{1}{2}\right)\)

(ii) (\(\frac{3}{4}\), -2), (\(\frac{-5}{2}\), 1)
Solution:
Mid-point of the line segment joining the points. (\(\frac{3}{4}\), -2) and (\(\frac{-5}{2}\), 1) is,
\(\left(\frac{\frac{3}{4}-\frac{5}{2}}{2}, \frac{-2+1}{2}\right)=\left(\frac{-7}{8}, \frac{-1}{2}\right)\)

Question 5.
Find the area of the triangle whose vertices are (1, 2), (3, 4) (\(\frac{1}{2}\), \(\frac{1}{4}\))
Solution:
Area of the triangle whose vertices are (1, 2), (3, 4) and (\(\frac{1}{2}\), \(\frac{1}{4}\)) is

Question 6.
If the area of the triangle with vertices (0, 0), (1, 0), (0, a) is 10 units, find the value of a.
Solution:
Area of the triangle with vertices (0, 0),(1,0), (0, a), is \(\frac{1}{2}\) × 1 × a = \(\frac{a}{2}\)
∴ \(\frac{a}{2}\) = 10 or a = 20

Question 7.
Find the value of a so that the points (1, 4), (2, 7), (3, a) are collinear.
Solution:
As points (1, 4), (2, 7), (3, a) are collinear, we have the area of the triangle with vertices (1, 4), (2, 7), and (3, a) is zero.
∴ \(\frac{1}{2}\) {1(7 – a) + 2(a – 4) + 3 (4 – 7)} = 0
or, 7 – a + 2a – 8 + 12 – 21 =0
⇒ a = 10

Question 8.
Find the slope of the lines whose inclinations are given.
(i) 30°
Solution:
The slope of the line whose inclination is 30°.
tan 30° = \(\frac{1}{\sqrt{3}}\)

(ii) 45°
Solution:
Slope = tan 45° = +1

(iii) 60°
Solution:
Slope = tan 60° = √3

(iv) 135°
Solution:
Slope = tan 135° = – 1

Question 9.
Find the inclination of the lines whose slopes are given below.
(i) \(\frac{1}{\sqrt{3}}\)
Solution:
The slope of the line is \(\frac{1}{\sqrt{3}}\)
∴ tan θ = \(\frac{1}{\sqrt{3}}\) or, θ = 30°
∴ The inclination of the line is 30°

(ii) 1
Solution:
Slope = 1 = tan 45°
∴ The inclination of the line is 45°.

(iii) √3
Solution:
Slope = √3 = tan 60°  ∴ θ = 60°
∴ Inclination = 60°

(iv) – 1
Solution:
Slope = – 1 = tan 135°
∴ Inclination = 135°

Question 10.
Find the angle between the pair of lines whose slopes are ;
(i) \(\frac{1}{\sqrt{3}}\), 1
Solution:

(ii) √3, -1
Solution:

Question 11.
(a) Show that the points (0, -1), (-2, 3), (6, 7), and (8, 3) are vertices of a rectangle.
Solution:


∴ The opposite sides are equal and two consecutive sides are perpendicular. So it is a rectangle.

(b) Show that the points (1, 1), (-1, -1), and (-√3, √3) are the vertices of an equilateral triangle.
Solution:

Question 12.
Find the coordinates of the point P(x, y) which is equidistant from (0, 0), (32, 10), and (42, 0).
Solution:

Question 13.
If the points (x, y) are equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

Question 14.
The coordinate of the vertices of a triangle are (α₁, β₁), (α₂, β₂), and (α₃, β₃). Prove that the coordinates of its centroid is \(\left(\frac{\alpha_1+\alpha_2+\alpha_3}{3}, \frac{\beta_1+\beta_2+\beta_3}{3}\right)\)
Solution:

Question 15.
Two vertices of a triangle are (0, -4) and  (6, 0). If the medians meet at the point (2, 0), find the coordinates of the third vertex.
Solution:

∴ \(\frac{6+x}{3}\) = 2, \(\frac{-4+y}{3}\) = 0
⇒ x = 0, y = 4
∴ The coordinates of the 3rd vertex are (0, 4).

Question 16.
If the point (0, 4) divides the line segment joining(-4, 10) and (2, 1) internally, find the point which divides it externally in these same ratios.
Solution:

Question 17.
Find the ratios in which the line segment joining (-2, -3) arid (5, 4) is divided by the coordinate axes and hence find the coordinates of these points.
Solution:

Question 18.
In a triangle, one of the vertices is at (2, 5) and the centroid of the triangle is at (-1, 1). Find the coordinates of the midpoint of the side opposite to the given angular point.
Solution:

Question 19.
Find the coordinates of the vertices of a triangle whose sides have midpoints at (2, 1), (-1, 3), and (-2, 5).
Solution:

∴ x₂ + x₃ – 4 or, x₂ – 4 – 3 = 1
∴ x₁ = – 4 – x₂ = -4 – 1 = -5
Similarly y₁ + y₂ + y₃ = 5 + 1 + 3 = 9
As y₁ + y₂ = 10
we have y₃ = 9 – 10 = – 1
Again y₁ + y₃ = 6
or, y₁ = 6 – y₃ = 6 + 1 = 7
and y₂ = 10 – y₁ = 10 – 7 = 3
∴ The coordinates of A, B, and C are (-5, 7), (1, 3), and (3, -1).

Question 20.
If the vertices of a triangle have their coordinates given by rational numbers, prove that the triangle cannot be equilateral.
Solution:
Let us choose the contradiction method. Let the triangle is equilateral if the co¬ ordinate of the vertices is rational numbers.
Let ABC be an equilateral triangle with vertices A (a, 0), B (a, 0), and C (b, c) where a, b, c are rational.

⇒ a² = b² + c² = \(\frac{a^2}{4}\) + c²
⇒ c² =  a² – \(\frac{a^2}{4}\) = \(\frac{3a^2}{4}\) ⇒ c = \(\frac{\sqrt{3}}{2}\) a     ….(2)
Now b = \(\frac{a}{2}\), c = \(\frac{\sqrt{3}}{2}\) a
If a is rational then b is rational but c is irrational, i.e., the coordinates of the vertices are not rational, which contradicts the assumption.
Hence assumption is wrong.
So the triangle cannot be equilateral if the coordinate of the vertices is rational numbers.

Question 21.
Prove that the area of any triangle is equal to four times the area of the triangle formed by joining the midpoints of its sides.
Solution:

∴ The area of triangle ABC is four times the area of triangle DEF. (Proved)

Question 22.
Find the condition that the point (x, y) may lie on the line joining (1, 2) and (5, -3).
Solution:

∴ As points A, B, and C are collinear, we have the area of the triangle ABC as 0.
∴ \(\frac{1}{2}\) {1(-3 – y) + 5(y – 2) + x(2 + 3)} = 0
or, – 3 – y + 5y – 10 + 5x = 0
or, 5x + 4y = 13

Question 23.
Show that the three distinct points (a², a), (b², b), and (c², c) can never be collinear.
Solution:
Area of the triangle with vertices (a², a), (b², b) , and (c², c) is
\(\frac{1}{2}\) {a²(b – c) + b²(c – a) c²(a – b)}
= (a – b)(b – c)(a – c)
which is never equal to zero except when a = b = c, hence the points are not collinear.

Question 24.
If A, B, and C are points (-1, 2), (3, 1), and (-2, -3) respectively, then show that the points which divide BC, CA, and AB in the ratios (1: 3), (4: 3) and (-9: 4) respectively are collinear.
Solution:
Let the points P, Q, and R divides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\), in \(\overline{\mathrm{AB}}\) the ratio 1: 3, 4: 3 and -9: 4

Question 25.
Prove analytically :
(a) The line segment joining the midpoints of two sides of a triangle is parallel to the third and half of its length.

Solution:
Let the coordinates of the triangle ABC be (x₁, y₁), (x₂, y₂) and (x₃, y₃)
The points D and E are the midpoints of the sides \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AC}}\)

(b) The altitudes of a triangle are concurrent.
Solution:

(c) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:

(d) An angle in a semicircle is a right angle.
Solution:

∴ The angle in a semicircle is a right angle. (Proved)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 10 Sequences and Series Ex 10(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 10 Sequences and Series Exercise 10(a)

Question 1.
Which of the following in a sequence?
(i) f(x) = [x], x ∈ R
(ii) f(x) = |x|, x ∈ R
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N
Solution:
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N is a sequence f(n) : N → X, X ⊂ R.

Question 2.
Determine if (t_n) is an arithmetic sequence if :
(i) t_n = an² + bn
Solution:
t_n = an² + bn
⇒ t_n+1 = a(n + 1)² + b(n – 1)
⇒ t_n+1 – t_n = a{(n + 1)² – n²} + b{n + 1 – n}
= a(2n + 1) + b
Which is not independent of n.
∴ (t_n) is not an A.P.

(ii) t_n = an + b
Solution:
t_n = an + b
⇒ t_n+1 = a(n + 1) + b
Now t_n+1 – t_n
= {a(n + 1) + b} – {an + b}
= a (constant)
∴ (t_n) is an arithmetic sequence.

(iii) t_n = an² + b
Solution:
t_n = an² + b
⇒ t_n+1 = a(n + 1)² + b
∴ t_n+1 – t_n = a[(n + 1)² – n²] + b – b
= a(2n + 1)
(does not independent of n)
∴ (t_n) is not an arithmetic sequence.

Question 3.
If a geometric series converges which of the following is true about its common ratio r?
(i) r > 1
(ii) -1 < r < 1
(iii) r > 0
Solution:
(ii) -1 < r < 1

Question 4.
If an arithmetic series ∑tn converges, which of the following is true about t_n?
(i) t_n < 1
(ii) |t_n| < 1
(iii) t_n = 0
(iv) t_n → 0
Solution:
(iii) t_n = 0

Question 5.
Which of the following is an arithmetic-geometric series?
(i) 1 + 3x + 7x² + 15x³+ ….
(ii) x + \(\frac{1}{2}\)x + \(\frac{1}{3}\)x² + ….
(iii) x + (1 + 2)x² + (1 + 2 + 3)x³ +…
(iv) x + 3x² + 5x³ + 7x⁴ + …
Solution:
(iv) x + 3x² + 5x³ + 7x⁴ + … is an arithmetic geometric series with a = 1, d = 2, r = x.

Question 6.
For an arithmetic sequence (t_n) t_p = q, t_q = p, (p ≠ q), find t_n.
Solution:
t_p = q ⇒ a + (p – 1)d = q    ……(1)
t_q = p ⇒ a + (q – 1)d = p    ……(2)
From (1) and (2) we have (p – q)d = q – p
⇒ d = (-1)
Putting d = (-1) in (1)
we have a = p + q – 1
∴ t_n = a + (n – 1)d
= (P + q – 1) + (n – 1) (-1)
= p + q – n

Question 7.
For an arithmetic series, ∑a_n S_p = q and S_q = p (p ≠ q) find S_p+q
Solution:
S_p = q and S_q = p

Question 8.
The sum of a geometric series is 3. The series of squares of its terms have a sum of 18. Find the series.
Solution:

Question 9.
The sum of a geometric series is 14, and the series of cubes of its terms have a sum of 392 find the series.
Solution:


∴ The series is \(\sum_{n=1}^{\infty} \frac{7}{2^{n-1}}\)

Question 10.
Find the sum as directed
(i) 1 + 2a + 3a² + 4a³ + …..(first n terms(a ≠ 1))
Solution:

(ii) 1 + (1 + x)y + (1 + x + x²)y² + …..(to infinity)
Solution:
Let S_∞ = 1 + (1 + x)y + (1x + x²)y² + …
⇒ S_n = 1 + (1 + 1 + x)y + (1 + x + x²)y² + ……+(1 + x + …. x^n-1)y^n-1

(iii) 1 + \(\frac{3}{5}\) + \(\frac{7}{25}\) + \(\frac{15}{125}\) + \(\frac{31}{625}\) + …..(to infinity)
Solution:

(iv) 1 + 4x + 8x² + 13x³ + 19x⁴ + …..(to infinity). Assuming that the series has a sum for |x| < 1.
Solution:

(v) 3.2 + 5.2² + 7.2³ + …..(first n terms)
Solution:
Sn = 3.2 + 5.22 + 7.3 + ….n terms = 2[3 + 5.2 + 7.22 + ….n terms]

Question 11.
Find the sum of the infinite series.
(i) \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots\)
Solution:

= 1 – \(\frac{1}{n+1}\)
∴ \(S_{\infty}=\lim _{n \rightarrow \infty} S_n=1\)

(ii) \(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)
Solution:
\(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)

(iii) \(\frac{1}{2 \cdot 5 \cdot 8}+\frac{1}{5 \cdot 8 \cdot 11}+\frac{1}{8 \cdot 11 \cdot 14}+\ldots\)
Solution:
Here t_n = \(\frac{1}{(3 n-1)(3 n+2)(3 n+5)}\)
The denominator of t_n is the product of 3 consecutive terms of A.P. Now multiplying and dividing by (3n + 5) – (3n – 1) we have
\(t_n=\frac{(3 n+5)-(3 n-1)}{6(3 n-1)(3 n+2)(3 n+5)}\)

(iv) \(\frac{3}{1^2 \cdot 2^2}+\frac{5}{2^2 \cdot 3^2}+\frac{7}{3^2 \cdot 4^2}+\ldots\) [Hint : take t_n = \(\frac{2 n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}\)]
Solution:

(v) \(\frac{1}{1 \cdot 5}+\frac{1}{3 \cdot 7}+\frac{1}{5 \cdot 9}+\ldots .\)
Solution:

Question 12.
Find S_n for the series.
(i) 1.2 + 2.3 + 3.4 + ….
Solution:

(ii) 1.2.3 + 2.3.4 + 3.4.5 + …
Solution:

(iii) 2.5.8 + 5.8.11 + 8.11.14 +…
Solution:

(iv) 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + …
[Hint : t_n = (3n – 1) (3n + 2)(3n + 5)]
Solution:

(v) 1.5 + 2.6 + 3.7 + …
[Hint: t_n = n(n + 4) is not a product of two successive terms of an A.P. for the term following n should be n+1, not n+4. So the method of previous exercises is not applicable. Instead, write t_n = n² + 4n and find S_n = \(\sum_{k=1}^n k^2+4 \sum_{k=1}^n k\) applying formulae]
Solution:

(vi) 2.3 + 3.6 + 4.11 + …
[Hint : Take t_n = (n + 1) (n² + 2)]
Solution:

(vii) 1.3² + 2.5² + 3.7² + ….
Solution:

Question 13.
Find the sum of the first n terms of the series:
(i) 5 + 6 + 8 + 12 + 20 + …
Solution:

(ii) 4 + 5 + 8 + 13 + 20 + …
Solution:

Question 14.
(i) Find the sum of the product of 1,2,3….20 taken two at a time. [Hint: Required sum = \(\frac{1}{2}\left\{\left(\sum_{k=1}^{20} k\right)^2-\sum_{k=1}^{20} k^2\right\}\)]
Solution:
We know that
(x₁ + x₂ + x₃ + …. + x_n)²
= (x²₁ + xⁿ₂ + … + x²_n) + 2 (Sum of all possible Products taken two at a time)
∴ 2 (Sum of products of 1. 2, 3,…… 20 taken two at a time)
= (1 + 2 + 3 + … 20)² – (1² + 2² + … + 20²)
\(=\left(\frac{20 \times 21}{2}\right)^2-\frac{20(20+1)(40+1)}{6}\)
= (210)² – 70 × 41
= 44100 – 2870 = 41230
∴ The required sum = \(\frac{41230}{2}\) = 20615

(ii) Do the same for 1, 3, 5, 7,….19.
Solution:
Here the required sum

Question 15.
If a = 1 + x + x² + ….. and b = 1 + y + y² + ….|x| <  1 and |y| <  1, then prove that 1 – xy + x²y² + x³y³ + ….. =  \(\frac{a b}{a+b-1}\)
Solution:

Question 16.
If a, b, c are respectively the pth, qth, rth terms of an A.P., then prove that a(q – r) + b(r – p) + c(p – q) = 0
Solution:
Let the first term of an A.P. = A and the common difference = D.
According to the question
t_P = a, t_q = b, t_r = c
⇒ A + (p – q)D = a      …..(1)
A + (q – 1)D = b           …..(2)
A + (r – 1)D = c            …..(3)
L.H.S = a(q – r) + b (r – p) + c (p – q)
= (A + (p – 1)D) (q – r) + (A + (q – 1)D)
(r- p) + (A + (r – 1)D) (p – q)
= A (q – r + r – p + p – q) + D [(p – 1)
(q – r) + (q – 1) (r – p) + (r – 1) (p – q)]
= D (pq – pr + qr – pq + pr – qr) – D (q – r + r – p + p – q) = 0

Question 17.
If \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P. and a + b + c ≠ 0, prove that \(\frac{\mathbf{b}+\mathbf{c}}{\mathbf{a}}, \frac{\mathbf{c}+\mathbf{a}}{\mathbf{b}}, \frac{\mathbf{a}+\mathbf{b}}{\mathbf{c}}\) are in A.P.
Solution:

Question 18.
If a², b², c² are in A.P. Prove that \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P.
Solution:

Question 19.
If \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{\mathbf{a}+\mathbf{b}}{c}\) are in A.P.,prove that \(\frac{\mathbf{1}}{\mathbf{a}}, \frac{\mathbf{1}}{\mathbf{b}}, \frac{\mathbf{1}}{\mathbf{c}}\) are inA.P.given a + b + c ≠ 0
Solution:

Question 20.
If (b – c)², (c – a)², (a – b)² are in A.P., prove that \(\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}\) are in A.P.
Solution:

Question 21.
If a, b, c are respectively the sum of p, q, r terms of an A.P., prove that \(\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\) = 0
Solution:

Question 22.
If a, b, c,d are in G.P., prove that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².
Solution:
Let a, b, c, and d are in G.P.
Let the common ratio = r
⇒ b = ar, c = ar², d = ar³
LHS = (a² + b² + c²) (b² + c² + d²)
= (a² + a²r² + a²r⁴) (a²r² + a²r⁴ + a²r⁶)
= a⁴r² (1 + r² + r⁴)²
= (a²r + a²r³ + a²r⁵)²
= (a.ar + ar.ar² + ar².ar³)²
= (ab + bc + cd)² = R.H.S. (proved)

 

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(c)

Question 1.
Compute the following :
(i) ¹²C₃
Solution:
¹²C₃ = \(\frac{(12) !}{3 ! 9 !}=\frac{12 \cdot 11 \cdot 10}{3 \cdot 2}\) = 220

(ii) ¹⁵C₁₂
Solution:
¹⁵C₁₂ = \(\frac{(15) !}{(12) ! 3 !}=\frac{15 \cdot 14 \cdot 13}{3 \cdot 2}\)
= 5.7.13 = 455

(iii) ⁹C₄ + ⁹C₅
Solution:
⁹C₄ + ⁹C₅ = \(\frac{9 !}{4 ! 5 !}+\frac{9 !}{5 ! 4 !}\)
\(=\frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1}\) × 2 = 252

(iv) ⁷C₃ + ⁶C₄ + ⁶C₃
Solution:
⁷C₃ + ⁶C₄ + ⁶C₃ = ⁷C₃ + ⁶C₄ + ⁶C4-1
= ⁷C₃ + ⁶⁺¹C₄ = ⁷C₃ + ⁷C₄
(∴ ⁿc_r + ⁿC_r-1– = ⁿ⁺¹c_r)
= ⁷C₄ + ⁷C_4-1 = ⁷⁺¹C₄
= ⁸C₄ = \(\frac{8 !}{4 !(8-4) !}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}\) = 70

(v) ⁸C₀ + ⁸C₁ + …….. + ⁸c₈
Solution:
⁸C₀ + ⁸C₁ + …….. + ⁸c₈ = 2⁸ = 256

Question 2.
Solve :
(i) ⁿC₄ = ⁿC₁₁ ;
Solution:
ⁿC₄ = ⁿC₁₁ ;  (∴ n = 4 + 11 = 15)

(ii) ²ⁿC₃ : ⁿC₃ = 44: 5
Solution:
²ⁿC₃ : ⁿC₃ = \(\frac{44}{5}\)
⇒ \(\frac{2 n !}{2 n-3 !} / \frac{n !}{n-3 !}=\frac{44}{5}\)

Question 3.
Find n and r if ⁿP_r = 1680, ⁿC_r = 70.
Solution:
ⁿP_r = 1680, ⁿC_r = 70
∴ \(\frac{{ }^n \mathrm{P}_r}{{ }^n \mathrm{C}_r}=\frac{1680}{70}\)
or, r ! = 24 = 4!
∴ r = 4
Again, ⁿC_r = 70 or ⁿC₄ = 70
or, \(\frac{n !}{4 !(n-4) !}=70\)
or, n(n – 1) (n – 2) (n – 3)
= 70 × 4! = 7 × 10 × 4 × 3 × 2
= 8 × 7 × 6 × 5
or, n(n – 1) (n – 2) (n – 3)
or, 8(8 – 1) (8 – 2) (8 – 3)
∴ n = 8

Question 4.
How many diagonals can an n-gon(a polygon with n sides) have?
Solution:
A polygon of n – sides has n vertices.
∴ The number of st. lines joining the n-vertices is ⁿC₂.
∴ The number of diagonals is ⁿC₂ – n

Question 5.
If a set A has n elements and another set B has m elements, what is the number of relations from A to B?
Solution:
If |A| = n, |B| = m
then |A × B| = mn
∴ The number of possible subsets of
A × B = 2^mn
∴ The number of relations from A to B is 2^mn.

Question 6.
From five consonants and four vowels, how many words consist of three consonants and two vowels?
Solution:
Words of consisting of 3 consonants and 2 vowels are to be formed from five consonants and 4 vowels.
∴ The number of ways = ⁵C₃ × ⁴C₂
Again, 5 letters can be arranged among themselves in 5! ways.
∴ The total number of ways
= ⁵C₃ × ⁴C₂ × 5! = 10 × 6 × 120 = 7200.

Question 7.
In how many ways can a committee of four gentlemen and three ladies be formed out of seven gentlemen and six ladies?
Solution:
A committee of 4 gentlemen and 3 ladies is to be formed out of 7 gentlemen and 6 ladies.
∴ The number of ways in which the committee can be formed.
⁷C₄ × ⁶C₃ = \(\frac{7 \cdot 6 \cdot 5}{3.2} \times \frac{6 \cdot 5 \cdot 4}{3 \cdot 2}\) = 700

Question 8.
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily. In how many ways can this be done? Find also the number of ways such that at least 3 black balls can be drawn.
Solution:
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily.
∴ The number of ways in which balls are drawn \({ }^9 \mathrm{C}_6=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}\) = 84 as the total number of balls is 9. If at least 3 black balls are drawn, then the drawing can be made as follows.

Black(4)	White(5)
3	3
4	2

The number of ways in which at least 3 black balls are drawn
= (⁴C₃ × ⁵C₃) + (⁴C₄ × ⁵C₂)
= (4 × 10) + (1 × 10) = 50

Question 9.
How many triangles can be drawn by joining the vertices of a decagon?
Solution:
A decagon has 10 vertices and 3 noncollinear points are required to be a triangle.
∴ The number of triangles formed by the joining of the vertices of a decagon is
¹⁰C₃ = \(\frac{10 !}{3 ! 7 !}=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}\) = 120

Question 9.
How many triangles can be drawn by joining the vertices and the center of a regular hexagon?
Solution:
A regular hexagon has six vertices. Triangles are to be formed by joining the vertices and center of the hexagon. So there is a total of 7 points. So the number of triangles formed.
⁷C₃ = \(\frac{7 !}{3 ! 4 !}=\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}\) = 35
As the hexagon has 3 main diagonals, which pass through the center hence can not form 3 triangles.
∴ The required number of triangles 35 – 3 = 32

Question 11.
Sixty points lie on a plane, out of which no three points are collinear. How many straight lines can be formed by joining pairs of points?
Solution:
Sixty points lie on a plane, out of which no. 3 points are collinear. A straight line required two points. The number of straight lines formed by joining 60 points is
⁶⁰C₂ = \(\frac{60 !}{2 \times 58 !}=\frac{60 \times 59}{2}\) = 1770

Question 12.
In how many ways can 10 boys and 10 girls sit in a row so that no two boys sit together?
Solution:
10 boys and 10 girls sit in a row so that no two boys sit together. So a boy is to be seated between two girls or at the two ends of the row. So the boys are to be sitted in 11 positions in ¹¹C₁₀ ways. Again 10 boys and 10 girls can be arranged among themselves in 10! and 10! ways respectively.
∴ The total number of ways = ¹¹C₁₀ × 10! × 10! = (11)! × (10)!

Question 13.
In how many ways can six men and seven girls sit in a row so that the girls always sit together?
Solution:
Six men and seven girls sit in a row so that the girls always sit together. Considering the 7 girls as one person, there are a total of 7 persons who can sit in 7! ways. Again the 7 girls can be arranged among themselves in 7! ways.
∴ The total number of arrangements
= 7! × 7!
= (7!)²

Question 14.
How many factors does 1155 have that are divisible by 3?
Solution:

∴ In order to be a factor of 1155 divisible by 3, we have to choose one or two of 5, 7, and 11 along with 3 or 3 alone.
∴ The number of ways = ³C₁ + ³C₂ + ³C₀ = 2³ – 1 = 7
∴ The number of factors is 7 excluding 1155 itself.

Question 15.
How many factors does 210 have?
Solution:

∴ We can choose at least one 2, 3, 5, or 7  to be a factor of 210.
∴ The number of factors.
= ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 2⁴ – 1 = 15
∴ The number of factors is 210 is 15. (Including 215 itself and excluding 1).

Question 16.
If n is a product of k distinct primes what is the total number of factors of n?
Solution:
n is a product of k distinct primes.
∴ In order to be a factor, of n, we have chosen at least one of k distinct primes.
∴ The number of ways = ^kC₁ + ^kC₂ + ……… ^kC_k-1 = 2^k  – 1 – 1
∴ The number of factors of n is 2^k – 2.
(Excluding 1 as 1 is not prime. It is also not include n.)

Question 17.
If m has the prime factor decomposition P₁^r1, P₂^r2 ….. P_n^rn, what is the total number of factors of m (excluding 1)?
Solution:
m has the prime factor decomposition P₁^r1, P₂^r2 ….. P_n^rn,
∴ m = P₁^r1, P₂^r2 ….. P_n^rn,
P₁ is a factor of m which occurs r₁ times. Each of the factors P₁^r1 will give rise to (r₁ + 1) factors.
Similarly
P₂^r2 gives (r₂ + 1) factors and so on.
∴ The total number of factors (r₁ + 1) (r₂ + 1) ….. (r_n + 1) – 1 (including m).

Question 18.
If 20! were multiplied out, how many consecutive zeros would it have on the right?
Solution:
If 20! were multiplied out, then the number of consecutive zeros on the right is 4. due to the presence of 4 x 5, 10, 14 x 15,20.

Question 19.
How many factors of 10,000 end with a 5 on the right?
Answer:
We have 1000 = 2⁴ × 5⁴ The factors of 10000 ending with 5 are 5, 5 × 5 = 25, 5 × 5 × 5 = 125
5 × 5 × 5 × 5 = 625
∴ There are 4 factors ending With 5.

Question 20.
A man has 6 friends. In how many ways can he invite two or more to a dinner party?
Solution:
A man has 6 friends. He can invite 2 or more of his friends to a dinner party.
∴ He can invite 2, 3, 4, 5, or 6 of his friends in
⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ = 2⁶ – ⁶C₀ – ⁶C₁
= 64 – 7 = 57 ways.

Question 21.
In how many ways can a student choose 5 courses out of 9 if 2 courses are compulsory?
Solution:
A student is to choose 5 courses out of 9 in which 2 courses are compulsory. as 2 courses are compulsory, he is to choose 3 courses out of 7 courses in ⁷C₃ = 35 ways.

Question 22.
In how many ways can a student choose five courses out of the courses? C₁, C₂, …………. C₉ if C₁, C₂ are compulsory and C₆, C₈ cannot be taken together?
Solution:
A student chooses five courses out of the courses? C₁, C₂, …………. C₉ if C₁, C₂ are compulsory and C₆, C₈ cannot be taken together.
∴ He is to choose 3 courses out of C₃, C₄, ………… ,C₈, C₉.
Without taking any restrictions 3 courses out of C₃, C₄, …… C₉, i.e. from 7 courses in ⁷C₃ ways. If C₆, C₈ are taken together then one course only to be choosen from C₃, C₄, C₅, C₇, C₉ by ⁵C₁ ways. Hence required number of ways.
= ⁷C₃ – ⁵C₁
= \(\frac{7 \times 6 \times 5}{3 \times 2 \times 1}\) – 5
= 35 – 5 = 30 ways.

Question 23.
A cricket team consisting of 11 players is to be chosen from 8 batsmen and 5 bowlers. In how many ways can the team be chosen so as to include at least 3 bowlers?
Solution:
A cricket team consisting of 1 1 player is to be chosen from 8 batsmen and 5 bowlers, as to include at least 3 bowlers.
The selection can be made as follows :

Batsmen(8)	Bowlers(5)
8	3
7	4
6	5

The number of selections is
(⁸C₈ × ⁵C₃) + (⁸C₇ × ⁵C₄) + (⁸C₆ × ⁵C₅)
= (1 × 10) + (8 × 5) + (28 × 1)
= 10 + 40 + 28 = 78

Question 24.
There are n + r points on a plane out of which n points lie on a straight line L and out of the remaining r points that lie outside L, no three points are collinear. What is the number of straight lines that can be formed by joining pairs of their points?
Solution:
There are n + r points on a plane out of which n points are collinear and out of which r points are not collinear.
∴ We can form a straight line by joining any two points.
n-collinear points form one line and r-non-collinear points form ^rC₂ lines.
Again, each of the r non-collinear points when joined to each of the noncollinear points, forms n lines.
∴ The number of such limes is r x n.
∴ The total number of lines

Question 25.
There are 10 books in a shelf with different titles; five of these have red covers and others have green covers. In how many ways can these be arranged so that the red books are placed together?
Solution:
There are 10 books in a shelf with different titles, 5 of these are red covers and others are green covers considering 5 red-covered books as one book, we have a total of 6 books which can be arranged in 6! ways. The five red cover books are arranged among themselves in 5! ways.
∴ The total number of arrangements
= 5! x 6!

Odisha State Board BSE Odisha 6th Class English Solutions Test-1(B) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-1(B)

BSE Odisha 6th Class English Test-1(B) Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.
1. Write the following Odia names of the persons in English. (Teacher will give the names of six persons in Odia.) [06]
Answer:
ଲାଲା ଲଜପାଟ ରୟ – Lala Lajpat Roy
B.R. ଆମ୍ବେଦକର | – B.R. Ambedkar
ବାଙ୍କିମ୍ ଚନ୍ଦ୍ର ପାଲ୍ – Bankim Chandra Pal.
ବୁକ୍ସି ଜଗବନ୍ଧୁ – Buxi Jagabandhu
ଫାକିର ମୋହନ ସେନାପତି – Fakir Mohan Senapati
ରାଧନାଥା ରାଧା – Radhanatha Ratha

2. Write the following place names in English. [06]
(Teacher will give names of six places in Odia.)
Answer:
ଶ୍ରୀନଗର – Srinagar
ହିମାଚଳ ପ୍ରଦେଶ – Himachal Pradesh
ପୁଡୁଚେରୀ – Puducheri
ଦ୍ୱାରିକା – Dwarika
ଚିଲିକା ହ୍ରଦ – Chilika lake
ନନ୍ଦନକାନନ୍ ପ୍ରାଣୀ ଉଦ୍ୟାନ – Nandankanan Zoo

3. Your teacher will give a dictation of ten words. Write them in the space given below. [05]
Answer :
Jagannatha     Moon
Saraswati        Capital
Ganesha         Chief-Minister
Earth /World  President
Sun                 Garden

4. Given below are some words. Your teacher will read aloud five of them.
Tick those which s/he reads aloud. [05]
temper, catch, squirrel, little, agile, climbed, thrash, burst, bamboo, curry, foolish.
[Listen to your teacher and tick those words s/he reads.]

5. Your teacher will read aloud a paragraph. You listen to him/her and fill in the gaps. (Question with Answer) [08]
Once there lived a greedy fat old man. One day he got up at 6 a.m. and brushed his teeth at 6.30 a.m. He took tea at 7 a.m. and breakfast at 8.30 a.m. Do you know how much tea he took?

6. Match the words which sound alike at the end. (Question with Answer)

7. Read the poem and answer the questions. [10]
Their ears are pink,
Their teeth are white,
They run about
The house at night
They nibble things
They shouldn’t touch
And no one seems
To like them much.

Question (i).
What is the color of their ears?
Answer:
The color of their ears is pink.

Question (ii).
What is the color of their teeth?
Answer:
The color of their teeth is white.

Question (iii).
Where do they run about at night?
Answer:
They run about the house at night.

Question (iv).
Which things do they nibble?
Answer:
They nibble things that they shouldn’t touch.

Question (v).
Do most people like mice?
Answer:
No, no one seems to like mice.

8. Read the following paragraph and answer the questions in complete sentences. [20]
After eating the dog, the man walked, walked, and walked till he met a little squirrel. The little squirrel asked the old man, “Old man, old man, what makes you so fat? The old man said, “I’ve taken a very heavy breakfast. In my breakfast, I took two mugs of tea, two liters of milk, three tins of biscuits, and five big pieces of cakes.” Then I ate a little boy and a small dog. I’ll also eat you up if I can catch you”.

“But you cannot catch me, old man,” said the active, agile, little squirrel. Then the squirrel jumped up the tree, the old man also climbed up the tree. The little squirrel jumped up to the main branch of the tree. The old man also climbed up to the main branch of the tree. Next, the little squirrel jumped up to a thin branch. The old man also climbed up to the thin branch. But thrash ! the small branch broke and the old man fell to the ground. His big belly burst out.

Question (i).
Whom did the old man meet in this section?
Answer:
In this section, the old man met a little squirrel.

Question (ii).
What did the squirrel ask the old man?
Answer:
The squirrel asked the old man, “What makes you so fat?”

Question (iii).
How many mugs of tea did the old man take?
Answer:
The old man took two mugs of tea.

Question (iv).
How much milk did he take?
Answer:
He took two liters of milk.

Question (v).
How many tins of biscuits did he take in his breakfast?
Answer:
For his breakfast, he took three tins of biscuits.

Question (vi).
How many pieces of cake did he take?
Answer:
He took five big pieces of cake.

Question (vii).
Where did the squirrel jump up first?
Answer:
The squirrel jumped up first to the tree.

Question (viii).
What did the old man do?
Answer:
The old man also climbed up the tree.

Question (ix).
Where did the squirrel jump up then?
Answer:
Then the squirrel jumped up to the main branch of the tree.

Question (x).
What happened to the old man when he fell down?
Answer:
When the old man fell down, his big belly burst out.

9. Read the following poem and answer the questions in complete sentences. [10]
“My story is said
The flowering plant is dead.
O flowering plant! Why did you die?
The black cow ate me up and made me lie.
O black cow! Why did the plant you eat?
Because the cowherd did not do well treat.
O, cowherd! Why didn’t you well the cow treat to eat?
The daughter-in-law did not give me food.”

Question (i).
Why is the flowering plant dead?
Answer:
The flowering plant is dead because the black cow ate it up.

Question (ii).
Why did the cow eat up the flowering plant?
Answer:
The cow ate up the flowering plant because the cowherd did not well her treat.

Question (iii).
Why didn’t the cowherd well the cow treat to eat?
Answer:
The cowherd didn’t well the cow treat to eat because the daughter-in-law did not give him food.

Question (iv).
What is said?
Answer:
Your story is said.

Question (v).
What type of plant is dead?
Answer:
The flowering plant is dead.

10. Read the following paragraph and answer the questions in a complete sentences. [20]
Once a foolish Santal son-in-law went to his in-law’s house. His mother-in-law cooked delicious dishes for her son-in-law. One of the dishes was a curry made out of the bamboo shoot. The son-in-law liked it very much and asked his mother-in-law, “Mother, the curry is extremely delicious. What is the curry made from? Instead of answering his question, she pointed at the bamboo door. The son-in-law asked, “Is it from bamboo ?” Yes, son, the curry is made from bamboo and is, therefore, called “Bamboo Curry”. The next day, the son-in-law was about to leave for his home. The bamboo curry came to his mind. He thought of cooking bamboo curry at home. But they did not have bamboo. So he carried home the bamboo door of his in-law’s house.

Question (i).
Who went to his father-in-law’s house?
Answer:
A foolish Santal son-in-law went to his father-in-law’s house.

Question (ii).
What did his mother-in-law cook for him?
Answer:
His mother-in-law cooked delicious dishes for him.

Question (iii).
What was one of the dishes made?
Answer:
One of the dishes was a curry made out of the bamboo shoot.

Question (iv).
How was the curry?
Answer:
The curry was extremely delicious.

Question (v).
What did the son-in-law ask his mother-in-law?
Answer:
The son-in-law asked his mother-in-law what the curry was made from.

Question (vi).
What did the mother-in-law answer?
Answer:
Instead of answering his question, she pointed at the bamboo door.

Question (vii).
Why is the curry called “Bamboo Curry”?
Answer:
The curry is called ‘Bamboo Curry’ because it is made from bamboo.

Question (viii).
When was the son-in-law returning to his home?
Answer:
The next day, the son-in-law was returning to his home.

Question (ix).
What did he think of doing?
Answer:
He thought of cooking bamboo curry at home.

Question (x).
What did not they have? What did he do for that?
Answer:
They did not have bamboo. So he carried home the bamboo door of his in-law’s house.

Odisha State Board BSE Odisha 6th Class English Solutions Test-1(A) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-1(A)

BSE Odisha 6th Class English Test-1(A) Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.

1. Write the following Odia names of the persons in English. [06]
(Teacher will give the names of six persons in Odia.)
Answer:
ଲାଲ ବାହାଦୂର ଶାସ୍ତ୍ରୀ | – Lal Bahadur Shastri
ବାଲ ଗଙ୍ଗାଧର ତିଲକ | – Bal Gangadhar Tilak
ସର୍ଦ୍ଦାର ଭାଲାଭଭାଇ ପଟେଲ | – Sardar Ballavabhai Patel
ହରେକୃଷ୍ଣ ମହାତ୍ମା – Harekrushna Mahatab
ଭୀମା ଭୋଇ – Bhima Bhoi
ପ୍ରାଣକୃଷ୍ଣ ପରଜା | – Pranakrushna Parija

2. Write the following place names in English. [06]
(Teacher will give names of six places in Odia.)
Answer:
ହିମାଳୟ – Himalaya
ପଞ୍ଜାବ – Punjab
ଅରୁଣାଚଳ ପ୍ରଦେଶ – Arunachal Pradesh
କୋଲକାତା – Kolkata
ବ୍ରହ୍ମପୁର – Berhampur
ବାଲାସୋର – Balasore

3. Your teacher will give a dictation of ten words. Write them in the space given below. [05]
Answer:
Grandfather      Ice-cream       Friend
Grandmother   Computer       Neighbour
Bread               Blackboard      Government
Mango

4. Given below are some words. Your teacher will read aloud five of them.
Tick those which s/he reads aloud. [05]
kite, paper, greedy, breakfast, tea, mugs, biscuits, cakes, conversation, heavey, squirrel.
[Listen to your teacher carefully and tick those words as he reads aloud.]

5. Your teacher will read aloud a paragraph. You listen to him/her and nil in the gaps. (Question with Answer) [08]
Answer:
The old man said, “I’ve taken two mugs of tea, and two liters of milk. I also took three tins of biscuits and five kilograms of cakes. And if I can catch you, I’ll eat you up .’’Then the old man caught the thin little boy and ate him up.

6. Match the words which sound alike at the end. (Question with Answer) [10]

7. Read the poem and answer the questions. [10]
I think mice
Are rather nice.
Their tails are long
Their faces small,
They haven’t any
Skins at all.

Question (i).
What is this poem about?
Answer:
This poem is about mice.

Question (ii).
Who is ‘I’ at the beginning of the poem?
Answer:
‘I’ at the beginning of the poem is the poet himself.

Question (iii).
What are the tails of mice like?
Answer:
The tails of mice are long.

Question (iv).
How are their faces?
Answer:
Their faces are small.

Question (v).
They haven’t any skint at all”- What does this line mean?
Answer:
‘‘They haven’t any skins at all”- This line means that mice have very thin skins.

8. Read the following paragraph and answer the questions in complete sentences. [20]
Once there lived a greedy fat old man. One day he got up at 6 a.m. and brushed his teeth at 6.30 a.m. He took tea at 7 a.m. and breakfast at 8.30 a.m. He took two mugs of tea and two liters of milk. Then he took three tins of biscuits and five big pieces of cake. After breakfast, he looked really very very fat.

Question (i).
What is this paragraph about?
Answer:
This paragraph is about a greedy fat old man.

Question (ii).
How was the old man?
Answer:
The old man was greedy and fat.

Question (iii).
When did he get up one day?
Answer:
One day he got up at 6 a.m.

Question (iv).
When did he brush his teeth?
Answer:
He brushed his teeth at 6.30 a.m.

Question (v).
What did he take at 7 a.m.?
Answer:
He took tea at 7 a.m.

Question (vi).
When did he take his breakfast?
Answer:
He took his breakfast at 8.30 a.m.

Question (vii).
How many mugs of tea did he take?
Answer:
He took two mugs of tea.

Question (viii).
How much milk did he take?
Answer:
He took two liters of milk.

Question (ix).
How many tins of biscuit did he take?
Answer:
He took three tins of biscuits.

Question (x).
How did he look like after breakfast?
Answer:
After breakfast, he looked really very very fat.

9. Read the following poem and answer the questions in complete sentences. [10]

For want of a nail,
the shoe was lost.
For want of a shoe,
the horse was lost.
For want of a horse,
the rider was lost.
For want of a rider,
the battle was lost.
For want of a battle,
the kingdom was lost.
And all for the want of a
horseshoe nail.

Question (i).
Why was the shoe lost?
Answer:
The shoe was lost for want of a nail.

Question (ii).
Why was the horse lost?
Answer:
The horse was lost for want of a shoe.

Question (iii).
Why was the rider lost?
Answer:
The rider was lost for want of a horse.

Question (iv).
Why was the battle lost?
Answer:
The battle was lost for want a rider.

Question (v).
Why was the kingdom lost?
Answer:
The kingdom was lost for want of a battle.

10. Read the following paragraph and answer the questions in a complete sentences. [20]
A son-in-law, after his marriage, was planning to visit his father-in-law’s house for the first time. A man from his village gave him the advice, “Use big and high-sounding words in your father-in-law’s house. Always sit on a high place. First say ‘No’ to any food given to you.”
He, therefore, used very long and high-sounding words. He told his mother-in-law, “You are the sweetest, kindest, greatest, and gentlest lady.” The mother-in-law was very much pleased to hear this. She praised her son-in-law in front of her neighbors for using high-sounding words and calling her the kindest and greatest lady.

Question (i).
What was the son-in-law’s plan?
Answer:
The son-in-law’s plan was to visit his father-in-law’s house.

Question (ii).
When was the son-in-law planning to visit his father-in-law’s house?
Answer:
After his marriage, the son-in-law was planning to visit his father-in-law’s house.

Question (iii).
Did he visit his father-in-law’s house before?
Answer:
No, he did not visit his father-in-law’s house before.

Question (iv).
Who advised him to do something at his in-law’s house?
Answer:
A man from his village advised him to do something at his in-law’s house.

Question (v).
What was the first advice?
Answer:
The first piece of advice was to use big and high-sounding words in his father-in-law’s house.

Question (vi).
What was the second advice for the son-in-law?
Answer:
The second piece of advice for the son-in-law was always to sit on a high place.

Question (vii).
What was the third piece of advice for the son-in-law?
Answer:
The third piece of advice for the son-in-law was first to say ‘No’ to any food given to him.

Question (viii).
What did he tell his mother-in-law?
Answer:
He told his mother-in-law that she was the sweetest, kindest, greatest and gentlest lady.

Question (ix).
How did the mother-in-law feel to hear his words?
Answer:
The mother-in-law was very much pleased to hear his words.

Question (x).
What did she do?
Answer:
She praised her son-in-law in front of her neighbors for calling her the kindest and greatest lady.

Odisha State Board BSE Odisha 6th Class English Solutions Test-1 Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-1

BSE Odisha 6th Class English Test -1 Text Book Questions and Answers

• The figures in the right-hand margin indicate the marks for each question.

1. Write the following Odia names of the persons in English.
(Teacher will give the names of six persons in Odia.) [06]

Answer:
ମହାତ୍ମା ଗାନ୍ଧୀ – Mahatma Gandhi
ଜବାହରଲାଲ ନେହେରୁ | – Jawaharlal Nehru
ସୁଭାଷ ବୋଷ | – Subhas Bose
ଗୋପାବନ୍ଧୁ ଦାସ – Gopabandhu Das
ମଧୁସୂଦନ ଦାସ – Madhusudan Das
ବିଜୁ ପଟ୍ଟନାୟକ – Biju Patnaik

2. Write the following place names in English.
(Teacher will give names of six places in Odia.) [06]

Answer:
ଦିଲ୍ଲୀ – Delhi
କାଶ୍ମୀର – Kashmir
ଅୟୋଧ୍ୟା – Ayodhya
ଆସାମ | – Assam
ଭୁବନେଶ୍ୱର – Bhubaneswar
କଟକ | – Cuttack

3. Your teacher will give a dictation of ten words. Write them in the space given below. [05]

Answer :
School     Mother     Town
Book        House      Temple
Teacher    Village      God
Father

4. Given below are some words. Your teacher will read aloud five of them. Tick those which s/he reads aloud. [05]
house, gentle, great, night, cotton, lemon, nibble, thief, delicious, curry, kingdom, nail, battle, ground.
[Listen to your teacher and tick those words your teacher reads aloud.]

5. Your teacher will read aloud a paragraph. You listen to him/her and fill in the gaps. [08]
(Question with Answer)
Next ____________the son-in-law was ___________to leave for his ______________ . The bamboo _____________came to his ___________________. He thought of_____________ bamboo curry at home. But ________________ did not have _______________________.
Answer:
Next day the son-in-law was about to leave for his home. The bamboo curry came to his mind. He thought of cooking bamboo curry at home. But they did not have bamboo.

6. Match the words which sound alike at the end. (Question with Answer) [10]
(i)

Answer:

(ii)

Answer:

7. Read the poem and answer the questions. [10]

A kite on the ground
Is just paper and string
But up in the air
Will dance and sing.
A kite in the air
Will dance and caper
But back on the ground
Is just string and paper.

Question (a).
What is this poem about?
Answer:
This poem is about a kite.

Question (b).
What does it do when up in the air?
Answer:
When Up in the air, it will dance and sing.

Question (c).
What is it when it is on the ground?
Answer:
When on the ground, it is just paper and string.

Question (d).
Do you like flying a kite?
Answer:
Yes, I like flying a kite very much.

Question (e).
What are kites normally made of?
Answer:
Normally, kites are made of paper and string.

8. Read the following paragraph and answer the questions in complete sentences. [20]
A son-in-law after his marriage was planning to visit his father-in-law for the first time. A man from his village gave him some advice, “Use big and high-sounding words in your father-in-law’s house. Always sit in a high place. First, say ‘no’ to any food given to you.’’

Question (a).
What is the paragraph about?
Answer:
The paragraph is about a son-in-law.

Question (b).
What was he planning?
Answer:
He was planning to visit his father-in-law’s house.

Question (c).
When was he planning?
Answer:
He was planning after his marriage.

Question (d).
Had he gone to his father-in-law’s house before?
Answer:
No, he had not gone to his father-in-law’s house before.

Question (e).
Who gave him some advice?
Answer:
A man from his village gave him some advice.

Question (f).
How many pieces of advice did he give?
Answer:
He gave three pieces of advice.

Question (g).
What was the first advice?
Answer:
The first piece of advice was to use big and high-sounding words in his father-in-law’s house.

Question (h).
What was the second piece of advice?
Answer:
The second piece of advice was always to sit in a high place.

Question (i).
What was the third piece of advice?
Answer:
The third piece of advice was first to say ‘no’ to any food given to him.

Question (j).
Will he carry out the advice?
Answer:
Yes, he will carry out the advice.

9. Read the following poem and answer the questions in complete sentences. [10]

I woke up this morning
And I got out of bed,
Then I took a cup of tea
And ate a slice of bread.
1 went to the bus stop
And caught the bus to school,
On my way back it rained
And the weather was cool.

Question (i).
Who is T in the poem?
Answer:
T in the poem is the poet.

Question (ii).
What did s/he take after getting up?
Answer:
After getting up, s/he took a cup of tea.

Question (iii).
What did s/he eat?
Answer:
S/he ate a slice of bread.

Question (iv).
How did s/he go to school?
Answer:
S/he went to school by bus.

Question (v).
What happened on his / her way back?
Answer:
On his/her way back, it rained and the weather was cool.

10. Read the following paragraph and answer the questions in complete sentences. [20]
There was a crow and there was a cuckoo. They lived together happily. The crow was hard-working. But the cuckoo was very lazy. The crow brought food. The cuckoo only ate. The crow built a nest. She laid her eggs there. The cuckoo also wanted to lay eggs. But she had
not built a nest. One day the crow was not in her nest. The cuckoo threw away the eggs and laid her own eggs there. The crow did not know this. She sat over the eggs for some days. Young ones came out. She took care of them. But when the young ones grew? up, they sang like cuckoos. So she drove them away. From that day the crow always drives away the cuckoo.

Question (i).
What is this paragraph about?
Answer:
This paragraph is about a crow and a cuckoo.

Question (ii).
Who was lazy?
Answer:
The cuckoo was very lazy.

Question (iii).
Who was hard-working?
Answer:
The crow was hard-working.

Question (iv).
Who brought food?
Answer:
The crow brought food.

Question (v).
Who ate it?
Answer:
The cuckoo only ate it.

Question (vi).
Who built a nest?
Answer:
The crow built a nest.

Question (vii).
What did the cuckoo do with the eggs of the crow?
Answer:
The cuckoo threw away the eggs of the crow and laid her own eggs there.

Question (viii).
Whose young ones the crow was taking care of?
Answer:
The crow was taking care of the young ones of the cuckoo.

Question (ix).
When did she come to know about this?
Answer:
When the young ones grew up and sang like cuckoos, she (crow) came to know about this.

Question (x).
Why could not she recognize them before?
Answer:
She (the crow) could not recognize them before because in her absence the cuckoo had thrown away her eggs and laid her (cuckoo’s) own eggs there. Further, the young ones of the crow and the cuckoo look alike.

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(a)

Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(a) ବିମ୍ନ ଟିତ୍ର (a) 6ର L₁ || L₂ || L₃ ଏର୍ଚ T₁ ଓ T₂ ଛେଦ ଦା |
(i) AB = 2 6ସ.ର୍ମି., BC = 3 6ସ.ର୍ମି. ଓ DE = 3 6ସ.ର୍ମି. 6ହ6ଲ EF = ……….|
(ii) DE = 6 6ସ.ର୍ମି., EF = 8 6ସ.ର୍ମି. ଓ BC = 6 6ସ.ର୍ମି. 6ହ6ଲ AC = ……….|

(b) ଗପଭୋକ୍ତ ତିତ୍ର (b) 6ର L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ଛେଦକ |
(i) AB = 1.5 × BC ହେଲେ, \(\frac { EF }{ FD }\) = ……………..
(ii) \(\overline{\mathrm{AC}}\) ର ମଧ୍ୟବିରୁ B 6ହ6ଲ, EF ର ……… ମୁଶ 6ହରଛି FD|
Solution:
(a) (i) L₁ || L₂ || L₃ ⇒ \(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\)
⇒ \(\frac { 2 }{ 3 }\) = \(\frac { 3 }{ EF }\) ⇒ EF = \(\frac{3 \times 3}{2}\) 6 ସ.ମି. = 4.5 6ସ.ମି.

(ii) L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ରୁକଚି 6ଛଦକ |
∴ \(\frac { DE }{ EF }\) = \(\frac { AB }{ BC }\) ⇒ \(\frac { 6 }{ 8 }\) = \(\frac { AB }{ 6 }\)
⇒ AB = \(\frac{6 \times 6}{8}\) = 4.5 6ସ.ମି. |
AC = AB + BC = 4.5 6ସ.ମି. + 6ସ.ମି. = 10.5 6ସ.ମି. |

(b) (i) ଏଠ।6ର L₁ || L₂ || L₃ ଏରଂ T₁ ଓ T₂ ଦୁଲଟି 6ଚ୍ଚଦକ |
\(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac { 3 }{ 2 }\) = \(\frac { DE }{ EF }\)
⇒ \(\frac{3+2}{2}\) = \(\frac{\mathrm{DE}+\mathrm{EF}}{\mathrm{EF}}\) ⇒ \(\frac { DE }{ EF }\) = \(\frac { 5 }{ 2 }\) ⇒ \(\frac { EF }{ DF }\) = \(\frac { 2 }{ 5 }\)

(ii) \(\overline{\mathrm{AC}}\) ର ମଧ୍ୟ ଦିନୁ B ଅର୍ଥ।ତ୍ AB = BC
\(\frac { EF }{ FD }\) = \(\frac { BC }{ AC }\) = \(\frac{\mathrm{BC}}{\mathrm{AB}+\mathrm{BC}}\) = \(\frac { BC }{ 2BC }\) = \(\frac { 1 }{ 2 }\) ⇒ 2EF = FD>
ଅର୍ଥାତ୍ EF ଭ 2 ଣ୍ଣଣ 6ଦ୍ରଣଛି FD |

Question 2.
ଟିତ୍ର6ର L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ଦୁଲଟି 6ଚ୍ଛଦକ | L₂ ଓ L₃ ରପ6ର ପ୍ଥଥ।କୁ6ର G ଓ H ଦିହୁ ରିଜିତ 6 ପ୍ପପତି BC = AD ଏବଂ CH = BE;

ପ୍ରମାଣ କର 6ଯ
(i) DG : EH = DE : EF
(ii) (DG + EH) : EH = DF : EF
Solution:
ଦଉ : L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ଦୁକଟି 6ଚ୍ଛଦକ ଯଥାସ୍ତ6ମ L₁, L₂ ଓ L₃ କୁ A,B,C ଓ D,E,F ଦିନ୍ଦ6ର 6ଚ୍ଛଦକ6ର | L₂ ରପ6ର G ଏକ ଦିବ୍ର 6ସ୍ ପରି BG = AD ଏର୍ବ L₃ ରପ6ର H ଏକ ଦିନ୍ଦୁ 6ନ୍ଦ୍ ପରି CH = BE |
ପ୍ରାମାଣ୍ୟ :
(i) DG : EH = DE : EF
(ii) (DG + EH) : EH = DF : EF

ଅକ୍ଳଜ : \(\overline{\mathrm{DG}}\) ଓ \(\overline{\mathrm{EH}}\) ଅଲ୍ଲବ ଦବାପାର |
ପ୍ତମାଣ : (i) AD = BG (ଦର) \((\overline{\mathrm{AD}} \| \overline{\mathrm{BG}})\), ∵ L₁ || L₂ (ଦର)
⇒ ABGD ଏକ ସାମାତ୍ରତିଦ ଚିତ୍ର | AB = DG
6ସଦୃପତି BEHC ଏକ ସାମାତ୍ର ଚିତ୍ର ଓ BC = EH
∴ \(\frac { DG }{ EH }\) = \(\frac { AB }{ BC }\) …(1)
ପୁନଣ୍ଡ \(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac { DG }{ EH }\) = \(\frac { DE }{ EF }\) (1ରୁ) (ପ୍ରମାଣିତ)

(ii) ପୁଫରୁ ପ୍ରମାଣିର \(\frac { DG }{ EH }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac{\mathrm{DG}+\mathrm{EH}}{\mathrm{EH}}\) = \(\frac{\mathrm{DE}+\mathrm{EF}}{\mathrm{EF}}\) (Componendo ଦ୍ଦାରା )
⇒ \(\frac{\mathrm{DG}+\mathrm{EH}}{\mathrm{EH}}\) = \(\frac { DF }{ EF }\) (ପ୍ରମାଣିତ)

Question 3.
ନିମ୍ନ ଚିତ୍ରରେ L₁ || L₂ || L₃ ଏବଂ T₁, ଓ T₂, ଦୁଇଟି ଛେଦକ । ଯଦି AB = BC ହୁଏ, ପ୍ରମାଣ କର ଯେ 2 BE = AD + CF |
Solution:
ଦତ୍ତ : L₁ || L₂ || L₃ ଛେଦକ T₁ ଓ T₂ , L₁, L₂, ଓ L₃, କୁ ଯଥାକ୍ରମେ A, B, C ଓ D, E, F ବିନ୍ଦୁରେ ଛେଦ କରେ ଏବଂ AB = BC |
ପ୍ରାମାଣ୍ୟ : 2BE = AD + CF
ଅକନ : E ଦିଦୁ ମଧ୍ୟ6ଦକ AC ସମାତ ତା6ଦ ଅଜିତ ବେଡା L₁ || L₃ କୁ ଯଥାକୁ6ମ X ଓ Y ଦିଦ6ର 6ଛଦକ୍ଳରୁ |

ଅଙ୍କନ : E ବିନ୍ଦୁ ମଧ୍ୟଦେଇ \(\overline{\mathrm{AC}}\)ସହ ସମାନ୍ତର ଭାବେ ଅଙ୍କିତ ରେଖା L₁ ଓ L₃, କୁ ଯଥାକ୍ରମେ X ଓ Y ବିନ୍ଦୁରେ ଛେଦକରୁ ।

Question 4.
ଚିତ୍ରରେ L₁ || L₂ || L₃, ଏବଂ T₁, ଓ T₂, ଦୁଇଟି ଛେଦକ । L₁, L₂, ଓ L₃, କୁ ଛେଦକ T₁, ଯଥାକ୍ରମେ A, B ଓ C ବିନ୍ଦୁରେ ଛେଦକରେ ଏବଂ L₁, L₂, ଓ L₃, କୁ ଛେଦକ T₂, ଯଥାକ୍ରମେ D, E ଓ F ବିନ୍ଦୁରେ ଛେଦ କରେ । DE = EF ହେଲେ, ପ୍ରମାଣ କର ଯେ, CF – AD = 2 EB । (ସୂଚନା : AF ଅଙ୍କନ କର)

Solution:
ଦତ୍ତ : L₁ || L₂ || L₃ ଏବଂ T₁, ଓ T₂, ଦୁଇଟି ଛେଦକ ଯଥାକ୍ରମେ L₁, L₂, ଓ L₃, କୁ A, B, C ଓ D, E, F ବିନ୍ଦୁରେ ଛେଦକରେ ।
ପ୍ରାମାଣ୍ୟ : CF – AD = 2EB.
ଅଙ୍କନ : \(\overline{\mathrm{AF}}\) ଅଙ୍କନ କର । B ବିନ୍ଦୁରୁ \(\overline{\mathrm{AF}}\) ସହ ସମାନ୍ତର ଭାବେ ଅଙ୍କିତ ରେଖା \(\overline{\mathrm{FC}}\) କୁ ଓ ବିନ୍ଦୁରେ ଛେଦକରୁ ।
\(\overline{\mathrm{AF}}\), L₂, କୁ H ବିନ୍ଦୁରେ ଛେଦକରୁ ।

ପ୍ରମାଣ : DE = EF (ଦଣ) ଓ \(\stackrel{\leftrightarrow}{\mathrm{BE}}\) || \(\overline{\mathrm{AD}}\) (∵ L₂ || L₁)
⇒ AH = HF,
ଦର୍ମାବ AH = HF ଓ \(\overline{\mathrm{BH}}\) || \(\overline{\mathrm{CF}}\) ⇒ AB = BC
\(\overline{\mathrm{BG}}\) || \(\overline{\mathrm{AF}}\) (ଅକo) ⇒ CG = GF ⇒ CF = 2GF △AFD 6ର H ଓ E ଯଥାକ୍6ମ \(\overline{\mathrm{AF}}\) ଓ \(\overline{\mathrm{DF}}\) ର ମଧ୍ୟଦିଦୁ
⇒ \(\overline{\mathrm{EH}}\) || \(\overline{\mathrm{AD}}\)
∴ ଦଘଣପଷ୍ଟ = CF – AD = 2GF – 2HE
= 2BH – 2HE (∵ GF = BH)
= 2(BH – HE) = 2BE = ଦାମପଖ (ପ୍ରମାଣିତ)

Question 5.

(i) ଭପରିସ୍ଥ ଚିତ୍ର (a) 6ର A – D – B ଏବଂ A – E – C | m∠DAE = 50°, m ∠AED = m∠ABC = 65° | AD = 3 6ସ.ମି., AE : EC = 2 : 1 6ଦୃ6କ, \(\overline{\mathrm{DB}}\) ଓ \(\overline{\mathrm{AB}}\) ର 6ଦିଘ୍ୟ ବ୍ଶଷଯ ଦର |
(ii) ଉପରିସ୍ଥ ବିତ୍ର (b) 6ର \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{QR}}\), NR = \(\frac { 2 }{ 5 }\) PR ଏବଂ PQ = 10 6ସ. ମି. 6ଦୃ6କ, PM ଓ QM କିଣପ୍ର କର |
(iii) କପଟାସ୍ଟ ଚିତୃ (b) 6ର PM = \(\frac { 1 }{ 2 }\) PQ, NR = 1.2 6ସ.ମି. ଓ \(\overline{\mathbf{M N}} \| \overline{\mathbf{Q R}}\) 6 ଦୁ6 କ, PR ଷ୍ଟିର କର
Solution:
(i) △ ADE 6ର m∠AD = 180° – (50° + 65°) = 65°
∠ADE ≅ ∠ABC ⇒ \(\overline{\mathrm{DE}} \| \overline{\mathrm{BC}}\)

⇒ \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\) = \(\frac { 2 }{ 1 }\)
⇒ \(\frac{3}{DB}\) = \(\frac { 2 }{ 1 }\) ⇒ DB = \(\frac { 3 }{ 2 }\) 6ପ.ମି. = 1.5 6ସ.ମି
AB = AD + DB = 3 6ସ.ମି. + 1.5 6ସ.ମି = 4.5 6ସ.ମି. |

(ii) \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{QR}}\) (ଦଉ)
\(\frac { PM }{ MQ }\) = \(\frac { PN }{ NR }\) ⇒ \(\frac{P M+M Q}{M Q}\) = \(\frac{\mathrm{PN}+\mathrm{NR}}{\mathrm{NR}}\)

⇒ \(\frac { PQ }{ MQ }\) = \(\frac { PR }{ NR }\) ⇒ \(\frac { 10 }{ MQ }\) = \(\frac{\mathrm{PR}}{\frac{2}{5} \mathrm{PR}}\)
⇒ \(\frac { 10 }{ MQ }\) = \(\frac { 5 }{ 2 }\)
⇒ MQ = \(\frac{10 \times 2}{5}\) 6ସ.ମି. = 4 6ସ.ମି. |
PM = PQ – MQ = 10 6ସ.ମି. – 4 6ସ.ମି. = 6 6ସ.ମି. |
∴ \(\overline{\mathrm{PM}}\) \(\overline{\mathrm{QM}}\) ର ଦେଶ ଯଥାଦୃ6ମ 6 6ସ.ମି. ଓ 4 6ସ.ମି.|

(iii) \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{QR}}\) , PM = \(\frac { 2 }{ 3 }\) PQ
⇒ MQ = PQ – PM = PQ – \(\frac { 2 }{ 3 }\) PQ = \(\frac { 1 }{ 3 }\) PQ

⇒ \(\frac { 1.2 }{ PR }\) = \(\frac{\frac{1}{3} \mathrm{PQ}}{\mathrm{PQ}}\) ⇒ \(\frac { 1.2 }{ PR }\) = \(\frac { 1 }{ 3 }\)
⇒ PR = 1.2 × 3 6ସ.ମି. = 3.6 ସେ.ମି. |
∴ \(\overline{\mathrm{PR}}\) ର 6ଦଣu = 3.6 ପେ.ପି. |

Question 6.
(i) △ABC 6ର, \(\overline{\mathbf{AB}}\) ଓ \(\overline{\mathbf{AC}}\) ର ମଧତିହୁ ଯଥାକ୍ର6ମ X ଓ Y 6ହ6କ, ଦଶାଥ 6ଯ \(\overline{\mathbf{X Y}} \| \overline{\mathbf{B C}}\) |
(ii) ଏକ ତ୍ରିସ୍ତକର 6ଣାଟିଏ ତାଦୃତ ମଧ୍ୟଝିହୁ 6ଦକ ଅଠ୍ୟ ଏଜ ଚାହପ୍ତତି ଅଜାଡ ସମାତର 6ରଖା, ତୃତୀୟ ବlନୁ ସମକ୍ରିଖଶ୍ନ କ6ର |
(iii) 6ଣାଟିଏ ପମ6କାଣା ତ୍ରିହ୍ନର ଦଶାବ ମଧ୍ୟବିଦୁ ଅଠ୍ୟ ଏବ୍ଲ ଗ୍ଲାସ୍ଥଗି ସମସ୍ତିଖଣ୍ଡକ6ର , ପ୍ରମାଣ କର |
Solution:

Question 7.
△PQR ରେ, \(\overline{\mathbf{PQ}}\) ଓ \(\overline{\mathbf{QR}}\) ବାହୁଦ୍ୱୟର ମଧ୍ୟବିନ୍ଦୁ ଯଥାକ୍ରମେ M ଓ N । \(\overline{\mathbf{PR}}\) ଉପରିସ୍ଥ S ଯେକୌଣସି ଏକ ବିନ୍ଦୁ ହେଲେ, ପ୍ରମାଣ କର ଯେ \(\overline{\mathbf{MN}}\), \(\overline{\mathbf{QS}}\) କୁ ସମର୍ଦ୍ଦିଖଣ୍ଡ କରିବ ।
Solution:
ଦତ : △PQR 6ର \(\overline{\mathbf{PQ}}\) ଓ \(\overline{\mathbf{QR}}\) ର ମଧ୍ୟବିଦୁ ଯଥାଦୃ6ମ M ଓ N | \(\overline{\mathbf{PR}}\) ଭପରିସ୍ଥ ‘S’ ଏକ ବିଦୁ | \(\overline{\mathbf{QS}}\) ଓ \(\overline{\mathbf{MN}}\) ର 6ଛଦଦିଦୁ O |
ସ୍ତାମାଶ୍ୟ : OQ = OS

ପ୍ରମାଣ : \(\frac { QM }{ MP }\) = 1 (∵ QM = MP (ଦର) )
\(\frac { QN }{ NR }\) = 1 (∵ QN = NR (ଦର) )
⇒ \(\frac { QM }{ MP }\) = \(\frac { QN }{ NR }\) ⇒ \(\overline{\mathrm{MN}} \| \overline{\mathrm{PR}}\)
\(\overline{\mathrm{MO}} \| \overline{\mathrm{PS}}\) (∵ \(\overline{\mathrm{MN}} \| \overline{\mathrm{PR}}\) )
⇒ \(\frac { QM }{ MP }\) = \(\frac { QO }{ OS }\) ⇒ 1 = \(\frac { QO }{ OS }\) (∵ \(\frac { QM }{ MP }\) = 1)
⇒ OQ = OS (ପ୍ତମାଣିତ)

Question 8.
ABCD ଣ୍ଠାପିଲିଷ୍ଟମ6ର \(\overline{\mathbf{AB}} \| \overline{\mathbf{CD}}\) | ଲଣ \(\overline{\mathbf{AC}}\) ଓ \(\overline{\mathbf{BD}}\) ର 6ବ୍ଳଦଦିନ୍ଦି P 6ଦୃ6କ , ପ୍ତମାଶା କର 6ମ
(i) AP : PC = BP : PD
(ii) CP : AC = DP : BD |
Solution:

Question 9.
ABCD ଗ୍ରାପିଦିକ୍ଷ୍ମମ6ର \(\overline{\mathbf{AB}}\) || \(\overline{\mathbf{DC}}\) ଏର \(\overline{\mathbf{AD}}\) ର ମଧ୍ୟତିଦୁ P | \(\overline{\mathbf{AB}}\) ସଦୃ ସାପବ୍ଦର ଉ6ର ଅବିର \(\overleftrightarrow{\mathbf{P Q}}\) \(\overline{\mathbf{BC}}\) କ Q ଦିହି6ର 6ଛଦନ6କ , ପ୍ରମାଣ ଦ୍ଦର 6ଯ Q 6ଦୃରଚି \(\overline{\mathbf{BC}}\) ର ମଧ୍ୟନିନ୍ଦୁ |
Solution:

Question 10.
ABCD ଉପରୋକ୍ତ \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{BC}}\) , \(\overline{\mathrm{CD}}\) ,ଓ \(\overline{\mathrm{DA}}\) ର ମଧ୍ୟବିନ୍ଦୁ ସର୍ବସମ P, Q, R ଓ S |
(a) ପ୍ରମାଣ କର ଯେ PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର ।
(b) ଉପରୋକ୍ତ ଚତୁର୍ଭୁଜ ABCD ର କର୍ଣ୍ଣଦ୍ଵୟ ପରସ୍ପର ପ୍ରତି ଲମ୍ବ ହେଲେ, ପ୍ରମାଣକର ଯେ PORS ଏକ ଆୟତ ଚିତ୍ର ।
Solution:
(a) ଦତ୍ତ : ABCD ଉପରୋକ୍ତ \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{BC}}\) , \(\overline{\mathrm{CD}}\) , ଓ \(\overline{\mathrm{DA}}\) ର ମଧ୍ୟବିନ୍ଦୁ ସର୍ବସମ P, Q, R ଓ S |
ପ୍ରାମାଣ୍ୟ : PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର ।
ଅଙ୍କନ : \(\overline{\mathrm{BD}}\) ଅଙ୍କନ କରାଯାଉ ।
ପ୍ରମାଣ : △ABD ରେ \(\overline{\mathrm{AB}}\) ର ମଧ୍ୟବିନ୍ଦୁ P ଓ \(\overline{\mathrm{AD}}\) ର ମଧ୍ୟବିନ୍ଦୁ S ।
\(\frac { AP }{ BP }\) = 1 (∵ AP = BP (ଦତ୍ତ))
ସେହିପରି ଧୂ) \(\frac { AS }{ SD }\) = 1 (∵ AS = SD (ଦତ୍ତ))

∴\(\frac { AP }{ BP }\) = \(\frac { AS }{ SD }\) ⇒ \(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{BD}}\)
ସେହିପରି ପ୍ରମାଣ କରାଯାଇ ପାରିବ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{BD}}\) |
\(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{BD}}\) ଓ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{BD}}\)
\(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{QR}}\)
\(\overline{\mathrm{AC}}\) ଅଙ୍କନ କରି ସେହିପରି ପ୍ରମାଣ କରାଯାଇ ପାରେ ଯେ \(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{SR}}\) |
∴ PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର |

(b) ଦତ୍ତ : ABCD ଚତୁର୍ଭୁଜରେ \(\overline{\mathrm{AC}}\) ⊥ \(\overline{\mathrm{BD}}\) | \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CD}}\) ଓ \(\overline{\mathrm{DA}}\) ର ମଧ୍ୟବିନ୍ଦୁ ଯଥାକ୍ରମେ P, Q, R, S |
ପ୍ରାମାଣ୍ୟ : PORS ଏକ ଆୟତଚିତ୍ର ।
ପ୍ରମାଣ : ମନେକର \(\overline{\mathrm{AC}}\) ଓ \(\overline{\mathrm{BD}}\) ଛେଦବିନ୍ଦୁ ( ଏବଂ \(\overline{\mathrm{PS}}\) ଓ \(\overline{\mathrm{AC}}\) ର ଛେଦବିନ୍ଦୁ X ।

(a) ର ଅନୁରୂପ ପ୍ରମାଣ ଦ୍ଵାରା PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର 6ଚିତ୍ର ।
m∠AOB = 90° ⇒ m∠PXO = 90°
(∵ \(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{BD}}\))
m∠XPXQ = 90° (∵ \(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{AC}}\))
∴ PQRS ଏକ ସାମାନ୍ତରିକ |

Question 11.
ନିମ୍ନ ଚିତ୍ରରେ △ABC ର \(\overline{\mathrm{BA}}\) ବାହୁ ସହ \(\overline{\mathrm{CM}}\) ସମାନ୍ତର, \(\overline{\mathrm{AB}}\) ର ପ୍ରାମାବିନ୍ଦୁ P | \(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{AC}}\), \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{CM}}\); ସାମାନ୍ତରିକ ମେ
\(\overline{\mathrm{PR}}\) || \(\overline{\mathrm{AM}}\) |
Solution:
ଦତ୍ତ : △ABC ରେ \(\overline{\mathrm{BA}}\) || \(\overline{\mathrm{CM}}\), AP = BP,
\(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{AC}}\) ଓ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{CM}}\)
ପ୍ରାମାଣ୍ୟ : \(\overline{\mathrm{PR}}\)||\(\overline{\mathrm{AM}}\)
ପ୍ରମାଣ : △ABC ରେ \(\overline{\mathrm{PQ}}\) ||\(\overline{\mathrm{AC}}\) (ଦତ୍ତ) ।

⇒ \(\frac { BP }{ AP }\) = \(\frac { BQ }{ QC }\)
△BCM ରେ \(\overline{\mathrm{QR}}\) ||\(\overline{\mathrm{CM}}\) (ଦତ୍ତ) ।
⇒ \(\frac { BQ }{ QC }\) = \(\frac { BR }{ RM }\) ……(i)
∴ (i) ଓ (ii)ରୁ \(\frac { BP }{ AP }\) = \(\frac { BR }{ RM }\)
△BCM ରେ \(\frac { BP }{ AP }\) = \(\frac { BR }{ RM }\) ⇒ \(\overline{\mathrm{PR}}\) ||\(\overline{\mathrm{AM}}\) (ପ୍ରମାଣିତ)
ତ୍ରିଭୁଜର କୋଣର ସମଦ୍ବିଖଣ୍ଡକ ସମ୍ବନ୍ଧୀୟ ଆଲୋଚନା

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(c) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(c)

Question 1.
ଦୁଇଟି ସଂଖ୍ୟାର ଯୋଗଫଳ 137 ଓ ସେମାନଙ୍କର ବିୟୋଗଫଳ 43। ତେବେ ସଂଖ୍ୟାଦ୍ୱୟ ନିରୂପଣ କର ।
ସମାଧାନ:
ମନେକର ସଂଖ୍ୟାଦ୍ଵୟ x ଓ y ।
ପ୍ରଶ୍ନନୁସାରେ x + y = 137 ……..(i) ଏବଂ x – y = 43 …….(ii)
ସମୀକରଣ (i) ଓ (ii) କୁ ପ୍ରୟୋଗକଲେ,

x ର ମାନ ସମୀକରଣ (i) ରେ ସ୍ଥାପନ କଲେ, x + y = 137
⇒ 90 + y = 137 ⇒ y = 137 – 90 ⇒ y = 47
∴ ସଂଖ୍ୟାଦ୍ଵୟ 90 ଓ 47।

Question 2.
ସମବାହୁ ତ୍ରିଭୁଜର ବାହୁ ତ୍ରୟର ଦୈର୍ଘ୍ୟ x + 4 ସେ.ମି., 4x – y ସେ.ମି. ଓ y + 2 ସେ.ମି. ହେଲେ ବାହୁର ଦୈର୍ଘ୍ୟ ସ୍ଥିର କର ।
ସମାଧାନ :
ସମବାହୁ ତ୍ରିଭୁଜର ବାହୂତ୍ରୟର ଦୈର୍ଘ୍ୟ ସମାନ । ଅର୍ଥାତ୍ x + 4 = 4x – y = y + 2
⇒ x + 4 = y + 2 = x – y = 2 – 4
⇒ x – y = -2 … (i)
ପୁନଶ୍ଚ 4x – y = y + 2 ⇒ 4x – y – y = 2 ⇒ 4x – 2y = 2
⇒ 2(2x – y) = 2 ⇒ 2x – y = 1 …… (ii)
ସମୀକରଣ (ii)ରୁ ସମୀକରଣ (i)କୁ ବିୟୋଗ କଲେ,

x ର ମାନ ସମୀକରଣ (i) ରେ ସ୍ଥାପନ କଲେ, 3 – y = – 2
⇒ – y = -2 – 3 = -5 ⇒ y = 5
∴ ସମବାହୁ ତ୍ରିଭୁଜର ଏକ ବାହୁର ଦୈର୍ଘ୍ୟ = y + 2 = 5 + 2 = 7 ସେ.ମି.

Question 3.
ABCD ଅ।ୟତକ୍ଷେତ୍ରର AB = 3x + y ସେ.ମି, BC = 3x + 2 ସେ.ମି, CD = 3y – 2x ସେ.ମି, ଓ DA = y + 3 ସେ.ମି. ହେଲେ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ନିରୂପଣ କର ।
ସମାଧାନ :
ଆୟତଚିତ୍ରର ବିପରୀତ ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ପରସ୍ପର ସମାନ ।
ତେଣୁ ABCD ଆୟତଚିତ୍ରରେ AB = CD ଏବଂ BC = AD ହେବ ।
AB = CD ହେଲେ, 3x + y = 3y – 2x ⇒ 3x + 2x = 3y – y ⇒ 5x – 2y=0 (i)
ସେହିପରି BC = AD ହେଲେ, 3x + 2 = y + 3
⇒ 3x – y = 3 – 2 ⇒ 3x – y =1 …..(ii)

‘x’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, 3 × 2 – y = 1 ⇒ 6 – 1 = y ⇒ y = 5
∴ AB = 3x + y = 3 × 2 + 5 = 11 ସେ.ମି
BC = 3x + 2 = 3 × 2 + 2 = 8 ସେ.ମି
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = AB × BC = 11 × 8 = 88

Question 4.
ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ଗୋଟିଏ ସଂଖ୍ୟା, ତାହାର ଅଙ୍କଦ୍ଵୟର ଯୋଗଫଳର 4 ଗୁଣ । କିନ୍ତୁ ସଂଖ୍ୟାଟିରେ 18 ଯୋଗ କଲେ ଅଙ୍କଦ୍ଵୟର ସ୍ଥାନ ବଦଳିଯାଏ । ତେବେ ସଂଖ୍ୟାଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାଟିର ଦଶକ ଏବଂ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କଦ୍ଵୟ ଯଥାକ୍ରମେ x ଏବଂ y l .:. ସଂଖ୍ୟାଟି = 10x + y
ପ୍ରଶ୍ନନୁସାରେ, ସଂଖ୍ୟାଟି ତାହାର ଅଙ୍କ ଦ୍ବୟର ଯୋଗଫଳର 4 ଗୁଣ ।
⇒ 10x + y = 4(x + y) ⇒ 10x + y = 4x + 4y ⇒ 10x – 4x + y – 4y = 0
⇒ 6x – 3y = 0 ⇒ 3(2x – y) = 0 … (i)
⇒2x – y = 0
ପ୍ରଶ୍ନନୁସାରେ ସଂଖ୍ୟାଟିରେ 18 ଯୋଗକଲେ ଅଙ୍କ ଦୁଇଟିର ସ୍ଥାନ ବଦଳିଯାଏ ।
⇒ 10x + y + 18 = 10y + x ⇒ 10x – x + y – 10y = -18
⇒ 9(x- y) = -18 ⇒ x – y = \(\frac{-18}{9}=-2\)

x ର ମାନ ସମୀକରଣ (i)ରେ ସଂସ୍ଥାପନ କଲେ, 2x – y = 0
⇒ 4 – y = 0 ⇒ y=4
∴ ସଂଖ୍ୟାଟି = 10x + y = 10 × 2 + 4 = 24

Question 5.
ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ଗୋଟିଏ ସଂଖ୍ୟା ଓ ତାହାର ଅଙ୍କଦ୍ଵୟର ସ୍ଥାନ ବଦଳାଇ ଲେଖୁଲେ ଯେଉଁ ସଂଖ୍ୟା ମିଳିବ, ସେ ଦୁହିଁଙ୍କର ଯୋଗଫଳ 99 ଓ ଅଙ୍କଦ୍ୱୟର ଅନ୍ତର 3 ହେଲେ ସଂଖ୍ୟାଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଦଶକ ଓ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କ ଦ୍ବୟ ଯଥାକ୍ରମେ x ଓ y
∴ ସଂଖ୍ୟାଟି = 10x + y
ଅଙ୍କଦ୍ୱୟର ସ୍ଥାନ ବଦଳାଇ ଲେଖୁଲେ ସଂଖ୍ୟାଟି ହେବ = 10y + x
ପ୍ରଶ୍ନନୁସାରେ, 10x + y + 10y + x = 99
⇒ 11x + 11y = 99 ⇒ x + y = \(\frac{99}{11}=9\) ………(i)
ଅଙ୍କଦ୍ୱୟର ଅନ୍ତର 3 ।
ଅର୍ଥାତ୍ x – y = 3 ବା y – x = 3 … (ii)
ଯଦି x – y = 3 ହୁଏ,
∴ x = \(\frac{x+y+x-y}{2}=\frac{9+3}{2}=\frac{12}{2}=6\)
y = 9 – x = 9 – 6 = 3
∴ ସଂଖ୍ୟାଟି = 10x + y = 10 × 6 + 3 = 63
ଯଦି ଆମେ ଦ୍ବିତୀୟ ସମୀକରଣକୁ y – x = 3 ନେବା ତେବେ ସଂଖ୍ୟାଟି 36 ହେବ ।
∴ ସଂଖ୍ୟାଟି 63 ବା 36 ।

Question 6.
ଦୁଇଟି ସଂଖ୍ୟାର ସମଷ୍ଟି, ସେମାନଙ୍କ ବିୟୋଗଫଳର 4 ଗୁଣ ଏବଂ ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ 8 । ତେବେ ସଂଖ୍ୟା ଦୁଇଟି କେତେ ?
ସମାଧାନ :
ମନେକର ସଂଖ୍ୟାଦ୍ଵୟ x ଓ y
ପ୍ରଶ୍ନନୁସାରେ, x + y = 4 (x – y)
⇒ 4x – 4y – x – y = 0 ⇒ 3x – 5y = 0 … (i)
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ x + y = 8 …… (ii)

x ର ମାନ ସମୀକରଣ (ii)ରେ ସଂସ୍ଥାପନ କଲେ, x + y = 8
⇒ y = 8 – x = 8 – 5 = 3
∴ ସଂଖ୍ୟାଦ୍ୱୟ 5 ଓ 3 ।

Question 7.
ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଅଙ୍କମାନଙ୍କର ସମଷ୍ଟି 10; କିନ୍ତୁ ଅଙ୍କଗୁଡ଼ିକର ସ୍ଥାନ ବଦଳାଇ ଲେଖୁଲେ ଉତ୍ପନ୍ନ ସଂଖ୍ୟାଟି ମୂଳ ସଂଖ୍ୟାର ଦୁଇ ଗୁଣରୁ 1 ଊଣା ହୁଏ, ସଂଖ୍ୟାଟି ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ସଂଖ୍ୟାଟିର ଦଶକ ସ୍ଥାନୀୟ ଅଙ୍କ ଏବଂ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କ ଯଥାକ୍ରମେ x ଏବଂ y ।
∴ ସଂଖ୍ୟାଟି 10x + y ହେବ ।
ଅଙ୍କଗୁଡ଼ିକର ସ୍ଥାନ ବଦଳିଗଲେ ଉତ୍ପନ୍ନ ସଂଖ୍ୟାଟି 10y + x ହେବ ।
ପ୍ରଶ୍ନନୁସାରେ, ଅଙ୍କଗୁଡ଼ିକର ଯୋଗଫଳ 10 । x + y = 10 … (i)
ପୁନଶ୍ଚ ଉତ୍ପନ୍ନ ସଂଖ୍ୟାଟି, ମୂଳ ସଂଖ୍ୟାର ଦୁଇଗୁଣରୁ 1 କମ୍ ।
ଅର୍ଥାତ୍ 10y + x = 2(10x + y) – 1 = 10y + x = 20x + 2y – 1
⇒ 8y – 19x = -1 ⇒ 19x – 8y = 1 … (ii)

‘x’ ର, ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ, 3 + y = 10 ⇒ y = 10 -3 = 7
∴ ସଂଖ୍ୟାଟି = 10x + y = 10 × 3 + 7 = 37

Question 8.
ଦୁଇଟି ସଂଖ୍ୟା ମଧ୍ୟରୁ ପ୍ରଥମଟିର 3 ଗୁଣରୁ ଦ୍ବିତୀୟଟିର 2 ଗୁଣ ବିୟୋଗ କଲେ ବିୟୋଗଫଳ 2 ହେବ ଏବଂ ଦ୍ଵିତୀୟଟିରେ 7 ଯୋଗ କଲେ ଯୋଗଫଳ ପ୍ରଥମଟିର 2 ଗୁଣ ହେବ। ସଂଖ୍ୟାଦ୍ଵୟ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ପ୍ରଥମ ସଂଖ୍ୟା ଓ ଦ୍ୱିତୀୟ ସଂଖ୍ୟା ଯଥାକ୍ରମେ x ଏବଂ y ହେଉ ।
ପ୍ରଥମଟିର 3 ଗୁଣରୁ ଦ୍ବିତୀୟଟିର 2 ଗୁଣ ବିୟୋଗକଲେ ବିୟୋଗଫଳ 2 ହେବ ।
ଅର୍ଥାତ୍ 3x – 2y = 2 …(i)
ସେହିପରି ଦ୍ବିତୀୟ ସଂଖ୍ୟାରେ 7 ଯୋଗକଲେ, ଯୋଗଫଳ ପ୍ରଥମ ସଂଖ୍ୟାର 2 ଗୁଣ ସଙ୍ଗେ ସମାନ ହେବ ।
ଅର୍ଥାତ୍ y + 7 = 2x = 2x – y = 7 …(ii)

‘x’ ର, ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ, 3 × 12 – 2y = 2 ⇒ 36 – 2y = 2
⇒ 2y = 34 ⇒ y = 17
∴ ପ୍ରଥମ ସଂଖ୍ୟାଟି 12 ଏବଂ ଦ୍ବିତୀୟ ସଂଖ୍ୟାଟି 17 ।

Question 9.
ଗୋଟିଏ ଭଗ୍ନାଂଶର ଲବ ଓ ହରରେ 2 ଯୋଗକଲେ ତାହା \(\frac{9}{11}\) ହୁଏ । ମାତ୍ର ଲବ ଓ ହରରେ 3 ଯୋଗକଲେ \(\frac{5}{6}\)
ହୁଏ । ତେବେ ଭଗ୍ନାଂଶଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଭଗ୍ନସଂଖ୍ୟାର ଲବ x ଓ ହର y
ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{x}{y}\)
ପ୍ରଶ୍ବାନୁସାରେ, \(\frac{x+2}{y+2}=\frac{9}{11}\)
⇒ 11x + 22 = 9y + 18 ⇒ 11x – 9y = – 4 (i)
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ, \(\frac{x+3}{y+3}=\frac{5}{6}\)
⇒ 6x + 18 = 5y + 15 ⇒ 6x – 5y = – 3 … (ii)

x ର ମାନ ସମୀକରଣ (i)ରେ ସଂସ୍ଥାପନ କଲେ, 11 × 7 – 9y = -4
⇒77 – 9y = 4 ⇒ – 9y = – 4 – 77 = – 81 ⇒ y =\(\frac{81}{9}=9\)
∴ ଭଗ୍ନ ସଂଖ୍ୟାଟି \(\frac{x}{y}=\frac{7}{9}\)

Question 10.
ଗୋଟିଏ ଭଗ୍ନାଂଶର ଲବର 3 ଗୁଣ ଓ ହରରୁ 3 ବିୟୋଗ କଲେ ଭଗ୍ନାଂଶଟି \(\frac{18}{11}\) ହୁଏ । ମାତ୍ର ଲବରେ 8 ଯୋଗକଲେ ଓ ହରକୁ 2 ଗୁଣ କଲେ ତାହା \(\frac{2}{5}\) ହୁଏ । ତେବେ ଭଗ୍ନାଂଶ କେତେ ?
ସମାଧାନ :
ମନେକର ଭଗ୍ନସଂଖ୍ୟାର ଲବ x ଓ ହର y
∴ ଭଗ୍ନ ସଂଖ୍ୟାଟି \(\frac{x}{y}\)
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{3x}{y-3}=\frac{18}{11}\)
⇒ 33x= 18y – 54 ⇒ 33x – 18y = – 54 ……..(i)
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ, \(\frac{x+8}{2y}=\frac{2}{5}\)
⇒ 5x + 40 = 4y ⇒ 5x – 4y = -40 ……….(ii)

x ର ମାନ ସମୀକରଣ (i)ରେ ବସାଇଲେ,
⇒ 33 × 12 – 18y = 54 ⇒ -18y = -54 – 396
⇒ 18y = 450 ⇒ y = \(\frac{450}{18}=25\)
∴ ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{x}{y}=\frac{12}{25}\)

Question 11.
5ଟି କଲମ ଓ ଟି ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ ମିଶି ୨ ଟଙ୍କା ଏବଂ 3ଟି କଲମ ଓ 2ଟି ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ ମିଶି 5 ଟଙ୍କା ହୁଏ । ତେବେ ଗୋଟିଏ କଲମ ଓ ଗୋଟିଏ ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ କେତେ ?
ସମାଧାନ :
ମନେକର ଗୋଟିଏ କଲମର ଦାମ୍ x ଟଙ୍କା ଓ ଗୋଟିଏ ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ y ଟଙ୍କା ।
ପ୍ରଶ୍ନନୁସାରେ, 5x + 6y = 9 …… (i)
3x + 2y = 5 …….. (ii)

xର ମାନ ସମୀକରଣ (ii)ରେ ବସାଇଲେ, 3 × \(\frac{3}{2}\) + 2y = 5
⇒ 2y = 5 – \(\frac{9}{2}\) ⇒ 2y = \(\frac{10-9}{2}\)
⇒ y = \(\frac{1}{4}\) ଟଙ୍କା ।
∴ ଗୋଟିଏ କଲମର ଦାମ୍ \(\frac{3}{2}\) ଟଙ୍କା ଓ ଗୋଟିଏ ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ \(\frac{1}{4}\) ଟଙ୍କା ।

Question 12.
ପିତାଙ୍କ ବୟସ ପୁତ୍ର ବୟସର 3 ଗୁଣ । 12 ବର୍ଷ ପରେ ପିତାଙ୍କ ବୟସ ପୁତ୍ର ବୟସର 2 ଗୁଣ ହେବ । ତେବେ ପିତା ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ କେତେ ?
ସମାଧାନ :
ମନେକର ପିତାଙ୍କର ବୟସ x ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ y ବର୍ଷ |
ପ୍ରଶ୍ନନୁସାରେ, x = 3y … (i)
ପୁନଶ୍ଚ 12 ବର୍ଷ ପରେ ପିତାଙ୍କର ବୟସ (x + 12) ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ (y + 12) ବର୍ଷ |
ପ୍ରଶ୍ନନୁସାରେ x + 12 = 2 (y + 12) ……..(ii)
ସମୀକରଣ (i)ର ମାନ ସମୀକରଣ (ii)ରେ ସଂସ୍ଥାପନ କଲେ,
3y +12 = 2 (y + 12) ⇒ 3y + 12 = 2y + 24
⇒3y – 2y = 24 – 12
⇒ y = 12 ବର୍ଷ |
ସମୀକରଣ (i)ରେ y = 12 ପ୍ରୟୋଗ କଲେ
x = 3y = 3 × 12 = 36 ବର୍ଷ |
∴ ପିତାଙ୍କର ବର୍ତ୍ତମାନ ବୟସ 36 ବର୍ଷ ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ 12 ବର୍ଷ ।

Question 13.
ଏକ ଆୟତ କ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟକୁ 5 ସେ.ମି. କମାଇ ପ୍ରସ୍ଥକୁ 3 ସେ.ମି. ବଢ଼ାଇବା ଦ୍ଵାରା ଏହାର କ୍ଷେତ୍ରଫଳ 9 ବର୍ଗ ସେ.ମି. କମିଯାଏ । ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟକୁ 3 ସେ.ମି. ଓ ପ୍ରସ୍ଥକୁ 2 ସେ.ମି. ବଢ଼ାଇବା ଦ୍ୱାରା କ୍ଷେତ୍ରଫଳ 67 ବର୍ଗ ସେ.ମି. ବଢ଼ିଯାଏ । ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ x ସେ.ମି. ଓ ପ୍ରସ୍ଥ y ସେ.ମି. ।
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = xy ବର୍ଗ ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, (x – 5) (y + 3) = xy – 9 ⇒ 3x – 5y – 6 = 0 … (i)
ପୁନଶ୍ଚ (x + 3) (y + 2) = xy + 67 = 2x +3y – 61 = 0 …….(ii)

⇒ \(\frac{x}{323}=\frac{y}{171}=\frac{1}{19}\)
⇒ x = \(\frac{323}{19}\) = 17 ଏବଂ y = \(\frac{171}{19}\) = 9
∴ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ 17 ସେ.ମି. ଓ ପ୍ରସ୍ଥ 9 ସେ.ମି. ।
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = ଦୈର୍ଘ୍ୟ × ପ୍ରସ୍ଥ = 17 × 9 = 153 ବର୍ଗ ସେ.ମି. ।

Question 14.
2 ଜଣ ପୁରୁଷ ଓ 3 ଜଣ ସ୍ତ୍ରୀ ଲୋକ ଏକତ୍ର ଗୋଟିଏ କାର୍ଯ୍ୟକୁ 5 ଦିନରେ ଶେଷ କରିପାରନ୍ତି । ସେହି କାର୍ଯ୍ୟକୁ 4 ଜଣ ପୁରୁଷ ଓ ୨ ଜଣ ସ୍ତ୍ରୀ ଲୋକ ଏକତ୍ର 2 ଦିନରେ ଶେଷ କରିପାରନ୍ତି । ତେବେ ଜଣେ ସ୍ତ୍ରୀ ଲୋକ କିମ୍ବା ଜଣେ ପୁରୁଷ ସେହି କାର୍ଯ୍ୟକୁ କେତେ ଦିନରେ ଶେଷ କରିପାରବେ ?
ସମାଧାନ :
ମନେକର ଜଣେ ପୁରୁଷ ଏବଂ ଜଣେ ସ୍ତ୍ରୀ ଗୋଟିଏ କାର୍ଯ୍ୟକୁ ଯଥାକ୍ରମେ x ଓ y ଦିନରେ ଶେଷ କରିପାରିବେ ।
∴ ଜଣେ ପୁରୁଷ ଏବଂ ଜଣେ ସ୍ତ୍ରୀ 1 ଦିନରେ କାର୍ଯ୍ୟର ଯଥାକ୍ରମେ \(\frac{1}{x}\) ଅଂଶ ଏବଂ \(\frac{1}{y}\) ଅଂଶ କରିବେ ।
2 ଜଣ ପୁରୁଷ ଓ 3 ଜଣ ସ୍ତ୍ରୀ 1 ଦିନରେ କାର୍ଯ୍ୟର = \(\frac{2}{x}\) + \(\frac{3}{y}\) ଅଂଶ କରିବେ ।
କିନ୍ତୁ 2 ଜଣ ପୁରୁଷ ଓ 3 ଜଣ ସ୍ତ୍ରୀ କାର୍ଯ୍ୟଟିକୁ 5 ଦିନରେ କରନ୍ତି ।
1 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବେ = \(\frac{1}{5}\) ଅଂଶ ।
ପ୍ରଶାନୁସାରେ, \(\frac{2}{x}+\frac{3}{y}=\frac{1}{5}\) ……….(i)
ପୁନଶ୍ଚ, 4 ଜଣ ପୁରୁଷ ଓ 9 ଜଣ ସ୍ତ୍ରୀ 1 ଦିନରେ କାର୍ଯ୍ୟର କରିବେ = \(\frac{4}{x}\) + \(\frac{9}{y}\) ଅଂଶ ।
କିନ୍ତୁ 4 ଜଣ ପୁରୁଷ ଓ 9 ଜଣ ସ୍ତ୍ରୀ କାର୍ଯ୍ୟଟିକୁ 2 ଦିନରେ କରନ୍ତି ।
1 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବେ = \(\frac{1}{2}\) ଅଂଶ ।
ପୁନଶ୍ଚ, ପ୍ରଶ୍ନନୁସାରେ, \(\frac{4}{x}+\frac{9}{y}=\frac{1}{2}\)

⇒ \(– \frac{3}{y}=\frac{4-5}{10}=\frac{-1}{10}\) ⇒ \(\frac{3}{y}=\frac{1}{10}\)
⇒ y = 30
y ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
⇒ \(\frac{2}{x}+\frac{3}{30}=\frac{1}{5}\) ⇒ \(\frac{2}{x}+\frac{1}{10}=\frac{1}{5}\)
\(\frac{2}{x}+\frac{1}{5}-\frac{1}{10}\) ⇒ \(\frac{2}{x}=\frac{1}{10}\) ⇒ x = 20
∴ ଜଣେ ପୁରୁଷ କିମ୍ବା ଜଣେ ସ୍ତ୍ରୀ ସେହି କାର୍ଯ୍ୟକୁ ଯଥାକ୍ରମେ 20 ଦିନରେ କିମ୍ବା 30 ଦିନରେ କରିବେ ।

Question 15.
A ଓ B ଏକତ୍ର କାମ କରି ଗୋଟିଏ କାର୍ଯ୍ୟକୁ 8 ଦିନରେ ଶେଷ କରିପାରନ୍ତି । ସେମାନେ ଏକତ୍ର କାର୍ଯ୍ୟ ଆରମ୍ଭ କରି 3 ଦିନ କାର୍ଯ୍ୟ କରିବା ପରେ A ଚାଲିଗଲା ଓ ଅବଶିଷ୍ଟ କାର୍ଯ୍ୟକୁ B ଏକା ଆଉ 15 ଦିନରେ ଶେଷ କଲା । ପ୍ରତ୍ୟେକ ଏକାକୀ କାମ କଲେ କେତେ ଦିନରେ କାର୍ଯ୍ୟକୁ ଶେଷ କରିପାରିବେ ?
ସମାଧାନ :
ମନେକର A ଓ B କାର୍ଯ୍ୟକୁ ଯଥାକ୍ରମେ x ଓ y ଦିନରେ କରିପାରିବେ ।
A ଓ B 1 ଦିନରେ କାର୍ଯ୍ୟଟିର ଯଥାକ୍ରମେ \(\frac{1}{x}\) ଅଂଶ ଏବଂ \(\frac{1}{y}\) ଅଂଶ କରିପାରିବେ ।
A ଓ B ମିଶି 1 ଦିନରେ କାର୍ଯ୍ୟଟିର \(\frac{1}{x}+\frac{1}{y}\) ଅଂଶ କରିପାରିବେ ।
କିନ୍ତୁ ପ୍ରଶ୍ନ ଅଛି A ଓ B କାର୍ଯ୍ୟଟିକୁ 8 ଦିନରେ କରିପାରନ୍ତି ।
∴ 1 ଦିନରେ କରିବେ = \(\frac{1}{8}\) ଅଂଶ ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{1}{x}+\frac{1}{y}=\frac{1}{8}\) …….. (i)
A ଓ B ଏକତ୍ର କାର୍ଯ୍ୟ ଆରମ୍ଭ କରିବାର 3 ଦିନ ପରେ A ଚାଲିଗଲା ।
ଅବଶିଷ୍ଟ କାର୍ଯ୍ୟଟିକୁ B ଆଉ 15 ଦିନରେ ଶେଷକଲା ।
ଅର୍ଥାତ୍ A, 3 ଦିନ ଓ B (15 + 3) = 18 ଦିନ କାର୍ଯ୍ୟ କଲାପରେ କାର୍ଯ୍ୟଟି ସମ୍ପୂର୍ଣ ହେଲା ।
∴ A, 3 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବ = \(\frac{3}{x}\) ଅଂଶ । B, 18 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବ \(\frac{18}{y}\) ଅଂଶ ।
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ, \(\frac{3}{x}+\frac{18}{y}=1\) ……….(ii)

⇒ \(– \frac{15}{y}=\frac{3-8}{8}\) ⇒ \(– \frac{15}{y}=\frac{-5}{8}\)
⇒ \(\frac{3}{y}=\frac{1}{8}\) ⇒ y = 24
y ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
⇒ \(\frac{1}{x}+\frac{1}{24}=\frac{1}{8}\) ⇒ \(\frac{1}{x}=\frac{1}{8}-\frac{1}{24}\)
\(\frac{1}{x}=\frac{3-1}{24}=\frac{2}{24}=\frac{1}{12}\) ⇒ x = 12
∴ A ଏକାକୀ କାର୍ଯ୍ୟଟିକୁ 12 ଦିନରେ ଓ B ଏକାକୀ କାର୍ଯ୍ୟଟିକୁ 24 ଦିନରେ ଶେଷ କରିପାରିବେ ।

Question 16.
A ଓ Bର ଆୟର ଅନୁପାତ 8 : 7 ଓ ବ୍ୟୟର ଅନୁପାତ 19 : 16 । ଯଦି ଉଭୟେ 1250 ଟଙ୍କା ସଞ୍ଚୟ କରିପାରନ୍ତି, ତେବେ ସେମାନଙ୍କର ଆୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
A ଓ B ର ଆୟର ଅନୁପାତ 8 : 7 ।
ମନେକର A ଓ B ର ଆୟ ଯଥାକ୍ରମେ 8x ଟଙ୍କା ଏବଂ 7x ଟଙ୍କା ।
ସେହିପରି A ଓ B ର ବ୍ୟୟର ଅନୁପାତ 19 : 16 |
ତେଣୁ ମନେକର A ଓ B ର ବ୍ୟୟ ଯଥାକ୍ରମେ 19y ଟଙ୍କା ଏବଂ 16y ଟଙ୍କା ।
ପ୍ରଶ୍ନନୁସାରେ A ର ସଞ୍ଚୟ = 1250 ଟଙ୍କା ⇒ 8x – 19y = 1250 ……..(i)
ସେହିପରି B ର ସଞ୍ଚୟ = 1250 ଟଙ୍କା ⇒ 7x – 16y = 1250 ……….(ii)

∴ A = 8x = 8 × 750 = 6000 ଟଙ୍କା । ଏବଂ Bର ଆୟ = 7x = 7 × 750 = 5250 ଟଙ୍କା ।

Question 17.
5 ବର୍ଷ ପରେ ପିତାର ବୟସ ପୁତ୍ରର ବୟସର ତିନିଗୁଣ ହେବ ଓ 5 ବର୍ଷ ପୂର୍ବେ ପିତାର ବୟସ ପୁତ୍ର ବୟସର ସାତଗୁଣ ଥିଲା । ତେବେ ସେମାନଙ୍କର ବର୍ତ୍ତମାନ ବୟସ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ପିତାର ବର୍ତ୍ତମାନ ବୟସ x ବର୍ଷ ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ y ବର୍ଷ ।
5 ବର୍ଷ ପରେ ପିତାର ବୟସ ହେବ = (x + 5) ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ ହେବ = (y + 5) ବର୍ଷ
ପ୍ରଶ୍ନନୁସାରେ, x + 5 = 3(y + 5)
⇒ x + 5 = 3y + 15 ⇒ x – 3y = 10 …(i)
ପୁନଶ୍ଚ, 5 ବର୍ଷ ପୂର୍ବେ ପିତାର ବୟସ ଥିଲା = (x – 5) ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ ଥିଲା = (y – 5) ବର୍ଷ
ପ୍ରଶ୍ନନୁସାରେ, (x – 5) = 7(y – 5)
⇒ x -5 = 7y – 35 ⇒ x – 7y = -30 …(ii)
ସମୀକରଣ (i)ରୁ ସମୀକରଣ (ii)କୁ ବିୟୋଗ କଲେ, (x – 3y) – (x – 7y) = 10 – (-30)
⇒ x – 3y – x + 7y = 40 ⇒ 4y = 40 ⇒ y = 10
ସମୀକରଣ (i) ରେ y = 10 ସ୍ଥାପନ କଲେ, x – 3 × 10 = 10 ⇒ x = 40
∴ ପିତାର ବର୍ତ୍ତମାନ ବୟସ 40 ବର୍ଷ ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ 10 ବର୍ଷ ।

Question 18.
ଗୋଟିଏ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ 2 ମି. ଅଧ୍ଵ ଓ ପ୍ରସ୍ଥ 2ମି. କମ୍ ହେଲେ, କ୍ଷେତ୍ରଫଳ 28 ବ.ମି. କମିଯାଏ; ମାତ୍ର ଦୈର୍ଘ୍ୟ 1 ମି. କମ୍ ଓ ପ୍ରସ୍ଥ 2 ମି. ଅଧ୍ବକ ହେଲେ କ୍ଷେତ୍ରଫଳ 33 ବ. ମି. ବଢ଼ିଯାଏ । ମୂଳ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ ଓ ପ୍ରସ୍ଥ ଯଥାକ୍ରମେ x ମି. ଓ Y ମି. ।
∴ କ୍ଷେତ୍ରଫଳ = ଦୈର୍ଘ୍ୟ × ପ୍ରସ୍ଥ = xy ବର୍ଗ ମି. ।
ପ୍ରଶ୍ନନୁସାରେ, ଦୈର୍ଘ୍ୟ 2 ମି. ଅଧୂକ ଏବଂ ପ୍ରସ୍ଥ 2 ମି. କମ୍ ହେଲେ କ୍ଷେତ୍ରଫଳ 28 ବର୍ଗ ମି. କମିଯାଏ ।
ତେଣୁ ପରିବର୍ତ୍ତିତ ଦୈର୍ଘ୍ୟ ଏବଂ ପ୍ରସ୍ଥ ଯଥାକ୍ରମେ (x + 2) ମି. ଏବଂ (y – 2) ମି. ହେବ ।
(x + 2) (y – 2) = xy – 28 ⇒ xy + 2y – 2x – 4 = xy – 28
⇒ 2y – 2x = -28 + 4 ⇒ y – x = \(\frac{-24}{2}\) ⇒ y – x = -12
⇒ x – y = 12 … (i)
ସେହିପରି ଦୈର୍ଘ୍ୟ 1 ମି. କମ୍ ଓ ପ୍ରସ୍ଥ 2 ମି. ଅଧ୍ଯକ ହେଲେ, କ୍ଷେତ୍ରଫଳ 33 ବର୍ଗ ମି. ବୃଦ୍ଧିପାଏ ।
∴ ପରିବର୍ତ୍ତିତ ଦୈର୍ଘ୍ୟ ଏବଂ ପ୍ରସ୍ଥ ଯଥାକ୍ରମେ (x – 1) ମି, ଏବଂ (y + 2) ମି. ହେବ ।
∴ (x – 1) (y + 2) = xy + 33 ⇒ xy + 2x – y – 2 = xy + 33
⇒ 2x – y = 33 +2 ⇒ 2x – y = 35 ……(ii)

‘x’ ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ 23 – y = 12 = y = 11
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = xy ବର୍ଗ ମି. = (23 × 11) ବର୍ଗ ମି. = 253 ବର୍ଗ ମି. ।

Question 19.
50କୁ ଏପରି ଦୁଇଟି ସଂଖ୍ୟାର ସମଷ୍ଟି ରୂପେ ପ୍ରକାଶ କର ଯେପରିକି ସଂଖ୍ୟା ଦ୍ଵୟର ବ୍ୟକ୍ରମର ସମଷ୍ଟି \(\frac{1}{12}\) ହେବ ।
ସମାଧାନ :
ମନେକର ସଂଖ୍ୟା ଦୁଇଟି x ଓ y ।
ସଂଖ୍ୟା ଦୁଇଟିର ସମଷ୍ଟି 50 । ⇒ x + y = 50 ….. (i)
x ଓ y ର ବ୍ୟକ୍ରମ ଯଥାକ୍ରମେ \(\frac{1}{x}\) ଓ \(\frac{1}{y}\) ।
ପ୍ରଶାନୁସାରେ, \(\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\) ……..(ii)
⇒ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\) ⇒ \(\frac{x+y}{xy}=\frac{1}{12}\) ⇒ \(\frac{50}{xy}=\frac{1}{12}\) (∵ x + y = 50)
⇒ xy = 600 ……..(iii)
ଆମେ ଜାଣିଛେ, x – y = \(\sqrt{(x+y)^2-4xy}\) = \(\sqrt{(50)^2-4×600}\) = \(\sqrt{100}\)
⇒ x – y = 10 ……(iv)
ସମୀକରଣ (i) ଓ (iv)କୁ ଯୋଗକଲେ, 2x = 60 ⇒ x = 30
ସମୀକରଣ (i)ରେ x = 30 ସ୍ଥାପନ କଲେ, 30 + y = 50 ⇒ y = 20
∴ ସଂଖ୍ୟାଦ୍ବୟ 30 ଓ 20 ।

Question 20.
ଗୋଟିଏ ଭଗ୍ନ ସଂଖ୍ୟାର ଲବ ଓ ହରକୁ ଯୋଗକରି ଯୋଗଫଳର ଏକ-ତୃତୀୟାଂଶ ନେଲେ, ତାହା ହରଠାରୁ 4 ଊଣା ହୁଏ ଓ ହରରେ 1 ଯୋଗକରି ଭଗ୍ନ ସଂଖ୍ୟାଟିକୁ ଲଘିଷ୍ଠ ଆକାରରେ ଲେଖୁଲେ ତାହା \(\frac{1}{4}\) ହୁଏ । ଭଗ୍ନ ସଂଖ୍ୟାଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{x}{y}\), ଯାହାର ଲବ x ଏବଂ ହର y ।
ପ୍ରଶ୍ନନୁସାରେ, ଲବ ଓ ହରକୁ ଯୋଗକରି ଯୋଗଫଳର ଏକତୃତୀୟାଂଶ ନେଲେ ତାହା ହରଠାରୁ 4 ଊଣା ହୁଏ ।
\(\frac{1}{3}\)(x+y) = y – 4 ⇒ x + y = 3y – 12 ⇒ x – 2y= – 12 …….. (i)
ପୁନଶ୍ଚ ହରରେ 1 ଯୋଗକରି ଭଗ୍ନାଂଶଟିକୁ ଲଘିଷ୍ଠ ଆକାରକୁ ଆଣିଲେ ତାହା \(\frac{1}{4}\) ହୁଏ ।
\(\frac{x}{y+1}=\frac{1}{4}\) ⇒ 4x = y + 1 ⇒ 4x – y = 1 ……….(ii)

‘y’ ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ, x – 2 × 7 = -12
⇒ x – 14 = -12 ⇒ x = 2
∴ ଭଗ୍ନାଂଶଟି \(\frac{x}{y}=\frac{2}{7}\)

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

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(b)

Question 1.
Fill in the blanks in each of the following, using the answers given against each of them :
(a) The slope and x-intercept of the line 3x – y + k = 0 are equal if k = _________ . (0, -1, 3, -9)
Solution:
-9

(b) The lines 2x – 3y + 1 = 0 and 3x + ky – 1=0 are perpendicular to each other if k = ___________ . (2, 3, -2, -3)
Solution:
2

(c) The lines 3x + ky – 4 = 0 and k – Ay – 3x = 0 are coincident if k = _____________. (1, -4, 4, -1)
Solution:
4

(d) The distance between the lines 3x – 1 = 0 and x + 3 = 0 is _________ units. (4, 2, \(\frac{8}{3}\), \(\frac{10}{3}\))
Solution:
\(\frac{10}{3}\)

(e) The angle between the lines x = 2 and x – √3y + 1 = 0 is _________. (30°, 60°, 120°, 150°)
Solution:
60°

Question 2.
State with reasons which of the following are true or false :
(a) The equation x = k represents a line parallel to x – axis for all real values of k.
Solution:
False. As the line x = k is parallel to y- axis for all values of k.

(b) The line, y + x + 1 = 0 makes an angle 45° with y – axis.
Solution:
y + x + 1 = 0
∴ Its slope = -1 = tan 135°
∴ It makes 45° with y – axis, as it makes 135° with x – axis. (True)

(c) The lines represented by 2x – 3y + 1 = 0 and 3x + 2y – k = 0 are perpendicular to each other for positive values of k only.
Solution:
2x – 3y + 1 = 0, 3x + 2y – k = 0
∴ \(m_1 m_2=\frac{2}{3} \times \frac{(-3)}{2}=-1\)
∴ The lines are perpendicular to each other for + ve values of k only. (False)

(d) The lines represented by px + 2y – 1 = 0 and 3x + py + 1 = 0 are not coincident for any value of ‘p’.
Solution:
px + 2y – 1 = 0, 3x + py + 1=0
∴ \(\frac{p}{3}=\frac{2}{p}=\frac{-1}{1} \Rightarrow p^2=6\)
and p = -3 or -2
There is no particular value of p for which \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) (True)

(e) The equation of the line whose x and y – intercepts are 1 and -1 respectively is x – y + 1 = 0.
Solution:
Equation of the line whose intercepts 1 and -1 is \(\frac{x}{1}+\frac{y}{-1}\) = 1
or, x – y = 1 (False)

(f) The point (-1, 2) lies on the line 2x + 3y – 4 = 0.
Solution:
Putting x = – 1, y = 2
we have 2 (- 1) + 3 × 2 – 4
= -2 + 6 – 4 = 0
∴ The point (-1, 2) lies on the line 2x + 3 – 4 = 0 (True)

(g) The equation of a line through (1, 1) and (-2, -2) is y = – 2x.
Solution:
The equation of the line through (1, 1) and (-2, -2) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
or, y – 1 = \(\frac{-2-1}{-2-1}\) (x – 1)
or, y – 1 = x – 1
or, x – y = 0 (False)

(h) The line through (1, 2) perpendicular to y = x is y + x – 2 =0.
Solution:
The slope of the line y = x is 1.
∴ The slope of the line perpendicular to the above line is -1.
∴ The equation of the line through (1, 2) having slope – 1 is y – y₁ = m(x – x₁)
or, y – 2 = -1 (x – 1)
or, y – 2= -x + 1
or, x + y = 3 (False)

(i) The lines \(\frac{x}{a}+\frac{y}{b}\) = 1 and \(\frac{y}{a}-\frac{x}{b}\) = 1 are intersecting but not perpendicular to each other.
Solution:
\(\frac{x}{a}+\frac{y}{b}\) = 1 and \(\frac{y}{a}-\frac{x}{b}\) = 1
∴ \(m_1 m_2=\frac{\left(-\frac{1}{a}\right)}{\frac{1}{b}} \times \frac{\left(-\frac{1}{b}\right)}{\left(-\frac{1}{a}\right)}=-1\)
∴ The lines intersect and are perpendicular to each other. (False)

(j) The points (1, 2) and (3, – 2) are on the opposite sides of the line 2x + y = 1.
Solution:
2x + y = 1
Putting x = 1, y = 2,
we have 2 × 1 + 2 = 4 > 1
Putting x – 3, y = -2,
we have 2 × 3 – 2 – 4 > 1
∴ Points (1, 2) and (3, – 2) lie on the same side of the line 2x + y = 1 (False)

Question 3.
A point P (x, y) is such that its distance from the fixed point (α, 0) is equal to its distance from the y – axis. Prove that the equation of the locus is given by, y² = α (2x – α).
Solution:

Question 4.
Find the locus of the point P (x, y) such that the area of the triangle PAB is 5, where A is the point (1, -1) and B is the tie point (5, 2).
Solution:

= \(\frac{1}{2}\) (-3x + 4y + 7) = 5
or, – 3x + 4y + 7 = 10
or, 3x – 4y + 3 =0 which is the locus of the point P (x, y).

Question 5.
A point is such that its distance from the point (3, 0) is twice its distance from the point (-3, 0). Find the equation of the locus.

Question 6.
Obtain the equation of straight lines:
(a) Passing through (1, – 1) and making an angle 150°.
Solution:
The slope of the line
= tan 150° = –\(\frac{1}{\sqrt{3}}\)
∴ The equation of the line is y – y₁ = m(x – x₁)
or, y + 1 = –\(\frac{1}{\sqrt{3}}\) (x – 1)
or, y√3 + √3 = -x + 1
or, x + y√3 + √3 – 1 = 0

(b) Passing through (-1, 2) and making intercept 2 on the y-axis.
Solution:
Let the equation of the line be
y – mx + c or, y = mx + 2
∴ As the line passes through (-1, 2)
we have 2 = – m + 2, or, m = 0
∴ Equation of the line is y = 2.

(c) Passing through the points (2, 3) and (-4, 1).
Solution:
The equation of the line is

(d) Passing through (- 2, 3) and a sum of whose intercepts in 2.
Solution:
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}\) = 1 where a + b = 2     …….(1)
Again, as the line passes through the point (-2, 3), we have \(\frac{-2}{a}+\frac{3}{b}\) = 1      ………(2)
From (1), we have a= 2 – b
∴ From (2) \(\rightarrow \frac{-2}{2-b}+\frac{3}{b}=1\)
or, – 2b + 6 – 3b = (2 – b)b
or, 6 – 5b = 2b – b²
or, b² – 7b + 6 = 0
or, (b – 6)(b – 1) = 0
∴ b = 6, 1
∴ a =2 – b = 2 – 6 = -4
or, 2 – 1 = 1
∴ Equation of the lines are \(\frac{x}{-4}+\frac{y}{6}\) = 1 or \(\frac{x}{1}+\frac{y}{1}\) = 1
i, e. -3x + 2y = 12 or, x + y = 1

(e) Whose perpendicular distance from the origin is 2 such that the perpendicular from the origin has indication 150°.
Solution:
Here p = 2, α = 150°
The equation of the line in normal form is x cos α + y sin α = p
or, x cos 150° + y sin 150° = 2
or, \(\frac{-x \sqrt{3}}{2}+y \cdot \frac{1}{2}\) = 2
or, -x √3 + y = 4
or, x√3 – y + 4 = 0

(f) Bisecting the line segment joining (3, – 4) and (1, 2) at right angles.
Solution:
The slope of the line \(\overline{\mathrm{AB}}\) is

(g) Bisecting the line segment joining, (a, 0) and (0, b) at right angles.
Solution:
Refer to (f)

(h) Bisecting the line segments joining (a, b), (a’, b’) and (-a, b), (a’, -b’).
Solution:

(i) Passing through the origin and the points of trisection of the portion of the line 3x + y – 12 = 0 intercepted between the coordinate axes.
Solution:

(j) Passing through (-4, 2) and parallel to the line 4x – 3y = 10.
Solution:
Slope of the line 4x – 3y = 10 is \(\frac{-4}{-3}=\frac{4}{3}\)
∴ The slope of the line parallel to the above line is \(\frac{4}{3}\).
∴ Equation of the line through (- 4, 2) and having slope \(\frac{4}{3}\) is y – y₁ = m(x – x₁)
or, y – 2 = \(\frac{4}{3}\) (x + 4)
or, 3y – 6 = 4x + 16
or, 4x – 3y + 22 = 0

(k) Passing through the point (a cos³ θ, a sin³ θ) and perpendicular to the straight line x sec θ + y cosec θ = α.
Solution:
The slope of the line x sec θ + y cosec θ = a is \(\frac{-\sec \theta}{{cosec} \theta}\) = -tanθ
∴ Slope of the required line  = cot θ
∴ Equation of the line through (a cos³ θ, a sin³ θ) is y – y₁ = m(x – x₁)
or, y – a sin³ θ = cot θ(x – a cos³ θ)
or, y – a sin³ θ = \(\frac{\cos \theta}{\sin \theta}\) (x – a cos³ θ)
or y sin θ – a sin⁴ θ = x cos  θ – a cos⁴ θ
or (x cos θ – y sin θ) + a(sin⁴ θ – cos⁴ θ) = 0

(l) Which passes through the point (3, -4) and is such that its portion between the axes is divided at this point internally in the ratio 2: 3.
Solution:

(m) which passes through the point (α, β) and is such that the given point bisects its portion between the coordinate axis.
Solution:

x = 2α , y = 2β
∴ Equation of the line \(\overleftrightarrow{\mathrm{AB}}\) is \(\frac{x}{2 \alpha}+\frac{y}{2 \beta}\) = 1(Intercept form)

Question 7.
(a) Find the equation of the lines that is parallel to the line 3x + 4y + 7 = 0 and is at a distance 2 from it.
Solution:
3x + 4y + 7 = 0
or, \(\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}\) = 0(Normal form)
∴ Equation of the lines parallel to the above line and 2 units away from it are \(\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}\) ± 2 = 0
or, 3x + 4y + 7 ± 10 = 0
∴ 3x + 4y + 17 = 0 and 3x + 4y – 3 = 0

(b) Find the equations of diagonals of the parallelogram formed by the lines ax + by = 0, ax + by + c = 0, lx + my = 0, and lx + my + n = 0. What is the condition that this will be a rhombus?
Solution:

(c) Find the equation of the line passing through the intersection of 2x – y – 1 = 0 and 3x – 4y + 6 = 0 and parallel to the line x + y – 2 = 0.
Solution:
Let the equation of the required line be (2x – y – 1) + λ(3x – 4y + 6) = 0
or, x(2 + 3λ) + λ(-1 – Aλ) + 6λ – 1 = 0
As this line is parallel to the line x + y – 2 = 0
we have their slopes are equal.
∴ \(-\left(\frac{2+3 \lambda}{-1-4 \lambda}\right)=\frac{-1}{1}\)
or, 2 + 3λ = -1 – 4λ
or, 7λ = -3 or, λ = \(\frac{-3}{7}\)
∴ Equation of the line is (2x – y – 1) – \(\frac{3}{7}\) (3x – 4y + 6) = 0
or, 14x – 7y – 1 – 9x + 12y – 18 = 0
or, 5x + 5y – 25= 0
or, x + y = 5

(d) Find the equation of the line passing through the point of intersection of lines x + 3y + 2 = 0 and x – 2y – 4 = 0 and perpendicular to the line 2y + 5x – 9 = 0.
Solution:
Let the equation of the line be (x + 3y + 2) + λ(x – 2y – 4) = 0
or, x(1 + λ) + y(3 – 2λ) + 2 – 4λ = 0
As this line is perpendicular to the line 2y + 5x – 9 = 0.
We have the product of their slopes is -1.
∴ \(\frac{1+\lambda}{3-2 \lambda} \times \frac{5}{2}\) = -1
or, 5 + 5λ = – 6 + 4λ
or, λ = -6 – 5 = -11.
∴ Equation of the required line is (x + 3y + 2) – 11(x – 2y – 4) = 0
or, x + 3y + 2- 11x + 22y + 44 = 0
or, – 10x + 25y + 46 = 0
or, 10x – 25y – 46 = 0

(e) Find the equation of the line passing through the intersection of the lines x + 3y – 1 = 0 and 3x – y + 1 = 0 and the centroid of the triangle whose vertices are the points (3, -1) (1, 3) and (2, 4).
Solution:
Let the equation of the required line (x + 3y – 1) + λ(3x – y + 1) = 0   … (1)
Again, the centroid of the triangle with vertices (3, – 1), (1, 3), and (2, 4) is \(\left(\frac{3+1+2}{3}, \frac{-1+3+4}{3}\right)\) = (2, 2)
As line (1) passes through (2, 2), we have (2 + 6 – 1) +1(6 – 2 + 1) = 0
or, 7 + 5λ = 0 or, λ = \(\frac{-7}{5}\)
∴ Equation of the line (x + 3y – 1) – \(\frac{7}{5}\) (3x – y + 1) = 0
or, 5x + 15y – 5 – 21x + 7y – 7 = 0
or, 22y – 16x – 12 = 0
or, 11y – 8x – 6 = 0
or, 8x – 11y + 6 = 0

Question 8.
If lx + my + 3 = 0 and 3x – 2y – 1 = 0 represent the same line, find the values of l and m.
Solution:
lx + my + 3 = 0 and 3x – 2y – 1 = 0 represents the same line
∴ \(\frac{l}{3}=\frac{m}{-2}=\frac{3}{-1}\)
∴ l = -9, m = 6

Question 9.
Find the equation of sides of a triangle whose vertices are at (1, 2), (2, 3), and (-3, -5).
Solution:
Equation of \(\overline{\mathrm{AB}}\) is \(y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\)
\(y-2=\frac{3-2}{2-1}(x-1)\)
or, y – 2 = x – 1
or, x – y + 1 = 0

Question 10.
Show that origin is within the triangle whose sides are given by equations, 3x – 2y = 1, 5x + 3y + 11 = 0, and x – 7y + 25 = 0.
Solution:



∴ The origin lies within the triangle ABC.

Question 11.
(a) Find the equations of straight lines passing through the point (3, -2) and making an angle 45° with the line 6x + 5y = 1.

or, 11x – y = 35, x + 11y + 19 = 0

(b) Two straight lines are drawn through the point (3, 4) inclined at an angle 45° to the line x – y – 2 = 0. Find their equations and obtain area included by the above three lines.
Solution:

Slope of L₂ = 0
Then as L₂ ⊥ L₃
Slope of L₃ = ∞
∴ Equation of L₂ is
y – y₁ =m(x – x₁)
or, y – 4 = 0(x – 3) = 0
or, y = 4
∴ Equation of L₃ is y – 4 = ∞ (x – 3)
or, x – 3 = 0 or, x = 3
∴ Sloving L₁ and L₂, we have
x – y – 2 = 0, y = 4
or, x = 6
The coordinates of A are (6, 4).
Again solving L1 and L3, we have
x – y – 2 = 0, x = 3
or, y = x – 2 = 3 – 2 = 1
∴ The coordinates of B are (3, 1).
Area of the triangle PAB is

(c) Show that the area of the triangle formed by the lines given by the equations y = m₁x + c₁,y = m₂x + c_2, and x = 0 is \(\frac{1}{2} \frac{\left(c_1-c_2\right)^2}{\left[m_2-m_1\right]}\)
Solution:

Question 12.
Find the equation of lines passing through the origin and perpendicular to the lines 3x + 2y – 5 = 0 and 4x + 3y = 7. Obtain the coordinates of the points where these perpendiculars meet the given lines. Prove that the equation of a line passing through these two points is 23x + 11y – 35 = 0.
Solution:
The slopes of the line 3x + 2y – 5 = 0 and 4x + 3y = 7 are \(\frac{-3}{2}\) and \(\frac{-4}{3}\)
∴ Slopes of the lines perpendicular to the above lines are \(\frac{2}{3}\) and \(\frac{3}{4}\)
∴ Equation of the lines through the origin and having slopes \(\frac{2}{3}\) and \(\frac{3}{4}\)
y = \(\frac{2x}{3}\) and y = \(\frac{3x}{4}\)
Now solving 3x + 2y – 5 = 0 and y = \(\frac{2x}{3}\)
we have 3x + \(\frac{4x}{3}\) – 5 = 0
or, 9x + 4x – 15 = 0
or, x = \(\frac{15}{13}\)
∴ y = \(\frac{2 x}{3}=\frac{2}{3} \times \frac{15}{13}=\frac{10}{13}\)
∴ The perpendicular y = \(\frac{2 x}{3}\) meets the line 3x + 2y – 5 = 0 at \(\left(\frac{15}{13}, \frac{10}{13}\right)\)
Again, solving 4x + 3y = 7 and y = \(\frac{3 x}{4}\)
we have 4x + 3 × \(\frac{3 x}{4}\) = 7
or, 16x + 9x = 28 or, x = \(\frac{28}{25}\)

Question 13.
(a) Find the length of a perpendicular drawn from the point (-3, -4) to the straight line whose equation is 12x – 5y + 65 = 0.
Solution:
The length of the perpendicular drawn from the point (- 3, -4) to the straight line 12x – 5y + 65 = 0 is

(b) Find the perpendicular distances of the point (2, 1) from the parallel lines 3x – 4y + 4 = 0 and 4y – 3x + 5 = 0. Hence find the distance between them.
Solution:
The distance of the point (2, 1) from the line 3x – 4y + 4 = 0 is \(\left|\frac{3 \times 2-4 \times 1+4}{\sqrt{9+16}}\right|=\frac{6}{5}\)
Again distance of the point (2, 1) from the line 4y – 3x + 5 = 0 is

(c) Find the distance of the point (3, 2) from, the line x + 3y – 1 = 0, measured parallel to the line 3x – 4y + 1 = 0
Solution:

Let the coordinates of M be (h, k).
As \(\overline{\mathrm{PM}} \| \mathrm{L}_1\), We have their slopes are equal.

(d) Find the distance of the point (-1, -2) from the line x + 3y – 7 = 0, measured parallel to the line 3x + 2y – 5 = 0
Solution:
Slope of the line 3x + 2y – 5 = 0 is \(\left(-\frac{3}{2}\right)\)
Equation of the line through (-1, -2) and parallel to this line is y + 2 = – \(\frac{3}{2}\) (x + 1)
⇒ 2y + 4 = -3x – 3
⇒ 3x + 2y + 7 = 0 …(1)
Given line is : x + 3y – 7 = 0    ….(2)
from (1) and (2) we get 7y – 28 = 0 Py = 4 and x = \(-\frac{35}{7}\) = -5
Thus the required distance is \(\sqrt{(-1+5)^2+(-2-4)^2} \quad=\sqrt{16+36}\)
= √52 = 2√3 units.

(e) Fine the distance of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) from the origin.
Solution:
The equation of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) x – x₁
or, y – a sin α

Question 14.
Find the length of perpendiculars drawn from the origin on the sides of the triangle whose vertices are A( 2, 1), B (3, 2), and C (- 1, -1).
Solution:

Question 15.
Show that the product of perpendicular from the points \(\left(\pm \sqrt{a^2-b^2}, 0\right)\) upon the straight line \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 is b²
Solution:

Question 16.
Show that the lengths of perpendiculars drawn from any point of the straight line 2x + 11y – 5 = 0 on the lines 24x + 7y – 20 = 0 and 4x – 3y – 2 = 0 are equal to each other.
Solution:
Let P(h, k) is any point on the line
2k + 11y – 5 = 0
∴ 2h + 11k – 5 = 0
Now the length of the perpendicular from P on the line 24x + 7y – 20 = 0 is

Clearly d₁ = d₂

Question 17.
If p and p’ are the length of perpendiculars drawn from the origin upon the lines x sec α + y cosec α = 0 and x cos α – y sin α – a cos 2α = 0
Prove that 4p² + p’² = a²
Solution:

Question 18.
Obtain the equation of the lines passing through the foot of the perpendicular from (h, k) on the line Ax + By + C = 0 and bisect the angle between the perpendicular and the given line.
Solution:

Slope of the line L is \(\frac{-\mathrm{A}}{\mathrm{B}}\)
∴ Slope of the line \(\overline{\mathrm{PM}} \text { is } \frac{\mathrm{B}}{\mathrm{A}}\)
∴ Equation of the line \(\overline{\mathrm{PM}}\) is y – y₁ = m(x – x₁)
or, y – k = \(\frac{\mathrm{B}}{\mathrm{A}}\) (x – h)
or, Ay – Ak =Bx – Bh
or, Bx – Ay + Ak – Bh = 0
∴ Equation of the bisectors of the angles between the lines L and \(\overline{\mathrm{PM}}\) is
\(\frac{\mathrm{A} x+\mathrm{B} y+\mathrm{C}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}=\pm \frac{\mathrm{B} x-\mathrm{A} y+\mathrm{A} k-\mathrm{B} h}{\sqrt{\mathrm{B}^2+\mathrm{A}^2}}\)
or, Ax + By + C = ± (Bx – Ay +Ak – Bh)

Question 19.
Find the direction in which a straight line must be drawn through the point(1, 2) such that its point of intersection with the line x + y – 4 = 0 is at a distant \(\frac{1}{3} \sqrt{6}\) from this point.
Solution:

Question 20.
A triangle has its three sides formed by the lines x + y = 3, x + 3y = 3, and 3x + 2y = 6. Without solving for the vertices, find the equation of its altitudes and also calculate the angles of the triangle.
Solution:

or 1 + 3λ = – 3 – 6λ
or 9λ = – 4 or , λ = \(\frac{-4}{9}\)
∴ Equation of \(\overline{\mathrm{AD}}\) is (x + y – 3) – \(\frac{4}{9}\) (3x + 2y – 6) = 0
or, 9x + 9y – 27 – 12x – 8y + 24 = 0
or, -3x + y – 3 = 0
or, 3x – y + 3 = 0
Let the equation of \(\overline{\mathrm{BE}}\) be (x + y – 3) + 1(x + 3y – 3) = 0
or, x(1 + λ) + y( 1 + 3λ) – 3 – 3λ = 0
As \(\overline{\mathrm{BE}} \perp \overline{\mathrm{AC}}\)
we have \(\frac{1+\lambda}{1+3 \lambda} \times \frac{3}{2}\) = -1
or, 3 + 3λ = -2 – 6λ
or, 9λ = – 5 or λ = \(\frac{-5}{9}\)
∴ Equation of \(\overline{\mathrm{BE}}\) is (x + y – 3) – \(\frac{5}{9}\) (x + 3y – 3) = 0
or 9x + 9y – 27 – 5x – 15y + 15 = 0
or, 4x – 6y – 12 = 0
or, 2x – 3y – 6 = 0
Let the equation of \(\overline{\mathrm{CE}}\) be (3x + 2y – 6) + λ (x + 3y – 3) = 0
x(3 + λ) + y (x + 3λ) – 6 – 3λ = 0
As \(\overline{\mathrm{CF}} \perp \overline{\mathrm{AB}} .\)
we have \(\frac{3+\lambda}{2+3 \lambda}\) × 1 = -1
or, 3 + λ = -2 – 3λ
or, 4λ = -2 – 3 = -5
or, λ = \(\frac{-5}{4}\)
∴ Equation of is \(\overline{\mathrm{CF}}\) (3x + 2y – 6) \(\frac{-5}{4}\) (x + 3y – 3) = 0
or, 12x + 8y – 24 – 5x – 15y + 15 =0
or, 7x – 7y – 9 = 0

Question 21.
A triangle has its vertices at P(1, -1), Q(3, 4) and R(2, 5). Find the equation of altitudes through P and Q and obtain the coordinates of their point of intersection. (This point is called the ortho-center of the triangle.)
Solution:

Question 22.
(a) Show that the line passing through (6, 0) and (-2, -4) is concurrent with the lines
2x – 3y – 11 = 0 and 3x – 4y = 16
Solution:
The equation of the line through (6,0) and (-2, -4) is

(b) Show that the lines lx + my + n = 0 mx + ny + 1 = 0 and nx + ly + m = 0 are concurrent, l + m + n = 0
Solution:
As the lines
lx + my + n = 0
mx + ny + 1 = 0
and nx + ly + m = 0 are concurrent.

Question 23.
Obtain the equation of the bisector of the acute angle between the pair of lines.
(a) x + 2y = 1, 2x + y + 3 = 0
Solution:
Equation ofthe bisectors ofthe angles between the lines x + 2y – 1 = 0 and 2x + y + 3 = 0 are \(\frac{x+2 y-1}{\sqrt{1^2+2^2}}=\pm \frac{2 x+y+3}{\sqrt{2^2+1^2}}\)
or, x + 2y – 1 = ± (2x + y + 3)
∴ x + 2y – 1 = 2x + y + 3 and
x + 2y – 1= -2x – y – 3
∴ x – y + 4 = 0 and 3x + 3y + 2 = 0
Let θ be the angle between x + 2y – 1 = 0 and x – y + 4 = 0
∴ tan θ = \(\frac{a_1 b_2-a_2 b_1}{a_1 a_2+b_1 b_2}\)
\(=\frac{1 \cdot(-1)-(+1) \cdot 2}{1 \cdot 1+2(-1)}\)
\(=\frac{-1-2}{1-2}=\frac{-3}{-1}=3\)
sec² θ = 1 + tan² θ = 1 + 9 = 10
cos² θ = 1/10
cos θ = \(\frac{1}{\sqrt{10}}<\frac{1}{\sqrt{2}}\) ⇒ θ > 45°
∴ x – y + 4 = 0 is the obtuse angle bisector.
⇒ 3x + 3y + 2 = 0 is acute angle bisector.

(b) 3x – 4y = 5, 12y – 5x = 2
Solution:
Given equation of lines are
3x – 4y – 5 = 0    …..(1) and  5x – 12y + 2 = 0     ……(2)
Equation of bisectors of angles between these Unes are:

Question 24.
(a) Find the coordinates of the center of the inscribed circle of the triangle formed by the line x cos α + y sin α = p with the coordinate axes.
Solution:


\(\left(\frac{p}{\sin \alpha+\cos \alpha+1}, \frac{p}{\sin \alpha+\cos \alpha+1}\right)\)

(b) Find the coordinates of the circumcentre and incentre of the triangle formed by the lines 3x – y = 5, x + 2y = 4, and 5x + 3y + 1 = 0.
Solution:

Question 25.
The vertices B, and C of a triangle ABC lie on the lines 3y = 4x and y = 0 respectively, and the side \(\overline{\mathbf{B C}}\) passes through the point (2/3, 2/3). If ABOC is a rhombus, where O is the origin, find the equation of \(\overline{\mathbf{B C}}\) and also the coordinates of A.
Answer:
Let the coordinates of C be (a, 0) so that the length of the side of the rhombus is ‘a’

Question 26.
Find the equation of the lines represented by the following equations.
(a) 4x² – y² = 0
Solution:
4x² – y² = 0
or, (2x + y)(2x – y) = 0
∴ 2x + y = 0 and 2x – y = 0 are the two separate lines.

(b) 2x² – 5xy – 3y²
Solution:
2x² – 5xy – 3y²
or, 2x² – 6xy + xy – 3y² = 0
or, 2x(x – 3y) + y(x – 3y) = 0
or, (x – 3y)(2x + y) = 0
∴ x – 3y = 0 and 2x + y = 0 are the two separate lines.

(c) x² + 2xy sec θ + y² = 0
Solution:
x² + 2xy sec θ + y² = 0
∴ a = 1, b = 2y sec θ, c = y²
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{2 y \sec \theta \pm \sqrt{4 y^2 \sec ^2 \theta-4 y^2}}{2}\)
= \(\frac{-2 y \sec \theta \pm 2 y \tan \theta}{2}\)
= y (- sec θ ± tan θ)
∴ x = y (- sec θ + tan θ) and x – y (- sec θ – tan θ) are the two separate lines.

(d) 3x² + 4xy = 0
Solution:
3x² + 4xy = 0
or x (3x + 4y) = 0
∴ x = 0 and 3x + 4y = 0 are the two separate lines.

Question 27.
From the equations which represent the following pair of lines.
(a) y = mx; y = nx
Solution:
y – mx = 0, y – nx = 0
or (y – mx) (y – nx) = 0
or, y² – nxy – mxy + mnx² = 0
or, y² – xy (m + n) + mnc² = 0 which is the equation of a pair of lines.

(b) y – 3x = 0 ; y + 3x = 0
Solution:
y – 3x = 0, y + 3x = 0
∴ (y – 3x) (y + 3x) = 0
or, y² – 9x² = 0 which is the equation of a pair of lines.

(c) 2x – 3y + 1 = 0 ; 2x + 3y + 1 = 0
Solution:
2x – 3y + 1 = 0; 2x + 3y + 1 =0
or, (2x – 3y + 1)(2x + 3y + 1) = 0
or, (2x + 1)² – 9y² = 0
or, 4x² + 1 + 4x – 9y² = 0
or, 4x² – 9y² + 4x + 1= 0 which represents a pair of lines.

(d) x = y. x + 2y + 5 = 0
Solution:
x = y, x + 2y + 5 = 0
∴ (x – y) (x + 2y + 5) = 0
or, x² + 2xy + 5x – xy – 2y² – 5y =0
or, x² – 2y² + xy + 5x – 5y = 0 which represents a pair of lines.

Question 28.
Which of the following equations represents a pair of lines?
(a) 2x² – 6y² + 3x +  y + 1 = 0
Solution:
a = 2, b = -6, 2g = 3
2f = 1, c = 1
∴ g = \(\frac{3}{2}\), f = \(\frac{1}{2}\), h = 0
∴ abc + 2fgh – ah² – bg² – ch²
= 2(-6). 1 + 2 × \(\frac{1}{2}\) × \(\frac{3}{2}\) – 0 – (-6) × \(\frac{9}{4}\) – 1 × 0
= -12 + \(\frac{3}{2}\) + \(\frac{27}{2}\) = \(\frac{6}{2}\) = 3 ≠ 0
∴ The given equation does not represent a pair or lines.

(b) 10x² – xy – 6y² – x + 5y – 1 = 0
Solution:
a = 10. 2h = 1
B = -6, 2g = -1
2f = 5. C= -1
∴ h = –\(\frac{1}{2}\), g = –\(\frac{1}{2}\) , f = \(\frac{5}{2}\)
∴ abc + 2fgh – ah² – bg² – ch²
= 10(-6)(-1) + 2 × \(\frac{5}{2}\) × (-\(\frac{1}{2}\)) × (-\(\frac{5}{2}\)) – 10 × \(\frac{2.5}{4}\) – (-6)\(\frac{1}{4}\) – (-1)\(\frac{1}{4}\)
= 60 + \(\frac{5}{4}\) – \(\frac{250}{4}\) + \(\frac{6}{4}\) + \(\frac{1}{4}\)
= \(\frac{240+5+6-250+1}{4}=\frac{2}{4}\)
∴ The given equation does not represent a pair of lines.

(c) xy + x + y + 1 = 0
Solution:
xy + x + y + 1= 0
or, x(y + 1) + 1(y + 1 ) = 0
or (y + 1 )(x + 1) =0
∴ x + 1 = 0
and y + 1 = 0 are the two separate lines,
∴ The given equation represents a pair of lines.

Question 29.
For what value of λ do the following equations represent pair of straight lines?
(a) λx² + 5xy – 2y² – 8x + 5y – λ = 0
Solution:
λx² + 5xy – 2y² – 8x + 5y – λ = 0
∴ a = λ, 2h = 5, b = -2, 2g = -8
2f = 5, c = -1
∴ h = \(\frac{5}{2}\), g = -4, f = \(\frac{5}{2}\)
As the given equation represent a pair of lines, we have abc + 2fgh – ah² – bg² – ch² = 0
or, λ(-2)(-λ) + 2. \(\frac{5}{2}\) (-4). \(\frac{5}{2}\) -λ × \(\frac{25}{4}\) – (-2) (-4)² – (-λ) × \(\frac{25}{4}\) = 0
or, 2λ² – 50 – \(\frac{25 λ }{4}\) + 32 + \(\frac{25 λ }{4}\) = 0
or, 2λ² = 18 or, λ² = 9
λ = ±3

(b) x² – 4xy – y² +6x + 8y + λ = 0
Solution:
Here a = 1, 2h = -1, b = -1, 2g = 6, 2f = 8, c = τ
As the given equation represent a pair of lines, we have
abc + 2fgh – af² – bg² – ch² = 0
⇒ (-1) τ + 2.4.3 (-2) – 1. 4² – (-1). ³2 – τ(-2)² = 0
⇒ -τ – 48 – 16 + 9 – 4τ = 0
⇒ -5τ – 55 = 0 ⇒ τ = -11

Question 30.
(a) Obtain the value of λ for which the pair of straight lines represented by 3x² – 8xy + λy² = 0 are perpendicular to each other.
Solution:
3x² – 8xy + λy² = 0
∴ a = 3. 2h = -8, b = λ
As the pair of lines are perpendicular to each other, we have a + b = 0.
or, 3 + λ = 0 – or, λ = -3

(b) Prove that a pair of lines through the origin perpendicular to the pair of lines represented by px² – 2qxy + ry² = 0 is given by rx² – 2qxy + py² = 0
Solution:
px² – 2qxy + ry² = 0
∴ a = p, b = 2qy, c = ry²

(c) Obtain the condition that a line of the pair of lines ax² + 2hxy + by² = 0,
(i) Coincides with
Solution:

(ii) is perpendicular to, a line of the pair of lines px² + 2qxy + ry² = 0
Solution:

Question 31.
Find the acute angle between the pair of lines given by :
(a) x² + 2xy – 4y² = 0
Solution:
x² + 2xy – 4y² = 0
∴ a=1, 2h = 2, b = -4
∴ tan θ = \(\frac{\pm 2 \sqrt{h^2-a b}}{a+b}=\frac{\pm 2 \sqrt{1+4}}{1-4}\)
\(=\pm \frac{2 \sqrt{5}}{-3}=\mp \frac{2 \sqrt{5}}{3}\)
∴ The acute angle between the pair of lines is tan^-1 \(\frac{2 \sqrt{5}}{3}\)

(b) 2x² + xy – 3y² + 3x + 2y + 1 = 0
Solution:
2x² + xy – 3y² + 3x + 2y + 1 = 0
∴ a = 2, 2h = 1, b = -3, 2g = 3 2f= 2, c = 1.
tan θ = \(\frac{\pm 2 \sqrt{h^2-a b}}{a+b}\)
\(=\pm \frac{2 \sqrt{\frac{1}{4}+6}}{2-3}=\pm \frac{2 \times 5}{2(-1)}\) = ± 5
∴ The acute angle is tan^-1 5

(c) x² + xy – 6y² – x – 8y – 2 = 0
Solution:
Given Equation is x² + xy – 6y² – x – 8y – 2 = 0
here a = 1, 2h = 1, b = -6 thus if 0 is the acute angle between two lines then
tan θ = \(=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right|\)
= \(\left|\frac{2 \times 5}{-10}\right|\) = 1
∴ θ = 45°

Question: 32.
Write down the equation of the pair of bisectors of the following pair of lines :
(a) x² – y² = 0 ;
Solution:
x² – y² = 0
∴ a = 1, b = -1, h = 0
∴ The equation of the bisectors of the angles between the pair of lines are \(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{1+1}=\frac{x y}{0}\)
or, xy = 0

(b) 4x² – xy – 3y² = 0
Solution:
4x² – xy – 3y² = 0
∴ a = 4, 2h = -1, b = -3
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{(a-b)}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{7}=\frac{x y}{\left(-\frac{1}{2}\right)}\)
or, x² – y² = -14xy
or, x² + 14xy – y² = 0

(c) x² cos θ + 2xy – y² sin θ = 0
Solution:
x² cos θ + 2xy – y² sin θ = 0
∴ a = cos θ, 2h = 2, b = – sin θ
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{\cos \theta+\sin \theta}=\frac{x y}{1}\)
or, x² – y² = xy(cos θ + sin θ)

(d) x² – 2xy tan θ – y² = 0
Solution:
x² – 2xy tan θ – y² = 0
∴ a = 1, 2h = -2 tan θ, b = -1
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{2}=\frac{x y}{-\tan \theta}\)
or, x² – y² = 2xy cot θ
or, x² + 2xy cot θ – y² = 0

Question 33.
If the pair of lines represented by x² – 2pxy – y² = 0 and x² – 2qxy – y² = 0 be such that each pair bisects the angle between the other pair, then prove that pq = -1.
Solution:

Question 34.
Transform the equation: x² + y² – 2x – 4y + 1 = 0 by shifting the origin to (1, 2) and keeping the axes parallel.
Solution:
x² + y² – 2x – 4y + 1 = 0     ……(1)
Let h = 1, k = 2
Taking x’ + h and y = y’ + k we have
(x’ + h)² + (y’ + k)² – 2(x’ + j) -4(y’ + k) + 1=0
or, (x + 1)² + (y’ + 2)² – 2(x’ + 1) – 4(y’+ 2) + 1=0
or, x^‘2 + 1 + 2x’ + y^‘2 + 4 – 4y’ – 2x’ – 2 – 4y’- 8 + 1 = 0
or, x^‘2 + y’² – 4 = 0
∴ The transformed equation is x² + y² = 4

Question 35.
Transform the equation: 2x² + 3y² + 4xy – 12x – 14y + 20 = 0. When referred to parallel axes through(2, 1).
Solution:
2x² + 3y² + 4xy – 12x – 14y + 20 = 0
Let h = 2, k = 1
Taking x = x’ + 1 and y = y’ + 1
we have
2(x’ + 2)² + 3(y’ + 1)² + 4(x’ + k)(y’ + 1) – 12 (x’ + 2)- 14 (y’ + 1) + 20 = 0
or, 2x^‘2 + 8 + 8x’ + 3 + 6y’ + 3y^’2 + 4x’y’ + 4x’ + 8y’ + 8 – 12x’ – 14y’ – 18 = 0
or, 2x^‘2 + 3y^’2 + 4x’y’ + 1=0
The transformed equation is
2x² + 3y² + 4xy + 1 = 0

Question 36.
Find the measure of rotation so that the equation x² – xy + y² = 5 when transformed does not contain xy- term.
Solution:
x² – xy + y² = 5
Taking x = x’ cos α – y’ sin α
y = x’ sin α – y’ cos α
We get (x’ cos α – y’ sin α)² – (x’ cos α – y’ cos α) (x’ sin α + y’ cos α) + (x’ sin α + y’ cos α)² = 5
⇒ x^‘2 cos² α + y^‘2 sin² α – 2x’y sin α.
cos α – x^‘2 sin α. cos α – x’y cos² α + x’y’ sin² α + y^‘2 sin α. cos α + x^‘2 sin² α + y^‘2 cos² α + 2x’y’ sin α cos α = 5
Given that the transformed equation does not xy term.
Hence the co-efficient of x’y’ is zero.
That is sin² α – cos² α = 0
⇒ sin² α = cos² α
⇒ tan² α = 1 ⇒ tan α = 1 ⇒ α= 45°

Question 37.
What does the equation x + 2y – 10 =0 become when the origin is changed to (4, 3)?
Solution:
x + 2y – 10 = 0
Let h = 4, k = 3
Taking x = x’ + 4, y = y’ + 3
we have x’ + 4 + 2 (y’ + 3) – 10 = 0
or, x + 2y’ = 0
∴ The transformed equation is x + 2y = 0.

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(d) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(d)

Question 1.
ବନ୍ଧନୀ ମଧ୍ଯରୁ ଠିକ୍ ଉତ୍ତର ବାଛି ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।

(i) ପାର୍ଶ୍ୱସ୍ଥ ଚିତ୍ରରେ ଥିବା △ABC ରେ ∠ABC = 90°
ଏବଂ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\), m∠ABD = ………
[m∠BAD, m∠DBC, m∠DCB, 2m∠BAD]

Solution:
m∠DCB

(ii) ଉପ6ର।କ୍ତ ରିତ୍ର6ର ଥିବା △ABC ରେ ∠ABC ସମ6କାଶ ଏବଂ BD AC ହେଲେ,
(a) AB2 = AD × ……….. [BC, CD, AC, BD]
(b) BC2 = AC × ……….. [DC, AD, BD, AB]
(c) BD2 = DC × ………. [AC, BC, AB, AD]
Solution:
(a) AC
(b) DC
(c) AD

Question 2.
ବମ୍ନ ଟିତ୍ର6ର ଥିର୍। △PQR ର m∠PQR = 90° ଏର୍ବ \(\overline{\mathbf{QM}}\) ⊥ \(\overline{\mathbf{PR}}\)
(i) QM = 12 6ସ.ମି. ଏବଂ PM = 6 6ସ.ମି. 6ଦ୍ର6କ , PR ନିଣ୍ଟଯ କର |
(ii) PQ = 6 6ସ.ମି. ଏବଂ PM = 3 6ସ.ମି. 6ଦ୍ର6କ , PR ନିଣ୍ଟଯ କର |
(iii) QR = 12 6ସ.ମି. ଏବଂ MR = 3 6ସ.ମି. 6ଦ୍ର6କ , PM ନିଣ୍ଟଯ କର |
(iv) PQ = 12 6ସ.ମି. ଓ RM = 3 6ସ.ମି. 6ଦ୍ର6କ , PM ନିଣ୍ଟଯ କର |
(v) PQ = 8 6ସ.ମି. ଓ QR = 15 6ସ.ମି. 6ଦ୍ର6କ , QM ଓ MR ନିଣ୍ଟଯ କର |

Solution:
(i) QM² = PM × RM ⇒ 12² = 6 × RM
⇒ RM = \(\frac{12^2}{6}\) 6ସ.ମି. = 24 6ସ.ମି.
PR = PM + RM = 6 6ସ.ମି. + 24 6ସ.ମି. = 30 6ସ.ମି.

(ii) PQ² = PM × PR
⇒ PR = \(\frac{\mathrm{PQ}^2}{\mathrm{PM}}\) = \(\frac{6^2}{3}\) 6ସ.ମି. = 12 6ସ.ମି.

(iii) QR² = MR × PR
⇒ PR = \(\frac{\mathrm{QR}^2}{\mathrm{MR}}\) = \(\frac{12^2}{9}\) 6ସ.ମି. = 16 6ସ.ମି.
PM = PR – MR = 16 6ସ.ମି. – 9 6ସ.ମି. = 7 6ସ.ମି. |

(iv) ମ6ନଜର PM = x 6ସ.ମି.
∴ PR = PM + MR = (x + 7) 6ସ.ମି. |
PQ² = PM . PR ⇒ 12² = x(x + 7)
⇒ x² + 7x – 144 = 0 ⇒ x² + 16x – 9x – 144 = 0
⇒ x(x + 16) -9(x + 16) = 0 ⇒ (x – 9) (x + 16) = 0
⇒ x = 9 ବ।, x = -16 (ଅସମ୍ମତ)
∴ PM = x 6ସ.ମି. = 9 6ସ.ମି.

(v) PR = \(\sqrt{\mathrm{PQ}^2+\mathrm{QR}^2}\) = \(\sqrt{8^2+15^2}\) = \(\sqrt{64+225}\) = \(\sqrt{289}\) = 17 6ସ.ମି.
QR² = MR × PR
⇒ MR = \(\frac{\mathrm{QR}^2}{\mathrm{PR}}\) = \(\frac{15 \times 15}{17}\) = \(\frac { 225 }{ 17 }\) = 13\(\frac { 4 }{ 17 }\) 6ସ.ମି.
QM² = PM × MR = (PR – MR) × MR
= (17 – \(\frac { 225 }{ 17 }\)) × \(\frac { 225 }{ 17 }\) = \(\frac { 64 }{ 17 }\) × \(\frac { 225 }{ 17 }\) ଦଙ୍ଗ6ସ.ମି.
∴ QM = \(\sqrt{\frac{64 \times 225}{17 \times 17}}\) = \(\frac{8 \times 15}{17}\) = \(\frac { 120 }{ 17 }\) = 7\(\frac { 1 }{ 17 }\) 6ସ.ମି. |

Question 3.
ବମ୍ନ ଟିତ୍ର6ର m∠ABC = m∠DCB = 90°, \(\overline{\mathbf{AC}}\) ଓ \(\overline{\mathbf{BD}}\) ର ଛେଦଦିନ୍ଦୁ O ଏବଂ \(\overline{\mathbf{AC}}\) ⊥ \(\overline{\mathbf{BD}}\) | OC = 6 6ସ.ମି. ଦଙ୍ଗ6ସ.ମି.
(i) BO ନିଶ୍ଚୟ କର;
(ii) OA ନିର୍ଣ୍ଣୟ କର;
(iii) BC ନିର୍ଣ୍ଣୟ କର;
(iv) AB ନିର୍ଣ୍ଣୟ କର; ଏବଂ
(v) CD ନିର୍ଣ୍ଣୟ କର ।

Solution:

Question 4.
△ABC ଭେ ∠ABC ସପ6କାଣ ଏବଂ \(\overline{\mathbf{B D}}\) ⊥ \(\overline{\mathbf{AC}}\) | AD = p ଏକକ ପ୍ରତେକ, ପ୍ତମାଣ କର : BC = \(\frac{q \sqrt{\mathbf{p}^2+q^2}}{p}\)
Solution:

Question 5.
△ABC ଭେ, m∠ABC = 90° ଏବଂ BD ⊥ AC ଦ୍ରେଲେ , ପ୍ରତମାଣ କର ଯେ , AB² : BC² = AD : DC |
Solution:

Question 6.
△ABC ଭେ, ∠ABC ସପ6କାଶ , \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ BC² = AC.BD କ୍ତେଲେ , ପ୍ରମାଣ ଦର ଯେ \(\overline{\mathrm{BD}}\) 6ହର୍ଛି ∠ABC ତ ସପଦ୍ଦିଖଣ୍ଡକ |
Solution:

ଦର : △ABC ରେ , ∠B = 90° | \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ BC² = AC.BD
ପ୍ରମ।ଣu : \(\overline{\mathrm{BD}}\) , ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡକ |
ପ୍ରମ।ଣ : △BDC ~ △ABC ⇒ BC² = AC . AD ….(i)
କକୁ ଦଣ BC² = AC . AD ….(ii)
(i) ଓ (ii) ରୁ AC . AD = AC . BD ⇒ CD = BD ⇒ m∠DBC = m∠DCB … (iii)
କନ୍ତି △ABD ~ △BCD ⇒ m∠ABD = m∠DCB …(iv)
(iii) ଓ (iv) ର m∠DBC = m∠ABD ⇒ \(\overline{\mathrm{BD}}\) , ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ କ | (ପ୍ରମାଣିତ)

Question 7.
ପାଣ୍ଡଣ୍ ଚିତ୍ରରେ ଥ୍ରନା ରହୁକ୍ତିଙ ABCD ରେ m∠ABC = m∠ADC = 90° ଏବଂ AB = AD | କଣ୍ତଦ୍ରଯତ 6ଛଦଦିନ୍ଦୁ M ଦ୍ତେକେ , ପ୍ରମାଣ କର ଯେ AM × MC = DM | (ପ୍ରମେଯ 1.4 ର ସ୍ତ ଯୋଗ କରି ପ୍ରମାଣ କମା | )
Solution:

Question 8.
△ABC ରେ m∠ABC = 90°, \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ \(\overline{\mathrm{AC}}\) କ୍ E ଦିନ୍ଦୁରେ 6ଛଦ କଲୋ ପ୍ରମାଣ କର ଯେ AE² : EC² = AD : DC |
Solution:
ଦର : △ABC ରେ m∠ABC = 90° , \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ \(\overline{\mathrm{AC}}\) କ୍ E ଦିନ୍ଦୁରେ 6ଛଦ କଲୋ

ପ୍ତାମାଣଧ୍ୟ : \(\frac{\mathrm{AE}^2}{\mathrm{EC}^2}\) = \(\frac { AD }{ DC }\)
ପ୍ତମାଣ : △ABC ରେ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ \(\overline{\mathrm{AC}}\) କ୍ E ଦିନ୍ଦୁରେ 6ଛଦ କଲୋ
⇒ \(\frac { AB }{ BC }\) = \(\frac { AE }{ EC }\) ⇒ \(\frac{\mathrm{AE}^2}{\mathrm{EC}^2}\) = \(\frac{\mathrm{AB}^2}{\mathrm{BC}^2}\)
△ABC ରେ m∠ABC = 90° ଓ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\)
⇒ AB² = AD . AC ଓ BC² = CD . AC
∴ \(\frac{\mathrm{AE}^2}{\mathrm{EC}^2}\) = \(\frac{\mathrm{AB}^2}{\mathrm{BC}^2}\) = \(\frac{\mathrm{AD} \times \mathrm{AC}}{\mathrm{CD} \times \mathrm{AC}}\) = \(\frac { AD }{ CD }\) (ପ୍ରମାଣିତ)

Question 9.
△ABC ରେ , m∠BAC = 90° ଏବଂ \(\overline{\mathrm{AD}}\) ⊥ \(\overline{\mathrm{BC}}\) | ପ୍ରମାଣ କର ଯେ △ADC ର ଷ୍ଟ୍ରେତୃଫକ = \(\frac{\mathbf{A B} \times \mathbf{A C} \mathbf{C}^3}{2 \mathbf{B C} C^2}\) |

Solution:
ଦଉ : △ABC 6ର m∠BCA = 90° ଏବଂ \(\overline{\mathrm{AD}}\) | \(\overline{\mathrm{BC}}\) |
ପ୍ର।ମାଣ୍ୟ : △ADC ର ଷ୍ଟେର୍ଫଳ = \(\frac{\mathrm{AB} \times \mathrm{AC}^3}{2 \mathrm{BC}^2}\) |
ପ୍ରମାଣ : ABC ରେ m∠BAC = 90° ଓ \(\overline{\mathrm{AD}}\) ⊥ \(\overline{\mathrm{BC}}\)
⇒ AC² = CD . BC
△ABC ର ଷ୍ଟେର୍ଫଳ = \(\frac { 1 }{ 2 }\) AC × AB = \(\frac { 1 }{ 2 }\) BC × AD 

Question 10.
△ABC ର m∠ABC ସମକୋଣ,, \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠BAC ର ଛେଦକରେ \(\overline{\mathrm{BD}}\) କୁ E ଦିନ୍ଦୁରେ ଛେଦକରେ | ପ୍ତପାଣ କର ଯେ BE² : DE² = AC : AD |
Solution:
ଦତ୍ତ : △ABC ରେ ∠ABC ସମକୋଣ, \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠BAC ର ସମଦ୍ବିଖଣ୍ଡକ \(\overline{\mathrm{BD}}\) କୁ E ବିନ୍ଦୁରେ ଛେଦକରେ ।
ପ୍ରାମାଣ୍ୟ : \(\frac{\mathrm{BE}^2}{\mathrm{DE}^2}\) : \(\frac{\mathrm{AC}}{\mathrm{AD}}\)
ପ୍ରମାଣ : △ABC ରେ ∠B ସମକୋଣ ଓ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) |
⇒ AB² = AD × AC
△ABD ରେ ∠BAD ର ସମଦ୍ଦିଖଣ୍ଡକ ରଶ୍ମି \(\overline{\mathrm{BD}}\) କୁ E ରେ ଛେଦକରେ ।
⇒ \(\frac { AB }{ AD }\) = \(\frac { BE }{ DE }\)
∴ \(\frac{\mathrm{BE}^2}{\mathrm{DE}^2}\) = \(\frac{\mathrm{AB}^2}{\mathrm{AD}^2}\) = \(\frac { AD × AC }{ AD × AD }\) = \(\frac { AC }{ AD }\) (∵ AB² = AD × AC) (ପ୍ରମାଣିତ)

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

BSE Odisha Class 10 Maths Solutions Algebra Chapter 2 ଦ୍ବିଘାତ ସମୀକରଣ Ex 2(b)

Question 1.
ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନମାନଙ୍କର ଉତ୍ତର ଦିଅ ।
(i) ଗୋଟିଏ ସଂଖ୍ୟା ଓ ଏହାର ବ୍ୟତ୍‌କ୍ରମର ସମଷ୍ଟି 2 । ସଂଖ୍ୟାଟିକୁ x ନେଇ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(ii) ଦୁଇଗୋଟି କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଗୁଣଫଳ 20 । ସଂଖ୍ୟାଦ୍ଵୟ ମଧ୍ୟରୁ ଗୋଟିକୁ y ନେଇ ଆବଶ୍ୟକୀୟ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(iii) ଦୁଇଟି ସଂଖ୍ୟାର ସମଷ୍ଟି 18 ଏବଂ ସେମାନଙ୍କର ଗୁଣଫଳ 72 । ଗୋଟିଏ ସଂଖ୍ୟାକୁ x ନେଇ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(iv) କୌଣସି ସଂଖ୍ୟା, ତାହାର ବର୍ଗ ସମାନ ହେଲେ ସଂଖ୍ୟାଟି ନିର୍ଣ୍ଣୟ କର ।
(v) ପ୍ରଥମ n ସଂଖ୍ୟକ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି S = \(\frac{n(n+1) }{2}\)। ଯଦି S = 120 ହୁଏ, ତେବେ nର ମୂଲ୍ୟ ନିରୂପଣ କରିବା ପାଇଁ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(vi) \(\sqrt{x}\) + x = 6 କୁ ଏକ ଦ୍ବିଘାତ ସମୀକରଣ ରୂପେ ପ୍ରକାଶ କର ।
(vii) \(\sqrt{x+9}\) + 3 = x କୁ ଏକ ସ୍ଵିଘାତ ସମୀକରଣ ରୂପେ ପ୍ରକାଶ କର ।
(viii) x – 2\(\sqrt{2}\) – 6 = 0 ସମୀକରଣକୁ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ରୂପେ ପ୍ରକାଶ କର ।
ସମାଧାନ :
(i) ମନେକର ସଂଖ୍ୟାଟି x । ସଂଖ୍ୟାଟିର ବ୍ୟକ୍ରମ \(\frac{1}{x}\)
ପ୍ରଶ୍ନନୁସାରେ, x + \(\frac{1}{x}\) = 2 ⇒ x² + 1 = 2x ⇒ x² – 2x + 1 = 0
∴ ନିଶ୍ଚେୟ ଦ୍ବିଘାତ ସମୀକରଣ x² – 2x + 1 = 0

(ii) ମନେକର ଗୋଟିଏ ସଂଖ୍ୟା y । ଅନ୍ୟଟି = (y + 1)
ପ୍ରଶ୍ନନୁସାରେ, y (y + 1) = 20 ⇒ y² + y – 20 = 0
∴ ନିଶ୍ଚେୟ ଦ୍ବିଘାତ ସମୀକରଣ y + y – 20 = 0

(iii) ମନେକର ପ୍ରଥମ ସଂଖ୍ୟାଟି x ଓ ଦ୍ବିତୀୟ ସଂଖ୍ୟାଟି 18 – x ।
ପ୍ରଶ୍ନନୁସାରେ, x (18 – x) = 72 ⇒ 18x – x² = 72 = x² – 18x + 72 = 0
∴ ନିଶ୍ଚେୟ ଦ୍ବିଘାତ ସମୀକରଣ, x² – 18x + 72 = 0

(iv) ମନେକର ସଂଖ୍ୟାଟି x।
ପ୍ରଶ୍ନନୁସାରେ ସଂଖ୍ୟାଟି ତା’ର ବର୍ଗ ସହ ସମାନ ହେବ । ଅର୍ଥାତ୍ x = x²
⇒ x² – x = 0 ⇒ x ( x – 1 ) = 0
⇒ x = 0 ବା x – 1 = 0 ⇒ x= 0 ବା x = 1
∴ ସଂଖ୍ୟାଟି 0 କିମ୍ବା 1 ।

(v) ପ୍ରଶ୍ନନୁସାରେ \(\frac{n(n+1) }{2}\) = 120 ⇒ n² + n = 240 ⇒ n² + n – 240 = 0

(vi) √x + x = 6 ⇒ √x = 6 – x
⇒ (√x)² =(6 – x)² ⇒ x = 36 + x² – 12x
⇒ x² – 12x – x + 36 = 0 ⇒ x² – 13x + 36 = 0

(vii) √(x + 9) + 3 = x ⇒ √(x + 9) = x – 3
⇒ (\(\sqrt{x+9}\))² = (x – 3)² ⇒ x² – 6x + 9 = x + 9
⇒ x² – 6x – x + 9 – 9 = 0 = x² – 7x = 0
ବିକଳ୍ପ ସମାଧାନ :
ମନେକର \(\sqrt{x+9}\) = t
⇒ x + 9 = t² ⇒ x = t² – 9
xର ମାନ ସମୀକରଣରେ ବସାଇଲେ k + 3 = t² – 9
⇒ t – t – 9 – 3 = 0 ⇒ t – t – 12 = 0

(viii) x – 2√2 – 6 = 0 ⇒ x – 6 = 2√2
⇒ (x – 6)² = (2√2)² ⇒ x² – 12x + 36 = 8
⇒ x² – 12x + 36 – 8 = 0 ⇒ x² – 12x + 28 = 0

Question 2.
ନିମ୍ନଲିଖ ପ୍ରଶ୍ନମାନଙ୍କର ଉତ୍ତର ଦିଅ ।
(i) ଗୋଟିଏ ଧନାତ୍ମକ ସଂଖ୍ୟା ତାହାର ବର୍ଗମୂଳ ଅପେକ୍ଷା 12 ଅଧିକ ହେଲେ, ସଂଖ୍ୟାଟି ନିରୂପଣ କର ।
(ii) ଗୋଟିଏ ସଂଖ୍ୟା ଏବଂ ତାହାର ବ୍ୟତ୍‌କ୍ରମର ସମଷ୍ଟି \(\frac{41}{20}\) ହେଲେ ସଂଖ୍ୟାଟି ସ୍ଥିର କର ।
(iii) ଦୁଇଟି କ୍ରମିକ ପୂର୍ଣ ସଂଖ୍ୟାର ବ୍ୟୁତ୍‌କ୍ରମ ଭଗ୍ନସଂଖ୍ୟା ଦ୍ବୟର ଯୋଗଫଳ \(\frac{11}{30}\) ଦ୍ଵୟକୁ ନିରୂପଣ କରିବା ପାଇଁ ଆବଶ୍ୟକ ଦ୍ଵିଘାତ ସମୀକରଣଟି ଗଠନ କରି ସଂଖ୍ୟାଦ୍ଵୟ ନିରୂପଣ କର ।
(iv) ଦୁଇଟି କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ବ୍ୟୁତ୍‌କ୍ରମର ସମଷ୍ଟି \(\frac{25}{123}\) ହେଲେ ସଂଖ୍ୟାଦ୍ଵୟ ନିର୍ଣ୍ଣୟ କର ।
(v) ଯଦି 51 କୁ ଦୁଇଭାଗ କଲେ ସେମାନଙ୍କର ଗୁଣଫଳ 378 ହୁଏ, ତେବେ ସଂଖ୍ୟା ଦ୍ଵୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
(i) ମନେକର ଧନାତ୍ମକ ସଂଖ୍ୟାଟି x² । ଏହାର ବର୍ଗମୂଳ = √x² = x
ପ୍ରଶ୍ନନୁସାରେ, x² = x + 12 = x² -x – 12 = 0
⇒ x² – 4x + 3x – 12 = 0 ⇒ x (x – 4) + 3 (x – 4) = 0
⇒ (x – 4) (x + 3) = 0 ⇒ x – 4 = 0 ବା x + 3 = 0
⇒ x = 4 ବା x = -3 (ଅସମ୍ଭବ)
∴ ଧନାତ୍ମକ ସଂଖ୍ୟାଟି x² = 4² = 16

(ii) ମନେକର ସଂଖ୍ୟାଟି x । ତାହାର ବ୍ୟତ୍‌କ୍ରମ \(\frac{1}{x}\)
ପ୍ରଶ୍ନନୁସାରେ, x + \(\frac{1}{x}=\frac{41}{20}\) = \(\frac{x²+1}{x}=\frac{41}{20}\)
⇒ 20x² + 20 = 41x ⇒ 20x² – 41x + 20 = 0
⇒ 20x² – 25x – 16x + 20 = 0 ⇒ 5x (4x – 5) – 4 (4x – 5) = 0
⇒ (4x – 5) (5x – 4)=0 = 4x – 5 = 0 ବା 5x – 4 = 0
⇒ x = \(\frac{5}{4}\)ବା x = \(\frac{4}{5}\)
∴ ସଂଖ୍ୟାଟି \(\frac{4}{5}\) ହେଲେ ଏହାର ବ୍ୟତ୍‌କ୍ରମ \(\frac{5}{4}\) ହେବ ।
ଅଥବା x = \(\frac{5}{4}\) ହେଲେ ଏହାର ବ୍ୟତ୍‌କ୍ରମ \(\frac{4}{5}\) ହେବ ।
∴ ସଂଖ୍ୟାଟି \(\frac{4}{5}\) ବା \(\frac{5}{4}\) ହେବ ।

(iii) ମନେକର କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାଦ୍ୱୟ x ଓ x + 1
x ର ବ୍ୟୁତ୍‌କ୍ରମ \(\frac{1}{x}\) ଓ x + 1 ର ବ୍ୟକ୍ରମ \(\frac{1}{x+1}\)
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{1}{x}+\frac{1}{x+1}=\frac{11}{30}\) ⇒ \(\frac{x+1+x}{x(x+1)}=\frac{11}{30}\)
⇒ \(\frac{2x+1}{x²+x)}=\frac{11}{30}\) ⇒ 11x² + 11x = 60x + 30
⇒ 11x² + 11x – 60x – 30 = 0 ⇒ 11x² – 49x – 30 = 0
⇒ 11x² – 55x + 6x – 30 = 0 ⇒ 11x (x – 5) + 6 (x – 5) = 0 ⇒ (x – 5) (11x + 6) = 0 ⇒ x – 5 = 0 ବା 11x + 6 = 0
⇒ x = 5 ବା x = \(\frac{-6}{11}\) (ଅସମ୍ଭବ)
∴ x = 5 ହେଲେ ଏହାର ପରବର୍ତ୍ତୀ
∴ ନିଶ୍ଚେୟ କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାଦ୍ଵୟ 5 ଓ 6 । କ୍ରମିକ ସଂଖ୍ୟା x + 1 = 6

(iv) ମନେକର କ୍ରମିକ ପୂର୍ବସଂଖ୍ୟାଦ୍ୱୟ x ଓ x + 1 ।
ପ୍ରଶାନୁସାରେ, \(\frac{1}{x}+\frac{1}{x+1}=\frac{23}{132}\) ⇒ \(\frac{x+1+x}{x(x+1)}=\frac{23}{132}\)
⇒ \(\frac{2x+1}{x²+x)}=\frac{23}{132}\) ⇒ 23x² + 23x = 264x + 132
⇒ 23x² + 23x – 264x – 132 = 0 ⇒ 23x² – 241x – 132 = 0
⇒23x² – 253x + 12x – 132 = 0 ⇒ 23x (x – 11) + 12 (x – 11) = 0
⇒ (x – 11) (23x + 12) = 0 ⇒ x – 11 = 0 ବା 23x + 12 = 0
⇒ x = 11 ବା x = \(– \frac{12}{23}\) (ଅସମ୍ଭବ) ⇒ x + 1 = 11 + 1 = 12
∴ ପୂର୍ଣ୍ଣସଂଖ୍ୟାଦ୍ଵୟ 11 ଓ 12 ।

(v) ମନେକର ଗୋଟିଏ ସଂଖ୍ୟା x ଓ ଅନ୍ୟଟି 51 – x ।
ପ୍ରଶ୍ନନୁସାରେ, x (51 – x) = 378 ⇒ 51x – x² = 378
⇒ x – 51x + 378 = 0 ⇒ x² – 42x – 9x + 378 = 0
⇒x (x-42) – 9 (x – 42) = 0 ⇒ (x – 42) (x – 9) = 0
⇒ x – 42 = 0 ବା x – 9 = 0 ⇒ x = 42 ବା x = 9
∴ ଯଦି x = 9 ହୁଏ ତେବେ ଅନ୍ୟ ଭାଗଟି 42 ହେବ ।
∴ ସଂଖ୍ୟାଦ୍ୱୟ 9 ଓ 42 ସେହିପରି x = 42 ହୁଏ ତେବେ ଅନ୍ୟ ଭାଗଟି 9 ହେବ ।

Question 3.
ଏକ ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା, ତାହାର ଅଙ୍କ ଦ୍ଵୟର ଗୁଣଫଳର 3 ଗୁଣ । ଏକକ ସ୍ଥାନରେ ଥ‌ିବା ଅଙ୍କଟି ଦଶକ ସ୍ଥାନରେ ଥ‌ିବା ଅଙ୍କଠାରୁ 2 ବୃହତ୍ତର । ସଂଖ୍ୟାଟି ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କ x ଓ ଦଶକ ସ୍ଥାନୀୟ ଅଙ୍କ y ।
∴ ସଂଖ୍ୟାଟି = 10y + x
ପ୍ରଶ୍ନନୁସାରେ 10y + x = 3xy …… (i) ଏବଂ x = y + 2 ……. (ii)
ସମୀକରଣ (ii)ରୁ x ର ମାନ ସମୀକରଣ (i) ରେ ବସାଇଲେ, 10y + y + 2 = 3y (y + 2)
⇒ 11y + 2 = 3y² + 6y ⇒ 3y² + 6y – 11y – 2 = 0
⇒ 3y² – 5y – 2 = 0 ⇒ 3y² – 6y+ y – 2 = 0
⇒ 3y (y – 2) + 1 (y – 2) = 0 ⇒ (y – 2) (3y + 1) = 0
⇒ y – 2 = 0 ବା 3y + 1 = 0 = y = 2 ବା y = (ଅସମ୍ଭବ)
∴ x = y + 2 = 2 + 2 = 4
ସଂଖ୍ୟାଟି = 10y + x = 10 × 2 + 4 = 24

Question 4.
ଗୋଟିଏ ପରିବାରରେ ଆଲ୍‌ଫାର ବୟସ, ବିଟା ଓ ଗାମାର ବୟସ୍‌ର ଗୁଣଫଳ ସହ ସମାନ । ଯଦି ବିଟା, ଗାମାଠାରୁ 1 ବର୍ଷ ବଡ଼ ହୁଏ ଏବଂ ଆଲ୍‌ଫାର ବୟସ 42 ହୁଏ, ତେବେ 5 ବର୍ଷ ପରେ ବିଟାର ବୟସ କେତେ ହେବ ?
ସମାଧାନ :
ମନେକର ପରିବାରରେ ଗାମାର ବୟସ = x ବର୍ଷ
∴ ବିଟାର ବୟସ = (x + 1) ବର୍ଷ । ଆଲ୍‌ଫାର ବୟସ = 42।
ପ୍ରଶ୍ନନୁସାରେ, x(x + 1) = 42 ⇒ x² + x = 42
⇒ x² + x – 42 = 0 ⇒ x² + 7x – 6x – 42 = 0
⇒ x(x + 7) – 6(x + 7) = 0 ⇒ (x + 7) (x – 6) = 0
⇒ x + 7 = 0 ବା x – 6 = 0 = x = -7 ବା x = 6
x = – 7 ବୟସ ଋଣାତ୍ମକ ହେବନାହିଁ ତେଣୁ ଏହା ଅସମ୍ଭବ। ତେବେ ଏଠାରେ x = 6 ଗ୍ରହଣୀୟ ।
∴ ଗାମାର ବୟସ = x ବର୍ଷ = 6 ବର୍ଷ ଏବଂ ବିଟାର ବୟସ = 6 + 1 = 7 ବର୍ଷ ।
∴ 5 ବର୍ଷ ପରେ ବିଟାର ବୟସ ହେବ = 7 + 5 = 12 ।

Question 5.
କୌଣସି ଏକ ଅରଣ୍ୟରେ ବାସ କରୁଥିବା ମର୍କଟମାନଙ୍କ ମଧ୍ୟରୁ ସେମାନଙ୍କ ସଂଖ୍ୟାର ଏକ ଅଷ୍ଟମାଂଶର ବର୍ଗ କ୍ରୀଡ଼ାରତ ଏବଂ ଅବଶିଷ୍ଟ ବାରଟି ମର୍କଟ ଏକ ଶୃଙ୍ଗ ଉପରେ ବସିଥିଲେ । ଅରଣ୍ୟରେ ସମ୍ଭବତଃ କେତେ ମର୍କଟ ଥିଲେ ?
ସମାଧାନ :
ମନେକର ଅରଣ୍ୟରେ ସମ୍ଭବତଃ xଟି ମର୍କଟ ଥିଲେ ।
ସେମାନଙ୍କ ଏକ- ଅଷ୍ଟମାଂଶର ବର୍ଗ କ୍ରିଡ଼ାରତ ଥିଲେ । କ୍ରୀଡ଼ାରତ ଥିବା ମର୍କଟଙ୍କ ସଂଖ୍ୟା = (\(\frac{x}{8}\))² = \(\frac{x²}{64}\)
ଅବଶିଷ୍ଟ 12 ଟି ମର୍କଟ ଶୃଙ୍ଗ ଉପରେ ବସିଥିଲେ ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{x²}{64}+12\) = x = \(\frac{x²+768}{64}=x\)
⇒ x² + 768x = 64 x ⇒ x² – 64x + 768 = 0
⇒ x² -48x – 16x + 768 = 0 ⇒ x (x – 48) – 16 (x – 48) = 0
⇒ (x – 48) (x – 16)=0 = x – 48 = 0 ବା x – 16 = 0
⇒ x = 48 ବା x = 16
∴ ଅରଣ୍ୟରେ ସମ୍ଭବତଃ 48 କିମ୍ବା 16ଟି ମର୍କଟ ଥିଲେ ।

Question 6.
ଗୋଟିଏ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ 30 ବ. ସେ.ମି. । ତ୍ରିଭୁଜର ଉଚ୍ଚତା, ଭୂମିର ଦୈର୍ଘ୍ୟଠାରୁ 7 ସେ.ମି. ଅଧ୍ଵ ହେଲେ, ଭୂମିର ଦୈର୍ଘ୍ୟ ସ୍ଥିର କର ।
ସମାଧାନ :
ଗୋଟିଏ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = 30 ବର୍ଗ ସେ.ମି.
ତ୍ରିଭୁଜର ଉଚ୍ଚତା, ଭୂମିର ଦୈର୍ଘ୍ୟଠାରୁ 7 ସେ.ମି. ଅଧ୍ଵ ।
ମନେକର ଭୂମିର ଦୈର୍ଘ୍ୟ X ସେ.ମି. । ଉଚ୍ଚତା = x + 7 ସେ.ମି.
∴ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) ଭୂମି × ଉଚ୍ଚତା = \(\frac{1}{2}\)x(x+7)
⇒ 30 = \(\frac{1}{2}\)(x² + 7x) ⇒ x² + 7x = 60
⇒ x²+7x-60 = 0 ⇒ x² + 12x – 5x – 60 = 0
⇒ x (x + 12) – 5 (x + 12) = 0 ⇒ (x + 12) (x – 5)= 0
⇒ x + 12 = 0 ବା x – 5 = 0 = x=- 12 (ଅସମ୍ଭବ) ବା x = 5
∴ ତ୍ରିଭୁଜର ଭୂମିର ଦୈର୍ଘ୍ୟ 5 ସେ.ମି. ।

Question 7.
ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜର ସମକୋଣର ସଂଲଗ୍ନ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟ 5x ସେ.ମି. ଓ (3x – 1) ସେ.ମି. ଓ କ୍ଷେତ୍ରଫଳ 60 ବର୍ଗ ସେ.ମି. । ତେବେ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ସମକୋଣ ସଂଲଗ୍ନ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟ 5x ସେ.ମି. ଏବଂ (3x – 1) ସେ.ମି. ।
ସମକୋଣୀ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = 60 ବର୍ଗ ସେ.ମି. (ଦତ୍ତ)
⇒ \(\frac{1}{2}\) × ସମକୋଣ ସଂଲଗ୍ନ ଦୈର୍ଘ୍ୟର ବାହୁଦ୍ୱୟର ଗଣଫଳ = 60
⇒ \(\frac{1}{2}\) . 5x . (3x – 1) = 60 ⇒ x(3x – 1) = 60 × \(\frac{2}{5}\)
⇒ 3x² – x = 24 ⇒ 3x² – x – 24 = 0
ଏଠାରେ a = 3, b = -1, c = – 24

∴ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁ ଦ୍ୱୟର ଦୈର୍ଘ୍ୟ 5x = 5 × 3 = 15 ସେ.ମି. ।
ଓ 3x – 1 = 3 × 3 – 1 = 8 ସେ.ମି. ।
ବିକଳ୍ପ ପ୍ରଣାଳୀ : 3x² – x – 24 = 0 ର ସମାଧାନ ଉତ୍ପାଦକୀକରଣ ଦ୍ଵାରା ସମ୍ଭବ ।

Question 8.
n ବାହୁ ବିଶିଷ୍ଟ ବହୁଭୁଜର କର୍ଡ ସଂଖ୍ୟା \(\frac{n}{2}\)n(n-3)। ଯଦି ବହୁଭୁଜର 54ଟି କର୍ଣ୍ଣ ରହିବ, ତେବେ ବହୁଭୁଜର ବାହୁର ସଂଖ୍ୟା କେତେ ?
ସମାଧାନ :
n ବାହୁ ବିଶିଷ୍ଟ ବହୁଭୁଜର କର୍ଣ୍ଣ ସଂଖ୍ୟା = \(\frac{n}{2}\)n(n-3)
ଏହାର କର୍ଣ୍ଣ ସଂଖ୍ୟା 54 1
ପ୍ରଶ୍ବାନୁସାରେ, \(\frac{n}{2}\)n(n-3) = 54 ⇒ n² – 3n = 108
→n² – 3n-108 = 0 ⇒ n² – 12n + 9n – 108 = 0
→ n (n – 12) + 9 (n – 12) = 0 → (n – 12) (n + 9) = 0
→ n – 12 = 0 ବା n + 9 = 0
⇒ n =12 ବା n = -9 (ଅସମ୍ଭବ)
∴ ବହୁଭୁଜର ବାହୁ ସଂଖ୍ୟା 12 ।

Question 9.
ଦୁଇଟି ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳର ସମଷ୍ଟି 468 ବ.ମି. ଏବଂ ପରିସୀମାଦ୍ବୟର ଅନ୍ତର 24 ମି. ହେଲେ ପ୍ରତ୍ୟେକର ବାହୁର ଦୈର୍ଘ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
ଦୁଇଟି ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳର ସମଷ୍ଟି 468 ବର୍ଗ ମି. । ପରିସୀମାଦ୍ବୟର ଅନ୍ତର = 24 ମିଟର
ମନେକର ଗୋଟିଏ ବର୍ଗକ୍ଷେତ୍ରର ବାହୁର ଦୈର୍ଘ୍ୟ a ମି. ଓ ଅନ୍ୟଟିର ଦୈର୍ଘ୍ୟ b ମି. ।
ପ୍ରଶ୍ବାନୁସାରେ, a² + b² = 468 ଏବଂ 4a – 4b = 24
⇒ 4(a – b) = 24 ⇒ a – b = \(\frac{24}{4}\) = 6 ………(i)
2ab = (a² + b²) – (a – b)² = 468 – 36 = 432
(a + b)² = a² + b² + 2ab = 468 + 432 = 900
⇒ a + b = 30 ……..(ii)
ସମୀକରଣ (i) ଓ (ii)କୁ ଯୋଗକଲେ a – b + a + b = 6 + 30 = 36
⇒ 2a = 36 ⇒ a = = 18 ମି., b = 30 a 30 18 = 12 ମି.
∴ ବହୁଭୁଜର ବାହୁ ଦୈର୍ଘ୍ୟ 18 ମି. ଓ 12 ମି. ।

ବିକଳ୍ପ ପ୍ରଣାଳୀ : a² + b² = 468 ଏବଂ a – b = 6
a – b = 6 = b = a – 6
∴ a² + (a – b)² = 468⇒ 2a² – 12a + 36 = 468
2a² – 12a – 432 = 0 = a² – 6a – 216 = 0
⇒ (a – 18) (a + 12) = 0 ⇒ a = 18 ବା -12
ଏଠାରେ a = 18 ଗ୍ରହଣୀୟ ଏବଂ a = -12 ଅଗ୍ରହଣୀୟ
ଯଦି a = 18 ହୁଏ, ଦେବେ b = 18 – 6 = 12
∴ ବର୍ଗକ୍ଷେତ୍ର ବାହୁର ଦୈର୍ଘ୍ୟ 18 ମି. ଓ ଅନ୍ୟଟିର ଦୈର୍ଘ୍ୟ 12 ମି. ।

Question 10.
ଜଣେ ବ୍ୟକ୍ତି ତାଙ୍କ ଚାଲିବାର ବେଗକୁ ଯଦି ଘଣ୍ଟା ପ୍ରତି 1 କି.ମି. ବୃଦ୍ଧି କରେ, ତେବେ 2 କି.ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବା ପାଇଁ 10 ମିନିଟ୍ କମ୍ ସମୟ ନେଇଥା’ନ୍ତା । ତେବେ ବ୍ୟକ୍ତିର ଚାଲିବାର ଘଣ୍ଟା ପ୍ରତି ବେଗ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ବ୍ୟକ୍ତିଙ୍କର ଚାଲିବାର ଘଣ୍ଟାପ୍ରତି ବେଗ = x କି.ମି. ।
ଏହି ବେଗରେ 2 କି.ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବାପାଇଁ ସମୟ ଲାଗିଲା = \(\frac{2}{x}\) ଘଣ୍ଟା ।
ଯଦି ତାଙ୍କର ଚାଲିବାର ଘଣ୍ଟାପ୍ରତି ବେଗ 1 କି.ମି. ବୃଦ୍ଧି ହୁଏ, ତେବେ ତାଙ୍କର ବେଗ = (x+1) କି.ମି. ।
ବେଗ ବୃଦ୍ଧି ହେଲେ 2 କି.ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବାକୁ ସମୟ ଲାଗିବ = \(\frac{2}{x+1}\) ଘଣ୍ଟା ।
ପ୍ରଶ୍ବାନୁସାରେ, \(\frac{2}{x}-\frac{2}{x+1}=\frac{10}{60}\) (∵ 10 ମିନିଟ = \(\frac{10}{60}\) ଘଣ୍ଟା ।)

x = – 4 ଗ୍ରହଣୀୟ ନୁହେଁ (କାରଣ ବେଗ ଋଣାତ୍ମକ ହେବା ଅସମ୍ଭବ) ।
∴ ବ୍ୟକ୍ତିଙ୍କର ଚାଲିବାର ଘଣ୍ଟାପ୍ରତି ବେଗ 3 କି.ମି. ।

Question 11.
ଏକ ନୌକାର ବେଗ ସ୍ଥିର ଜଳରେ 15 କି.ମି. ପ୍ରତି ଘଣ୍ଟା । ଏହା ସ୍ରୋତର ପ୍ରତିକୂଳରେ 30 କି.ମି. ଅତିକ୍ରମ କରି ପୁନଶ୍ଚ (ଅନୁକୂଳରେ) ଫେରି ଆସିବାକୁ 4 ଘଣ୍ଟା 30 ମି. ସମୟ ନେଲା । ତେବେ ସ୍ରୋତର ଘଣ୍ଟା ପ୍ରତି ବେଗ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଗୋଟିଏ ନୌକାର ବେଗ ସ୍ଥିର ଜଳରେ 15 କି.ମି.|ଘଣ୍ଟା ।
ମନେକର ସ୍ରୋତର ବେଗ x କି.ମି./ଘଣ୍ଟା ।
ସ୍ରୋତର ଅନୁକୂଳରେ ନୌକାଟି 1 ଘଣ୍ଟାରେ ଯିବ (15 + x) କି.ମି. ।
ସ୍ରୋତର ପ୍ରତିକୂଳରେ ନୌକାଟି 1 ଘଣ୍ଟାରେ ଯିବ (15 – x) କି.ମି. ।
ସ୍ରୋତର ଅନୁକୂଳରେ 30 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ = \(\frac{30}{15+x}\) ଘଣ୍ଟା । ଏବଂ
ସ୍ରୋତର ପ୍ରତିକୂଳରେ 30 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ = \(\frac{30}{15-x}\) ଘଣ୍ଟା ।
ପ୍ରଶ୍ନ ନୁସାରେ, \(\frac{30}{15+x}+\frac{30}{15-x}=4\frac{1}{2}\)
⇒ 30(\(\frac{15-x+15+x}{(15+x)(15-x)}\)) = \(\frac{9}{2}\)
⇒ \(\frac{30×30}{225-x²}=\frac{9}{2}\) ⇒ 1800 = 9 (225 – x²)
⇒ 225 – x² = \(\frac{1800}{9}\) = 200 ⇒ 225 – 200 = x² ⇒ x² = 25 ⇒ x= √25 = 5
∴ ସ୍ରୋତର ବେଗ 5 କି.ମି./ ଘଣ୍ଟା ।

Question 12.
ଗୋଟିଏ ଶ୍ରେଣୀର ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ସଂଖ୍ୟକ ଛାତ୍ରଙ୍କ ମଧ୍ୟରେ 250 ଟଙ୍କାକୁ ସମାନ ଭାଗରେ ବଣ୍ଟାଗଲା । ଯଦି 25 ଜଣ ଛାତ୍ର ଅଧ‌ିକ ହୋଇଥା’ନ୍ତେ, ତେବେ ପ୍ରତ୍ୟେକ 0.50 ଟଙ୍କା ଲେଖାଏଁ କମ୍ ପାଇଥା’ନ୍ତେ । ଶ୍ରେଣୀର ଛାତ୍ର ସଂଖ୍ୟା ସ୍ଥିର କର ।
ମନେକର ଶ୍ରେଣୀର ଛାତ୍ର ସଂଖ୍ୟା x ଜଣ ।
x ଜଣ ଛାତ୍ରଙ୍କ ମଧ୍ୟରେ 250 ଟଙ୍କାକୁ ସମାନ ଭାଗରେ ବାଣ୍ଟିଲେ ଜଣକା ପାଇବେ = \(\frac{250}{x}\) ଟଙ୍କା
25 ଜଣ ଛାତ୍ର ଅଧ୍ଵ ହେଲେ ଛାତ୍ର ସଂଖ୍ୟା = (x + 25) ଜଣ ।
25 ଜଣ ଛାତ୍ର ଅଧ୍ୟା ହେଲେ ଜଣକା ପାଇବେ = (\(\frac{250}{x}\) – 0.50) ଟଙ୍କା
ପ୍ରଶ୍ନନୁସାରେ, (x + 25)(\(\frac{250}{x}\) – 0.50) = 250
⇒ (x + 25)(500 – x) = 500x ⇒ 500x – x² + 12500 – 25x = 500x
⇒ x² + 25x – 12500 = 0

x = – 125 ଗ୍ରହଣୀୟ ନୁହେଁ, କାରଣ ଛାତ୍ର ସଂଖ୍ୟା ଋଣାତ୍ମକ ହେବନାହିଁ ।
⇒ x = 100 ∴ ଶ୍ରେଣୀର ଛାତ୍ରସଂଖ୍ୟା 100 ।

Question 13.
ଗୋଟିଏ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ, ପ୍ରସ୍ଥ ଅପେକ୍ଷା 8 ମିଟର ଅଧୀକ । କ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ 240 ବର୍ଗ ମିଟର ହେଲେ କ୍ଷେତ୍ରଟିର ପରିସୀମା କେତେ ?
ସମାଧାନ :
ଗୋଟିଏ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ, ପ୍ରସ୍ଥ ଅପେକ୍ଷା 8 ମି. ଅଧ୍ବକ ।
ମନେକର ଆୟତକ୍ଷେତ୍ରର ପ୍ରସ୍ଥ = x ମି., ଦୈର୍ଘ୍ୟ = (x + 8) ମିଟର,
ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = (x + 8) x ବର୍ଗ ମିଟର ।
ପ୍ରଶ୍ନନୁସାରେ, (x + 8) x = 240
⇒x² + 8x = 240 ⇒ x² + 8x – 240 = 0
⇒ x² + 20x – 12x – 240 = 0
⇒ x(x + 20) – 12(x + 20) = 0
⇒ (x + 20) (x – 12) = 0
⇒ x + 20 = 0 ବା x – 12 = 0
⇒x = -20 (ଅସମ୍ଭବ) ବା x = 12
∴ ଆୟତକ୍ଷେତ୍ରର ପ୍ରସ୍ଥ = 12 ମି., ଦୈର୍ଘ୍ୟ = x + 8 = 12 + 8 = 20 ମି.
ଆୟତକ୍ଷେତ୍ରର ପରିସୀମା = 2 (ଦୈର୍ଘ୍ୟ + ପ୍ରସ୍ଥ) = 2 (20 + 12) = 2 × 32 = 64 ମିଟର ।

Question 14.
ଏକ ରେଳଗାଡ଼ି 300 କି.ମି. ଦୀର୍ଘ ଯାତ୍ରା ପଥରେ ସମାନ ବେଗରେ ଗତି କରୁଥିଲା । ଯଦି ଗାଡ଼ିର ବେଗ ଘଣ୍ଟାପ୍ରତି 5 କି.ମି. ଅଧ୍ଵ ହୋଇଥା’ନ୍ତା, ତେବେ ଗାଡ଼ିଟି ନିର୍ଦ୍ଦିଷ୍ଟ ସମୟର 2 ଘଣ୍ଟା ପୂର୍ବରୁ ଯଥା ସ୍ଥାନରେ ପହଞ୍ଚିନ୍ତା । ତେବେ ଗାଡ଼ିର ଘଣ୍ଟାପ୍ରତି ବେଗ ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର ଗାଡ଼ିର ଘଣ୍ଟାପ୍ରତି ବେଗ x କି.ମି. ।
ଏହି ବେଗରେ ଗାଡ଼ିକୁ 300 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ \(\frac{300}{x}\) ଘଣ୍ଟା ।
ମାତ୍ର ଘଣ୍ଟାକୁ x + 5 କି.ମି. ବେଗରେ 300 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ \(\frac{300}{x+5}\) ଘଣ୍ଟା ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{300}{x}-\frac{300}{x+5}=2\) ⇒ 300(\(\frac{1}{x}-\frac{1}{x+5}\) = 2

Question 15.
ଏକ ଆୟତାକାର ପଡ଼ିଆର ଦୈର୍ଘ୍ୟ 25 ମିଟର, ପ୍ରସ୍ଥ 16 ମିଟର ଓ ପଡ଼ିଆର ଚତୁଃପାର୍ଶ୍ବରେ ସମାନ ଚଉଡ଼ାର ଏକ ରାସ୍ତା ଅଛି । ଯଦି ଚତୁଃପାର୍ଶ୍ଵରେ ଥ‌ିବା ରାସ୍ତାର କ୍ଷେତ୍ରଫଳ 230 ବର୍ଗମିଟର ହୁଏ, ତେବେ ରାସ୍ତାର ଚଉଡ଼ା ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର ରାସ୍ତାର ଚଉଡ଼ା = x ମିଟର
ଆୟତାକାର ପଡ଼ିଆର ଦୈର୍ଘ୍ୟ = 25 ମିଟର, ପ୍ରସ୍ଥ = 16 ମିଟର ।
.. ବାହାର ଆୟତାକାର ପଡ଼ିଆର ଦୈର୍ଘ୍ୟ = (25 + 2x) ମି. ଓ ପ୍ରସ୍ଥ = (16 + 2x) ମି.
ପ୍ରଶ୍ନନୁସାରେ, (25 + 2x) (16 + 2x) – 25 × 16 = 230
⇒ 400 + 50x + 32x + 4x² – 400 = 230
⇒ 4x² + 82x – 230 = 0 ⇒ 2x² + 41x – 115=0
ଏଠାରେ a = 2, b = 41, c = -115

∴ ରାସ୍ତାର ଚଉଡ଼ା 2.5 ମିଟର ।
ବି.ଦ୍ର. : ଉତ୍ପାଦକୀକରଣଦ୍ୱାରା 2x² + 41x – 115 = 0 ର ସମାଧାନ ମଧ୍ୟ ସମ୍ଭବ ।

Question 16.
କେତେକ ଛାତ୍ରଛାତ୍ରୀ ଏକ ବଣଭୋଜିର ଆୟୋଜନ କଲେ । ଖାଦ୍ୟ ଅଟକଳ (Budget) 480 ଟଙ୍କା ଥିଲା । ସେମାନଙ୍କ ମଧ୍ୟରୁ 8 ଜଣ ବଣଭୋଜିକୁ ଗଲେ ନାହିଁ; ଯାହା ଫଳରେ ଖାଦ୍ୟ ବାବଦ ଖର୍ଚ୍ଚ ଜଣପିଛା 10 ଟଙ୍କା ବଢ଼ିଗଲା । ତେବେ କେତେଜଣ ଛାତ୍ରଛାତ୍ରୀ ବଣଭୋଜିକୁ ଯାଇଥିଲେ ?
ସମାଧାନ :
ମନେକର x ଜଣ ଛାତ୍ରଛାତ୍ରୀ ବଣଭୋଜିର ଆୟୋଜନ କରିଥିଲେ ।
ଖାଦ୍ୟ ଅଟକଳ = 480 ଟଙ୍କା ଥିଲା ।
ଜଣେ ପିଲା ପିଛା ଖର୍ଚ୍ଚ ପଡ଼ିଥାନ୍ତା = \(\frac{480}{x}\)
8 ଜଣ ବଣଭୋଜିକୁ ନଯିବାରୁ ପିଲା ସଂଖ୍ୟା = x – 8
∴ ବଣଭୋଜିକୁ ଯାଇଥବା ଛାତ୍ରଛାତ୍ରୀ ସଂଖ୍ୟା = x – 8
x – 8 ଜଣ ପିଲାଙ୍କ ପାଇଁ ଖର୍ଚ୍ଚ 480 ଟଙ୍କା ହେଲେ, ଜଣେ ପିଲା ପିଛା ଖର୍ଚ୍ଚ ପଡ଼ିଲା = \(\frac{480}{x-8}\) ଟଙ୍କା ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{480}{x-8}-\frac{480}{x}=10\) ⇒ 480(\(\frac{x-x+8}{(x-8)x}\)) = 10
⇒ \(\frac{8}{x²-8x}=\frac{1}{48}\) ⇒ x² – 8x = 384
⇒ x² -8x – 384 = 0 ⇒ x² – 24x + 16x – 384 = 0
⇒ x (x – 24) + 16 (x – 24) = 0 ⇒ (x – 24) (x + 16) = 0 ⇒x-24=0 Ql x + 16 = 0
⇒x= 24 Q1 x = – 16
∴ ବଣଭୋଜିକୁ ଯାଇଥବା ଛାତ୍ରଛାତ୍ରୀ ସଂଖ୍ୟା = x – 8 = 24 – 8 = 16 ଜଣ।

Question 17.
ସମାଧାନ ଜର :
(i) (x + 1) (x + 2) (x + 3) (x + 4) = 120
(ii) \(5 \sqrt{\frac{3}{x}}+7 \sqrt{\frac{x}{3}}=22 \frac{2}{3}\)
(iii) 3x + \(\frac{5}{16x}\) – 2 = 0
(iv) \(\frac{2x+1}{x+1}^4-6\frac{2x+1}{x+1}^2+8=0\)
(v) (3x² – 8)² – 23 (3x² – 8) + 76 = 0
(vi) 5 (5^x + 5^-x) = 26
(vii) (x² – 2x)² – 4 (x² – 2x) + 3 = 0
(viii) x^-4 – 5x^-2 + 4 = 0
(ix) \(2\left(x^2+\frac{1}{x^2}\right)-3\left(x+\frac{1}{x}\right)-1=0\)
(x) \(\frac{3}{√2x}-\frac{√2x}{5}=5 \frac{9}{10}\)
(xi) \(\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}\) (x ≠ 0, x ≠ – 1)
(xii) x (2x + 1) (x – 2) (2x – 3) = 63
(xiii) \(\frac{x-3}{x+3}+\frac{x+3}{x-3}=6 \frac{6}{7}\) (x ≠ – 3, 3)
(xiv) \(3\left(x^2+\frac{1}{x^2}\right)+4\left(x-\frac{1}{x}\right)-6=0\)
(xv) (\(\frac{x+1}{x-1}\))² – (\(\frac{x+1}{x-1}\)) – 2 = 0
(xvi) \(\sqrt{2x+9}\) + x = 13
(xvii) \(\sqrt{2 x+\sqrt{2 x+4}}\) = 4
ସମାଧାନ :
(i) (x + 1) (x + 2) (x + 3) (x + 4) = 120
= (x+1) (x+4) (x + 2) (x + 3) = 120 ⇒ (x² + 5x + 4) (x² + 5x + 6) = 120
ମନେକର x² + 5x = p
ପ୍ରଭେ ସମୀକରଣରଟି (p + 4) (p + 6) = 120
⇒ p² + 10p+ 24 = 120 ⇒ p² + 10p + 24 – 120 = 0
⇒p² + 10p – 96 = 0 ⇒ p² + 16p – 6p – 96 = 0
⇒p(p + 16) – 6(p + 16) = 0 ⇒ (p + 16) (p – 6) = 0
⇒p + 16 = 0 କିମ୍ବା p – 6 = 0 ⇒ p = -16 କିମ୍ବା p=6
p ପରିବର୍ତ୍ତେ x + 5x ନେଲେ x² + 5x = – 16
⇒ x² + 5x + 16 = 0 ଏଠାରେ D = b² – 4ac = 25 – 64 = -39
⇒ D < 0 ବୀଜଦ୍ଵୟ ଅବାସ୍ତବ, ଗ୍ରହଣୀୟ ନୁହେଁ ।
ପୁନଶ୍ଚ x² + 5x = 6
⇒ x²+5x – 6 = 0, ଏଠାରେ a = 1, b = 5, c = – 6

∴ ନିର୍ଦେୟ ସମାଧାନ – 6 ଓ 1 ।

(ii) \(5 \sqrt{\frac{3}{x}}+7 \sqrt{\frac{x}{3}}=22 \frac{2}{3}\)
ମନେକର \(\sqrt{\frac{3}{x}}\) = p = \(\sqrt{\frac{x}{3}}=\frac{1}{p}\)
ପ୍ରଭେ ସମୀକରଣରଟି 5p + \(\frac{7}{p}\) = \(\frac{68}{3}\) ⇒ \(\frac{5p²+7}{p}=\frac{68}{3}\)
⇒ 15p² + 21 = 68p ⇒ 15p² – 68p + 21 = 0, ଏଠାରେ a = 15, b = -68, c = 21

ବି.ଦ୍ର. : x ର ମାନ \(\frac{25}{147}\) ଦେଇ ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉଛି କି ନାହିଁ ପରୀକ୍ଷା କରି ଦେଖ । ଯଦି ନ ହେଉଥାଏ ତେବେ ସମାଧାନ କେବଳ 27 ହେବ । ଏଠାରେ \(\frac{25}{147}\) ଅଗ୍ରହଣୀୟ ।

(iii) 3x + \(\frac{5}{16x}\) – 2 = 0
⇒ \(\frac{48x²+5-32x}{16x}=0\) ⇒ 48x² + 5 – 32x = 0

(iv) \(\frac{2x+1}{x+1}^4-6\frac{2x+1}{x+1}^2+8=0\)
ମନେକର (2x+1)= p ଦଉ ସମୀକରଣ p⁴ – 6p² + 8 = 0 => p⁴ – 4p² – 2p² + 8 = 0
⇒ p²(p² – 4) – 2(p² – 4) = 0 ⇒ (p² – 4)(p² – 2) = 0
⇒ p² – 4 = 0 ବା p² – 2 = 0 ⇒ p² = 4 ବା p² = 2 ⇒ p = ±2 ବା p = ±√2
\(\frac{2x+1}{x+1}=2\), x ର ମାନ ନିର୍ଦୟ ଅସମ୍ଭବ |

ବି.ଦ୍ର. :
(\(\frac{2x+1}{x+1}\))² = p ମନେକର
ଦଉ ସମୀକରଣ p² – 6p + 8 = 0 ⇒ p = 4 ବା p = 2
\(\frac{2x+1}{x+1}\) = 4 ⇒ \(\frac{2x+1}{x+1}\) = ±2 ନେଇ x ର ମାନ ସ୍ଥିର କରାଯାଇପାରେ ।
ପୁନଶ୍ଚ (\(\frac{2x+1}{x+1}\))² = 2 = \(\frac{2x+1}{x+1}\) = ±√2 ନେଇ
x ର ମାନ ସ୍ଥିର କରାଯାଇପାରେ ।

(v) ମନେକର 3x² – 8 = p
ପ୍ରଭେ ସମୀକରଣ p² – 23p + 76 = 0 । ଏଠାରେ a = 1, b = 23, c = 76

(vi) ମନେକର 5^x = p
ପ୍ରଭେ ସମୀକରଣ 5(p + \(\frac{1}{p}\)) = 26 ⇒ \(\frac{5(p²+1)}{p}\) = 26
⇒ 5p² + 5 = 26p ⇒ 5p² – 26p + 5 = 0
⇒ 5p² – 25p – p + 5 = 0 ⇒ 5p (p -5) – 1(p – 5) = 0
⇒ (p – 5) (5p – 1) = 0 ⇒ p – 5 = 0 କିମ୍ବା 5p – 1 = 0
⇒p = 5 କିମ୍ବା p = \(\frac{1}{5}\) ⇒5^x = 5¹ କିମ୍ବା 5^x = 5^-1
⇒ x = 1 କିମ୍ବା x = -1
∴ ନିର୍ଦେୟ ସମାଧାନ 1 ଓ -1 ।

(vii) ମନେକର x² – 2x = p
ପ୍ରଭେ ସମୀକରଣ, p² – 4p + 3 = 0
p² – 3p – p + 3 = 0 p(p – 3) – 1(p – 3) = 0
⇒(p – 3) (p – 1) = 0 ⇒ p – 3 = 0 ବା p – 1 = 0
⇒ p = 3 ବା p = 1
x² – 2x = 3 ⇒ x² – 2x – 3 = 0
⇒ x – 3x + x – 3 = 0 ⇒ x (x – 3) + 1 (x – 3) = 0
⇒ (x – 3) (x + 1) = 0 ⇒ x – 3 = 0 ବା x + 1 = 0
⇒ x = 3 ବା x = – 1
ପୁନଶ୍ଚ x² – 2x = 1 ⇒ x² – 2x – 1 = 0 । ଏଠାରେ a = 1, b = – 2, c = – 1 ବା

(viii) ମନେକର x^-2 = p। ଦଉ ସମୀକରଣ, p² – 5p + 4 = 0

(ix) \(2\left(x^2+\frac{1}{x^2}\right)-3\left(x+\frac{1}{x}\right)-1=0\)

ପୁନଶ୍ଚ x + \(\frac{1}{x}\) = -1 ⇒ \(\frac{x²+1}{x}\) = -1 ⇒ x² + 1 = -x ⇒ x² + x + 1 = 0
ସମୀକରଣ x² + x + 1 = 0 ରେ D < 0 ହେତୁ ବାସ୍ତବ ବୀଜ ସମ୍ଭବ ନୁହେଁ ।
∴ ନିର୍ଦେୟ ସମାଧାନ 2 ଓ \(\frac{1}{2}\) ।

(x) ମନେକର \(\frac{1}{√2x}\) = p; √2x = \(\frac{1}{p}\)
ପ୍ରଭେ ସମୀକରଣ, 3p – \(\frac{1}{5p}=\frac{59}{10}\)
⇒ \(\frac{15p²-1)}{5p}=\frac{59}{10}\) ⇒ \(\frac{15p²-1)}{p}=\frac{59}{2}\)
⇒ 30p² – 2 = 59p ⇒ 30p² – 59p – 2 = 0
⇒30p² – 60p + p – 2 = 0 ⇒ 30p (p – 2) + 1 (p – 2) = 0
⇒ (p – 2) (30p+1) = 0 ⇒ p – 2 = 0 ବା 30p + 1 = 0
⇒p = 2 ବା p = \(– \frac{1}{30}\)
p = 2 ⇒ \(\frac{1}{√2x}\) = 2 ⇒ \(\frac{1}{2x}\) = 4 ⇒ x = \(\frac{1}{8}\)
p = \(– \frac{1}{30}\) ⇒ \(\frac{1}{√2x}=- \frac{1}{30}\) = \(\frac{1}{2x}=\frac{1}{900}\)
⇒ 2x = 900 ⇒ x = 450
x = 450 ପୁନଶ୍ଚ ସମୀକରଣରେ ବସାଇଲେ ଏହା ସମୀକରଣକୁ ସିଦ୍ଧ କରିବ ନାହିଁ ।
ତେଣୁ ଏହା ଗ୍ରହଣୀୟ ନୁହେଁ ।
∴ xର ଏକମାତ୍ର ମୂଲ୍ୟ \(\frac{1}{8}\) ଅଟେ ।

(xi) \(\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}\)
⇒ \(\frac{x²+(x+1)²}{x(x+1)}=\frac{34}{15}\) ⇒ \(\frac{x²+x²+2x+1}{x²+x)}=\frac{34}{15}\)
⇒ \(\frac{2x²+2x+1}{x²+x}=\frac{34}{15}\) ⇒ 34 (x² + x) = 15 (2x² + 2x + 1)
⇒ 34x² + 34x = 30x² + 30x + 15
⇒ 34x² – 30×2 + 34x – 30x – 15 = 0 ⇒ 4x² + 4x – 15 = 0
⇒ 4x² + 10x – 6x – 15=0 ⇒ 2x (2x + 5) – 3 (2x + 5) = 0
⇒ (2x + 5) (2x-3)=0 ⇒ x = \(\frac{-5}{2}\) ବା x = \(\frac{3}{2}\)
∴ ନିର୍ଦେୟ ସମାଧାନ \(\frac{3}{2}\) ଓ \(\frac{-5}{2}\) ।

(xii) x (2x + 1) (x – 2) (2x – 3) = 63 ⇒ {x (2x – 3)} {(2x + 1) (x – 2)} = 63
⇒ (2x² – 3x) (2x² -3x – 2) = 63
ମନେକର 2x² – 3x = p
∴ p (p – 2) = 63 ⇒ p² – 2p – 63 = 0
⇒p² – 9p + 7p – 63 = 0 ⇒ p (p – 9) + 7 (p – 9) = 0
(p – 9) (p + 7) =0 p – 9 = 0 p + 7 = 0
⇒ p = 9 ବା p = -7
∴ p = 9 ⇒ 2x² – 3x = 9 ⇒ 2x² – 3x – 9 = 0
⇒ 2x² – 6x + 3x – 9 = 0 ⇒ 2x (x – 3) + 3 (x – 3) = 0
⇒ (x – 3) (2x + 3) = 0 ⇒ x = 3 ବା x = \(\frac{-3}{2}\)
p = -7 ⇒ 2x² – 3x + 7 = 0
ଏଠାରେ b² – 4ac = 9 – 56 = – 47
D < 0 xର ଅବାବ ମାନ ଗ୍ରହଣୀୟ ନୁହେଁ ।
∴ ନିର୍ଦେୟ ସମାଧାନ 3 ଓ \(\frac{3}{2}\) ।

(xiii) ମନେକର \(\frac{x-3}{x+3}=p\), ପ୍ରଭଦ ସମୀକରଣଟି p – \(\frac{1}{p}=\frac{48}{7}\)
⇒ \(\frac{p²-1}{p}=\frac{48}{7}\) ⇒ 7(p²-1) = 48p ⇒ 7p² – 48p – 7 = 0
ଏଠାରେ a = 7, b = -48, c = -7

∴ ନିର୍ଦେୟ ସମାଧାନ -4 ଓ \(\frac{9}{4}\) ।

(xiv) \(3\left(x^2+\frac{1}{x^2}\right)+4\left(x-\frac{1}{x}\right)-6=0\)

(xv) (\(\frac{x+1}{x-1}\))² – (\(\frac{x+1}{x-1}\)) – 2 = 0
ମନେକର \(\frac{x+1}{x-1}=p\)
ଦଉ ସମାଜରଣ p² – p – 2 = 0
ଏଠାରେ a = 1, b = -1, c = -2

ଯଦି p = 2 ⇒ \(\frac{x+1}{x-1}\) = 2 ⇒ x + 1 = 2(x – 1) ⇒ x + 1= 2x – 2
⇒ x – 2x = -2 -1 ⇒ -x = -3 ⇒ x = 3
ଯଦି p = 1 ⇒ \(\frac{x+1}{x-1}\) = -1 ⇒ x + 1 = -1(x – 1) ⇒ x+ 1 = -x + 1
⇒ x + x = 1 – 1 ⇒ 2x = 0 ⇒ x = 0

(xvi) \(\sqrt{2x+9}\) + x = 13

x = 8 ଦ୍ବାରା ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉଛି; ମାତ୍ର x = 20 ହେଲେ, \(\sqrt{2x+9}\) + x ≠ 13
∴ ଦତ୍ତ ସମୀକରଣର ସମାଧାନ x = 8 ।

(xvii) \(\sqrt{2 x+\sqrt{2 x+4}}\) = 4

x = 6 ଦ୍ଵାରା ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉଛି; ମାତ୍ର x = \(\frac{21}{2}\) ହେଲେ,
\(\sqrt{2 x+\sqrt{2 x+4}}\) ≠ 4 (ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉନାହିଁ)
∴ ନିଶ୍ଚେ ସମୀକରଣର ସମାଧାନ 6।