myhelper: RD Sharma solution class 7 chapter 20 Mensuration I Objective Type Question

RD Sharma solution class 7 chapter 20 Mensuration I Objective Type Question

Objective Type Question

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Question 1:

If the area of a square is 225 m², then its perimeter is

(a) 15 m (b) 60 m (c) 225 m (d) 30 m

Answer 1:

Let a be the side of the square. Then
Area of square = a²
$⇒$ 225 = a²
$⇒$ a² = 15²
$⇒$ a = 15 m
Perimeter of the square = 4a = 4 $×$ 15 = 60 m
Hence, the correct option is (b).

Question 2:

If the perimeter of a square is 16 cm, then its area is

(a) 4 cm² (b) 8 cm² (c) 16 cm² (d) 12 cm²

Answer 2:

Let a be the side of the square. Then
Perimeter = 4a
$⇒$ 16 = 4a
$⇒$ a = 4 cm
Area of the square = a² = 4² = 16 cm²
Hence, the correct option is (c).

Question 3:

The length of a rectangle is 8 cm and its area is 48 cm². The perimeter of the rectangle is

(a) 14 cm (b) 24 cm (c) 12 cm (d) 28 cm

Answer 3:

Let a and b be the length and breadth of the rectangle respectively. Then
Area of the rectangle = ab
$⇒$ 48 = a $×$ 8 (∵ b = 8 cm)
$⇒$ a = 6 cm
Perimeter of the rectangle = 2(a + b) = 2( 6 + 8) = 28 cm
Hence, the correct option is (d).

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Question 4:

The area of a square and that of a square drawn on its diagonal are in the ratio

(a) 1 : $\sqrt{2}$ (b) 1 : 2 (c) 1 : 3 (d) 1 : 4

Answer 4:

Let a be the side of the square. Then
Area of the square = a²
Area of the square drawn on the diagona = ${(a \sqrt{2})}^{2} = 2 a^{2}$
Required ratio = a² : 2a² = 1 : 2
Hence, the correct option is (b).

Question 5:

The length of the diagonal of a square is d. the area of the square is

(a) d² (b) $\frac{1}{2} d^{2}$ (c) $\frac{1}{4} d^{2}$ (d) 2d²

Answer 5:

Let a be the side of the square. Then
Diagonal of the square = $a \sqrt{2}$
Therefore
$a \sqrt{2} = d \Rightarrow a = \frac{d}{\sqrt{2}} ∴ Area of the square = {(\frac{d}{\sqrt{2}})}^{2} = \frac{d^{2}}{2}$
Hence, the correct option is (b).

Question 6:

The ratio of the areas of two squares, one having its diagonal double that of the other, is

(a) 2 : 1 (b) 3 : 1 (c) 3 : 2 (d) 4 : 1

Answer 6:

Let d be the diagonal of the second square. Then, the diagonal of the first square will be 2d.
∵ $Side of a square = \frac{Diagonal}{\sqrt{2}}$
∴ $Required ratio = \frac{{(\frac{2 d}{\sqrt{2}})}^{2}}{{(\frac{d}{\sqrt{2}})}^{2}} = \frac{2 d^{2}}{\frac{d^{2}}{2}} = 4 : 1$
Hence, the correct option is (d).

Question 7:

If the ratio of the areas of two squares is 9 : 1, then the ratio of their perimeters is

(a) 2 : 1 (b) 3 : 1 (c) 3 : 2 (d) 4 : 1

Answer 7:

Let a and b be the sides of the squares, then as per the question
$\frac{a^{2}}{b^{2}} = \frac{9}{1} \Rightarrow {(\frac{a}{b})}^{2} = \frac{3^{2}}{1} \Rightarrow \frac{a}{b} = \frac{3}{1}$
Therefore
$Ratio of the perimeters of the squares = \frac{4 a}{4 b} = \frac{a}{b} = \frac{3}{1}$
Thus, the of the required ratio is 3 : 1.
Hence, the correct option is (b).

Question 8:

The ratio of the area of a square of side a and that of an equilateral triangle of side a is

(a) 2 : 1 (b) 2 : $\sqrt{3}$ (c) 4 : 3 (d) 4 : $\sqrt{3}$

Answer 8:

Area of the square = a²
Area of the equilateral triangle = $\frac{a^{2} \sqrt{3}}{4}$
$\frac{Area of square}{Area of equilateral triangle} = \frac{a^{2}}{\frac{a^{2} \sqrt{3}}{4}} = \frac{4}{\sqrt{3}}$
Thus, the required ratio is 4 : $\sqrt{3}$ .
Hence, the correct option is (d).

Question 9:

On increasing each side of a square by 25%, the increase in area will be

(a) 25% (b) 55% (c) 55.5% (d) 56.25%

Answer 9:

Let a be the side of the square. Then
Side of the new square = a + 25% of a = $a + a \times \frac{25}{100} = \frac{5 a}{4}$
Old area = a²
New area = ${(\frac{5 a}{4})}^{2} = \frac{25 a^{2}}{16}$
% increase in the area = $\frac{Change in area}{Old area} \times 100 = \frac{\frac{25 a^{2}}{16} - a^{2}}{a^{2}} \times 100 = \frac{9}{16} \times 100 = 56.25$
Hence, the correct option is (d).

Question 10:

The area of a square is 50 cm². The length of its diagonal is

(a) 5 $\sqrt{2}$ cm (b) 10 cm (c) 10 $\sqrt{2}$ cm (d) 8 cm

Answer 10:

Let a be the side of the square. Then
Area of the square = a² = 50 cm² $\Rightarrow a = \sqrt{50} = 5 \sqrt{2} cm$
Now
Diagonal of the square = $a \sqrt{2} = 5 \sqrt{2} \times \sqrt{2} = 5 \times 2 = 10 cm$
Hence,the correct option is (b).

Question 11:

Each diagonal of a square is 14 cm. Its area is

(a) 196 cm² (b) 88 cm² (c) 98 cm² (d) 148 cm²

Answer 11:

Let a be the side of the square. Then
Diagonal of the square = $a \sqrt{2} = 14 \Rightarrow a = \frac{14}{\sqrt{2}} cm$
Now
Area of the square = $a^{2} = {(\frac{14}{\sqrt{2}})}^{2} = 98 {cm}^{2}$
Hence,the correct option is (c).

Question 12:

The area of a square filed is 64 m². A path of uniform width is laid around and outside of it.
If the area of the path is 17 m², then the width of the path is

(a) 1 m (b) 1.5 m (c) 0.5 m (d) 2 m

Answer 12:

Let a be the side of inner square. Then
$Area of inner square = a^{2} \Rightarrow 64 = a^{2} \Rightarrow a = \sqrt{64} = 8 cm$
Let x be the width of the path, then
Side of outer square = (a + x) cm = (8 + x) cm
Now
Area of path = Area of outer square − Area of inner square
17 = (8 + x)² − 64
$\Rightarrow {(8 + x)}^{2} = 64 + 17 = 81 \Rightarrow 8 + x = 9 \Rightarrow x = 1 m$
Thus, the width of the path is 1 m.
Hence, the correct option is (a).

Question 13:

A path of 1 m runs around and inside a square garden of side of 20 m. The cost of levelling the path
at the rate of ₹2.25 per square metre is

(a) ₹154 (b) ₹164 (c) ₹182 (d) ₹171

Answer 13:

Width of the path = 1 m
Side of the square garden = 20 m
Side of the inner square = (20 − 2) m = 18 m
∴ Area of the path = Area of square garden − Area of inner square
$= 20^{2} - 18^{2} = 400 - 324 = 76 m^{2}$
Cost of levelling = ₹2.25 $×$ 76 = ₹171
Thus, the required cost is ₹171.

Question 14:

The length of and breadth of a rectangle are (3x + 4) cm and (4x − 13) cm. If the perimeter of the rectangle is 94 cm, then x =

(a) 4 (b) 8 (c) 12 (d) 6

Answer 14:

Here, l = (3x + 4) cm and b = (4x − 13) cm.
Perimeter of rectangle = 2 (l + b)
= 2[(3x + 4) + (4x − 13)]
= 2(7x − 9) = 14x − 18
Now, as per the question
Perimeter of rectangle = 94 cm
$∴ 14 x - 18 = 94 \Rightarrow 14 x = 94 + 18 = 112 \Rightarrow x = \frac{112}{14} = 8$
Hence, the correct option is (b).

Question 15:

In Fig. 38, ABCD and PQRC are squares such that AD = 22 cm and PC = y cm. If the area of the shaded region is 403 cm²,
then the value of y is

(a) 3 (b) 6 (c) 9 (d) 10

Answer 15:

Here, AD = 22 cm.
Area of square ABCD = (22)² cm² = 484 cm²
Area of square PQRC = y² cm²
Now, as per the question
Area of shaded region = Area of square ABCD − Area of square PQRC
$403 = 484 - y^{2} \Rightarrow y^{2} = 484 - 403 = 81 \Rightarrow y = 9$
Hence, the correct option is (c).

Question 16:

The length and breadth of a rectangle are (3x + 4) cm and (4x − 13) cm respectively. If the
perimeter of the rectangle is 94 cm, then its area is
(a) 432 cm² (b) 512 cm² (c) 542 cm² (d) 532 cm²

Answer 16:

Here, l = (3x + 4) cm, b = (4x − 13) cm and Perimeter of rectangle = 94 cm.
Perimeter of rectangle = 2(l + b) = 2[(3x + 4) + (4x − 13)] = (14x − 18) cm
As per the question
14x − 18 = 94 $\Rightarrow x = \frac{94 + 18}{14} = 8$
Now
l = 3 $×$ 8 + 4 = 28 cm
b = 4 $×$ 8 − 13 = 19 cm
Area of rectangle = l $×$ b = 28 $×$ 19 = 532 cm²
Hence, the correct option is (d).

Question 17:

The length and breadth of a rectangle are in the ratio 3 : 2. If the area is 216 cm², then its perimeter is

(a) 60 cm (b) 30 cm (c) 40 cm (d) 120 cm

Answer 17:

Here, l = 3x cm, b = 2x cm and area of rectangle = 216 cm².
Area of rectangle = l $×$ b = 3x $×$ 2x = 6x² cm²
As per the question
216 = 6x² $\Rightarrow x^{2} = 36 \Rightarrow x = 6$
Now
l = 3x = 3 $×$ 6 = 18 cm
b = 2x = 2 $×$ 6 = 12 cm
Perimeter of rectangle = 2(l + b) = 2(18 + 12) = 60 cm
Hence, the correct option is (a).

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Question 18:

If the length of a diagonal of a rectangle of length 16 cm is 20 cm, then its area is

(a) 192 cm² (b) 320 cm² (c) 160 cm² (d) 156 cm²

Answer 18:

Here, l = 16 cm, Length of diagonal = 20 cm. Let b be the breadth of the rectangle.
In the right-angled triangle formed with the adjacent sides and the diagonal, using Pythagoras theorem, we get
$l^{2} + b^{2} = {(Diagonal)}^{2} \Rightarrow 16^{2} + b^{2} = 20^{2} \Rightarrow b^{2} = 20^{2} - 16^{2} = 144 \Rightarrow b = 12 cm$
Area of rectangle = l $×$ b = 16 $×$ 12 = 192 cm²
Hence, the correct option is (a).

Question 19:

The area of a rectangle 144 cm long is same as that of a square of side 84 cm. The width of the rectangle is

(a) 7 cm (b) 14 cm (c) 49 cm (d) 28 cm

Answer 19:

Here, Length of rectangle = 144 cm, Area of square = 84 cm².
Let b be the breadth of the rectangle, then as per the question
Area of rectangle = Area of square
$\Rightarrow 144 \times b = 84^{2} \Rightarrow b = \frac{84 \times 84}{144} = 49 cm$
Thus, the breadth of the rectangle is 49 cm.
Hence, the correct option is (c).

Question 20:

The length and breadth of a rectangular field are in the ratio 5 : 3 and its perimeter is 480 m.
The area of the field is

(a) 7200 m² (b) 13500 m² (c) 15000 m² (d) 54000 m²

Answer 20:

Let l = 5x and b = 3x be the length and breadth of the rectangular field. Here, perimeter = 480 m.
So, as per the question
Perimeter = 2 (l + b)
$\Rightarrow 480 = 2 (5 x + 3 x) \Rightarrow 16 x = 480 \Rightarrow x = 30$
l = 5 $×$ 30 = 150 m
b = 3 $×$ 30 = 90 m
Now
Area of the rectangular filed = l $×$ b = 150 $×$ 90 = 13500 m²
Hence, the correct option is (b).

Question 21:

The length of a rectangular field is thrice its breadth and its perimeter is 240 m. The length of the filed is

(a) 30 m (b) 120 m (c) 90 m (d) 80 m

Answer 21:

Let l and b be the length and breadth of the rectangular field, then l = 3b.
So, as per the question
Perimeter = 2 (l + b)
$\Rightarrow 240 = 2 (3 b + b) \Rightarrow 8 b = 240 \Rightarrow b = 30 m$
$l = 3 b = 3 \times 30 = 90 m$
Hence, the correct option is (c).

Question 22:

If the diagonal of a rectangle is 17 cm and its perimeter is 46 cm, the area of the rectangle is

(a) 100 cm² (b) 110 cm² (c) 120 cm² (d) 240 cm²

Answer 22:

Let l and b be the length and breadth of the rectangle, where diagonal = 17 cm and perimeter = 46 cm.
So, as per the question
Perimeter = 2 (l + b)
$\Rightarrow$ 46 = 2 (l + b)
$\Rightarrow$ l + b = 23 ..... (i)
Now, in the triangle formed by the adjacent sides and one diagonal of the rectangle, using Pythagoras theorem, we have
l² + b² = (diagonal)²
$\Rightarrow$ l² + b² = 17²
$\Rightarrow$ l² + (23 − l)² = 17² [From (i)]
$\Rightarrow$ l² +l² + 23² − 46l = 289
$\Rightarrow$ 2l² + 529 − 46l = 289
$\Rightarrow$ 2l² − 46l + 240 = 0
$\Rightarrow$ l² − 23l + 120 = 0
$\Rightarrow$ l² − 15l − 8l + 120 = 0
$\Rightarrow$ l(l − 15) − 8(l − 15) = 0
$\Rightarrow$ (l − 15) (l − 8) = 0
$\Rightarrow$ l = 15 cm or l = 8 cm
If l = 15 cm, then from (i), b = 23 − 15 = 8 cm.
If l = 8 cm, then from (i), b = 23 − 8 = 23 cm.
Therefore
Area of the rectangle = l $×$ b = 15 $×$ 8 = 120 cm2
Hence, the correct option is (c).

Question 23:

The length and breadth of a rectangular field are 4 m and 3 m respectively. The field is divided into two
parts by fencing diagonally. The cost of fencing at the rate of ₹10 per metre is

(a) ₹50 (b) ₹30 (c) ₹190 (d) ₹240

Answer 23:

Let l and b be the length and breadth of the rectangle respectively. Then
l = 4 m and b = 3 m
Now, in the triangle formed by the adjacent sides and one diagonal of the rectangle, using Pythagoras theorem, we have
l² + b² = (Diagonal)²
$\Rightarrow$ (Diagonal)² = 4² + 3² = 16 + 9 = 25
$\Rightarrow$ Diagonal = 5 m
Length of fencing = 2(l + b) + Length of diagonal
                              = 2(4 + 3) + 5
                              = 14 + 5
                              = 19 m
Cost of fencing = ₹10 $×$ 19 = ₹190
Hence, the correct option is (c).

Question 24:

The area of a parallelogram is 100 cm². If the base is 25 cm, then the corresponding height is

(a) 4 cm (b) 6 cm (c) 10 cm (d) 5 cm

Answer 24:

Let b = 25 cm and h be the base and the corresponding height of the parallelogram. Then
Area of parallelogram = b $×$ h
$⇒$ 100 = 25 $×$ h
$⇒$ h = 4 cm
Hence, the correct option is (a).

Question 25:

The base of a parallelogram is twice of its height. If its area is 512 cm², then the length of base is

(a) 16 cm (b) 32 cm (c) 48 cm (d) 64 cm

Answer 25:

Let b and h be the base and height, then b = 2h.
Area of parallelogram = b $×$ h
$⇒$ 512 = 2h $×$ h
$⇒$ 2h² = 512
$⇒$ h² = 256
$⇒$ h = 16 cm
$⇒$ b = 2 $×$ 16 = 32 cm
Hence, the correct option is (b).

Question 26:

The lengths of the diagonals of a rhombus are 36 cm and 22.5 cm. Its area is

(a) 8.10 cm² (b) 405 cm² (c) 202.5 cm² (d) 1620 cm²

Answer 26:

Here, d₁ = 36 cm and d2 = 22.5 cm.
Area of parallelogram = $\frac{1}{2} (d_{1} \times d_{2}) = \frac{1}{2} (36 \times 22.5) = 405 {cm}^{2}$
Hence, the correct option is (b).

Question 27:

The length of a diagonal of a rhombus is 16 cm. If its area is 96 cm², then the length of other
diagonal is

(a) 6 cm (b) 8 cm (c) 12 cm (d) 18 cm

Answer 27:

Let d₁ and d₂ be the diagonals of the rhombus, where d₁ = 16 and area of rhombus = 96 cm².
Area of parallelogram = $\frac{1}{2} (d_{1} \times d_{2})$
$\Rightarrow 96 = \frac{1}{2} (16 \times d_{2}) \Rightarrow d_{2} = \frac{96 \times 2}{16} = 12 cm$
Thus, the length of other diagonal is 12 cm.
Hence, the correct option is (c).

Question 28:

The length of the diagonals of a rhombus of a rhombus are 8 cm and 14 cm. The area of one
of the 4 triangles formed by the diagonals is

(a) 12 cm² (b) 8 cm² (c) 16 cm² (d) 14 cm²

Answer 28:

Let d₁ = 8 cm and d₂ = 14 cm.
Area of parallelogram = $\frac{1}{2} (d_{1} \times d_{2})$
$= \frac{1}{2} (8 \times 14) = 56 {cm}^{2}$
Since, the diagonals of a rhombus divides it into 4 equal parts, so
Area of the required triangle = $\frac{56}{4} = 14 {cm}^{2}$
Hence, the correct option is (d).

Question 29:

The length of a rectangle 8 cm more than the breadth. If the perimeter of the rectangle is 80 cm,
then the length of the rectangle is
(a) 16 cm (b) 24 cm (c) 28 cm (d) 18 cm

Answer 29:

Let l and b be the length and breadth of the rectangle, then l = b + 8.
Perimeter of rectangle = 2 (l + b)
= 2(l + l − 8)
= 4l − 16
$\Rightarrow 80 = 4 l - 16 \Rightarrow 4 l = 80 + 16 = 96 \Rightarrow l = 24 cm$
Hence, the correct option is (b).

Question 30:

The length of a rectangle 8 cm more than the breadth. If the perimeter of the rectangle is 80 cm,
then the area of the rectangle is
(a) 192 cm² (b) 364 cm² (c) 384 cm² (d) 382 cm²

Answer 30:

Question 31:

The area of a rectangle is 11.6 m². If its breadth is 46.4 cm, then the perimeter is
(a) 25.464 m (b) 50.928 m (c) 101.856 m (d) None of these

Answer 31:

Here, area of rectangle(A) = 11.6 m², breadth(b) = 46.4 cm = 0.464 m.
Let l be the length of the rectangle, then
Area of rectangle = l $×$ b
= l $×$ 0.464
Area of rectangle = 11.6
$\Rightarrow l \times 0.464 = 11.6 \Rightarrow l = \frac{11.6}{0.464} = \frac{11600}{464} = 25 m$
Now
Perimeter = 2(l + b)
$= 2 (25 + 0.464) = 2 (25.464) = 50.928 m$
Hence, the correct option is (b).

Question 32:

The area of a rhombus is 119 cm² and its perimeter is 56 cm. The height of the rhombus is

(a) 7.5 cm (b) 6.5 cm (c) 8.5 cm (d) 9.5 cm

Answer 32:

Let b the side of the rhombus and h be its height.
Perimeter of rhombus = 56 cm
$\Rightarrow 4 b = 56 \Rightarrow b = 14 cm$
Now
Area of rhombus = 119 cm²
$\Rightarrow b \times h = 119 \Rightarrow 14 \times h = 119 \Rightarrow h = \frac{17}{2} = 8.5 cm$
Hence, the correct option is (c).

Question 33:

Each side of an equilateral triangle is 8 cm. Its area is

(a) 16 $\sqrt{3}$ cm² (b) 32 $\sqrt{3}$ cm² (c) 24 $\sqrt{3}$ cm² (d) 8 $\sqrt{3}$ cm²

Answer 33:

$Area of equilateral triangle = \frac{{(Side)}^{2} \sqrt{3}}{4}$
$= \frac{{(8)}^{2} \sqrt{3}}{4} = 16 \sqrt{3} {cm}^{2}$
Hence, the correct option is (a).

Question 34:

The area of an equilateral triangle is 4 $\sqrt{3}$ cm². The length of each of its side is

(a) 3 cm (b) 4 cm (c) 2 $\sqrt{3}$ cm (d) $\frac{\sqrt{3}}{2}$ cm

Answer 34:

$Area of equilateral triangle = \frac{{(Side)}^{2} \sqrt{3}}{4} \Rightarrow 4 \sqrt{3} = \frac{{(Side)}^{2} \sqrt{3}}{4} \Rightarrow {(Side)}^{2} = 16 \Rightarrow Side = 4 cm$
Hence, the correct option is (b).

Question 35:

The height of an equilateral triangle is $\sqrt{6}$ cm. Its area is

(a) 3 $\sqrt{3}$ cm² (b) 2 $\sqrt{3}$ cm² (c) 2 $\sqrt{2}$ cm² (d) 6 $\sqrt{2}$ cm²

Answer 35:

Let a and h respectively be the side and height of the equilateral triangle.
$Area of equilateral triangle = \frac{a^{2} \sqrt{3}}{4} \Rightarrow \frac{a^{2} \sqrt{3}}{4} = \frac{1}{2} \times a \times h \Rightarrow \frac{a \sqrt{3}}{4} = \frac{1}{2} \times \sqrt{6} \Rightarrow a = 2 \sqrt{2}$
Therefore
$Area of equilateral triangle = \frac{{(2 \sqrt{2})}^{2} \sqrt{3}}{4} = 2 \sqrt{3} {cm}^{2}$
Hence, the correct option is (b).

Question 36:

If A is the area of an equilateral triangle of height h, then

(a) A = $\sqrt{3}$ h² (b) $\sqrt{3}$ A = h (c) $\sqrt{3}$ A = h² (d) 3A = h²

Answer 36:

Let a and h be the side and height of the equilateral triangle respectively. Then
$\frac{a^{2} \sqrt{3}}{4} = \frac{1}{2} \times a \times h \Rightarrow \frac{a \sqrt{3}}{4} = \frac{1}{2} \times h \Rightarrow a = \frac{2 h}{\sqrt{3}}$
Therefore
$Area of equilateral triangle (A) = \frac{a^{2} \sqrt{3}}{4} \Rightarrow A = \frac{{(\frac{2 h}{\sqrt{3}})}^{2} \sqrt{3}}{4} = \frac{h^{2}}{\sqrt{3}} \Rightarrow \sqrt{3} A = h^{2}$
Hence, the correct option is (c).

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Question 37:

If area of an equilateral triangle is 3 $\sqrt{3}$ cm², then its height is
(a) 3 cm (b) $\sqrt{3}$ cm (c) 6 cm (d) 2 $\sqrt{3}$ cm

Answer 37:

Let a and h be respectively the side and height of the equilateral triangle. Then
$\frac{a^{2} \sqrt{3}}{4} = \frac{1}{2} \times a \times h \Rightarrow \frac{a \sqrt{3}}{4} = \frac{1}{2} \times h \Rightarrow a = \frac{2 h}{\sqrt{3}}$
Therefore
$Area of equilateral triangle = \frac{1}{2} \times a \times h \Rightarrow 3 \sqrt{3} = \frac{1}{2} \times \frac{2 h}{\sqrt{3}} \times h \Rightarrow h^{2} = 9 \Rightarrow h = 3 cm$
Hence, the correct option is (a).

Question 38:

The area of a rhombus is 144 cm² and one of its diagonals is double the other. The length of the
longer diagonal is

(a) 12 cm (b) 16 cm (c) 18 cm (d) 24 cm

Answer 38:

Let d₁ and d₂ be the diagonals of the rhombus, where d₁ = 2d₂.
$Area of rhombus = \frac{1}{2} \times d_{1} \times d_{2} \Rightarrow 144 = \frac{1}{2} \times d_{1} \times \frac{d_{1}}{2} (∵ d_{1} = 2 d_{2}) \Rightarrow {d_{1}}^{2} = 4 \times 144 \Rightarrow d_{1} = 2 \times 12 = 24 cm$
Hence, the correct option is (d).

Question 39:

In fig. 38, the value of k is

(a) $\frac{77}{8}$ (b) $\frac{73}{8}$ (c) $\frac{71}{8}$ (d) $\frac{75}{8}$

Answer 39:

In triangle ABC, we have
$Area of ∆ A B C = \frac{1}{2} \times B C \times A D = \frac{1}{2} \times A B \times C E \Rightarrow B C \times A D = A B \times C E \Rightarrow 14 \times 11 = 16 \times k \Rightarrow k = \frac{14 \times 11}{16} = \frac{77}{8}$
Hence, the correct option is (a).

Question 40:

In fig. 40, ABCD is a parallelogram of area 144 cm², the value of x is

(a) 8 (b) 6 (c) 9 (d) 10

Answer 40:

$Area of parallelogram = Base \times Height = \Rightarrow 144 = 16 \times x \Rightarrow x = \frac{144}{16} = 9$
Hence, the correct option is (c).

Question 41:

In fig. 41, if ABCD is a parallelogram of area 273 cm², the value of h is

(a) 13 (b) 12 (c) 8 (d) 14

Answer 41:

The quadrilateral ABCD is a trapezium whose area is 273 cm². So
$Area of trapezium = \frac{1}{2} (Sum of parallel sides) \times Height \Rightarrow 273 = \frac{1}{2} (24 + 18) \times h \Rightarrow h = \frac{273 \times 2}{42} = 13$
Hence, the correct option is (a).

Question 42:

In Fig. 42, ABCD is a parallelogram in which AD = 21 cm, DH =18 cm and DK = 27 cm.
The length of AB is

(a) 63 cm (b) 63.5 cm (c) 31.5 cm (d) 31 cm

Answer 42:

Area of a parallelogram = Base $×$ Height
$A B \times D H = A D \times D K \Rightarrow A B \times 18 = 21 \times 27 \Rightarrow A B = \frac{21 \times 27}{18} = \frac{63}{2} = 31.5 cm$
Hence, the correct option is (c).

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Question 43:

In Fig. 42, ABCD is a parallelogram in which AD = 21 cm, DH =18 cm and DK = 27 cm.
The perimeter of the parallelogram is

(a) 105 cm (b) 84.5 cm (c) 169 cm (d) 52.5 cm

Answer 43:

Area of a parallelogram = Base $×$ Height
$A B \times D H = A D \times D K \Rightarrow A B \times 18 = 21 \times 27 \Rightarrow A B = \frac{21 \times 27}{18} = \frac{63}{2} = 31.5 cm$
Here, ABCD is a parallelogram, so AB = CD and AD = BC.
Therefore
Perimeter of parallelogram ABCD = 2(AB + AD)
= 2(31.5 + 21)
= 105 cm
Hence, the correct option is (a).

Question 44:

In Fig. 42, the area of the parallelogram is

(a) 516 cm² (b) 616 cm² (c) 416 cm² (d) 606 cm²

Answer 44:

Here, ABCD is a parallelogram, so AD = BC = 21 cm.
Therefore
Area of parallelogram = Base $×$ Height
                                    = BC $×$ DK
                                    = 21 $×$ 27
                                    = 567 cm²

Question 45:

A piece of wire of length 12 cm is bent to form a square. The area of the square is

(a) 36 cm² (b) 144 cm² (c) 9 cm² (d) 12 cm²

Answer 45:

Let a be the length of the side of the square. Then as per he question, we have
4a = 12
a = 3 cm
Therefore
Area of square = a²
= 3²
= 9 cm²
Thus, the area of the square is 9 cm².
Hence, the correct option is (c)

Question 46:

The area of a right isosceles triangle whose hypotenuse is 16 $\sqrt{2}$ cm is

(a) 125 cm² (b) 158 cm² (c) 128 cm² (d) 144 cm²

Answer 46:

Let a be the length of the equal sides of the right isosceles triangle whose hypotenuse is 16 $\sqrt{2}$ cm.
Then using Pythagoras theorem in the triangle, we get
a² + a² = (16 $\sqrt{2}$ )²
      2a² = 512
        a² = 256
Therefore
Area of the triangle $= \frac{1}{2} \times Base \times Height$
                               $= \frac{1}{2} \times a \times a = \frac{1}{2} \times a^{2} = \frac{1}{2} \times 256 = 128 {cm}^{2}$
Thus, the area of the square is 128 cm².
Hence, the correct option is (c)

Question 47:

A wire is in the form of a square of side 18 m. It is bent in the form of a rectangle, whose length
and breadth are in the ratio 3 : 1. The area of the rectangle is

(a) 81 m² (b) 243 m² (c) 144 m² (d) 324 m²

Answer 47:

Side of square (a) = 18 m
Let l = 3x and b = x be the length and breadth of the rectangle. Then
Perimeter of rectangle = Perimeter of square
                       2(l + b) = 4a
                     2(3x + x) = 4 $×$ 18
                                8x = 72
                                  x = 9 m
Thus
Length (l) = 3x = 3 $×$ 9 = 27 m
Breadth (b) = x = 9 m
Therefore
Area of the rectangle = l $×$ b
                                 = 27 $×$ 9 = 243 m²
Thus, the area of the rectangle is 243 m².
Hence, the correct option is (b)