myhelper: RS Aggarwal 2019,2020 solution class 7 chapter 20 Mensuration Exercise 20G

RS Aggarwal 2019,2020 solution class 7 chapter 20 Mensuration Exercise 20G

Exercise 20G

Page-253

Question 1:

The length of a rectangle is 16 cm and the length of its diagonal is 20 cm. The area of the rectangle is

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

Answer 1:

Let BC = x cm
From right triangle ABC:
AC² = AB²+ BC²
⇒ (20)² = (16)² + x²
⇒ x² = (20)² - (16)²⇒ {400 - 256} = 144
⇒ x =

\sqrt{144}

= 12
∴ BC = 12 cm
∴ Area of the plot = (16 × 12) cm² = 192 cm²

Question 2:

Each diagonal of a square is 12 cm long. Its area is

(a) 144 cm²
(b) 72 cm²
(c) 36 cm²
(d) none of these

Answer 2:

(b) 72 cm²

Given:
Diagonal of the square = 12 cm
∴ Area of the square =

\{\frac{1}{2} \times {(Diagonal)}^{2}\}

sq. units.
=

\{\frac{1}{2} \times {(12)}^{2}\}

cm²
= 72 cm²

Question 3:

The area of a square is 200 cm². The length of its diagonal is

(a) 10 cm
(b) 20 cm
(c)

10 \sqrt{2}

cm
(d) 14.1 cm

Answer 3:

(b) 20 cm

Area of the square =

\{\frac{1}{2} \times {(D iagonal)}^{2}\}

sq. units.
Area of the square field = 200 cm²

Diagonal of a square =

\sqrt{2 \times Area of the square}

(\sqrt{2 \times 200})

cm =

(\sqrt{400})

cm = 20 cm
∴ Length of the diagonal of the square = 20 cm

Question 4:

The area of a square field is 0.5 hectare. The length of its diagonal is

(a) 100 m
(b) 50 m
(c) 250 m
(d)

50 \sqrt{2}

Answer 4:

(a) 100 m

Area of the square =

\{\frac{1}{2} \times {(D iagonal)}^{2}\}

sq. units.
Given:
Area of square field = 0.5 hectare
=

(0.5 \times 10000)

m² [since 1 hectare = 10000 m²]
= 5000 m²

Diagonal of a square =

\sqrt{2 \times Area of the square}

(\sqrt{2 \times 5000})

m = 100 m
Hence, the length of the diagonal of a square field is 100 m.

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Question 5:

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

(a) 80 m
(b) 120 m
(c) 90 m
(d) none of these

Answer 5:

(c) 90 m

Let the breadth of the rectangular field be x m.
Length = 3x m
Perimeter of the rectangular field = 2(l + b)
⇒ 240 = 2( x + 3x)
⇒ 240 = 2(4x)
⇒ 240 = 8x ⇒ x = $(\frac{240}{8}) = 30$
∴ Length of the field = 3x = (3 × 30) m = 90 m

Question 6:

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

(a) 25%
(b) 55%
(c) 40.5%
(d) 56.25%

Answer 6:

(d) 56.25%

Let the side of the square be a cm.
Area of the square = (a)² cm²
Increased side = (a + 25% of a) cm
= $(a + \frac{25}{100} a)$ cm = $(a + \frac{1}{4} a) cm = (\frac{5}{4} a)$ cm
Area of the square = ${(\frac{5}{4} a)}^{2} c m^{2} = (\frac{25}{16} a^{2})$ cm²
Increase in the area = $[(\frac{25}{16} a^{2}) - a^{2}]$ cm²= $(\frac{25 a^{2} - 16 a^{2}}{16})$ cm² = $(\frac{9 a^{2}}{16})$ cm²
% increase in the area = $\frac{Increased area}{Old area} \times 100$
= $[\frac{(\frac{9}{16} a^{2})}{a^{2}} \times 100]$ = $(\frac{9 \times 100}{16}) = 56.25$

Question 7:

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 7:

(b) 1:2

Let the side of the square be a.
Length of its diagonal = $\sqrt{2} a$
∴ Required ratio = $\frac{a^{2}}{{(\sqrt{2} a)}^{2}} = \frac{a^{2}}{2 a^{2}} = \frac{1}{2} = 1 : 2$

Question 8:

The perimeters of a square and a rectangle are equal. If their areas be A m² and B m², then which of the following is a true statement?

(a) A < B
(b) A ≤ B
(c) A > B
(d) A ≥ B

Answer 8:

(c) A > B

We know that a square encloses more area even though its perimeter is the same as that of the rectangle.

∴ Area of a square > Area of a rectangle

Question 9:

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 9:

(b) 13500 m²

Let the length of the rectangular field be 5x.
Breadth = 3x
Perimeter of the field = 2(l + b) = 480 m (given)
⇒ 480 = 2(5x + 3x) ⇒ 480 = 16x
⇒ x = $\frac{480}{16}$ = 30
∴ Length = 5x = (5 × 30) = 150 m
Breadth = 3x = (3 × 30) = 90 m
∴ Area of the rectangular park = 150 m × 90 m = 13500 m²

Question 10:

The length of a room is 15 m. The cost of carpeting it with a carpet 75 cm wide at Rs 50 per metre is Rs 6000. The width of the room is

(a) 6 m
(b) 8 m
(c) 13.4 m
(d) 18 m

Answer 10:

(a) 6 m

Total cost of carpeting = Rs 6000
Rate of carpeting = Rs 50 per m
∴ Length of the carpet = $(\frac{6000}{50})$ m = 120 m
∴ Area of the carpet = $(120 \times \frac{75}{100})$ m² = 90 m² [since 75 cm = $\frac{75}{100} m$ ]
Area of the floor = Area of the carpet = 90 m²
∴ Width of the room = $(\frac{Area}{Length}) = (\frac{90}{15}) m = 6 m$

Question 11:

The sides of a triangle measure 13 cm, 14 cm and 15 cm. Its area is

(a) 84 cm²
(b) 91 cm²
(c) 168 cm²
(d) 182 cm²

Answer 11:

(a) 84 cm²

Let a = 13 cm, b = 14 cm and c = 15 cm
Then, s = $\frac{a + b + c}{2}$ = $(\frac{13 + 14 + 15}{2})$ cm = 21 cm
∴ Area of the triangle = $\sqrt{s (s - a) (s - b) (s - c)}$ sq. units
                               = $\sqrt{21 (21 - 13) (21 - 14) (21 - 15)}$ cm²
                               = $\sqrt{21 \times 8 \times 7 \times 6}$ cm²
                               = $\sqrt{3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3}$ cm²
                               = (2 × 2 × 3 × 7) cm²
                               = 84 cm²

Question 12:

The base and height of a triangle are 12 m and 8 m respectively. Its area is

(a) 96 m²
(b) 48 m²
(c) $16 \sqrt{3}$ m²
(d) $16 \sqrt{2}$ m²

Answer 12:

(b) 48 m²

Base = 12 m
Height = 8 m
Area of the triangle = $(\frac{1}{2} \times Base \times Height)$ sq. units
= $(\frac{1}{2} \times 12 \times 8)$ m²
= 48 m²

Question 13:

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

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

Answer 13:

(b) 4 cm

Area of the equilateral triangle = $4 \sqrt{3}$ cm²

We know:
Area of an equilateral triangle = $\frac{\sqrt{3}}{4} {(side)}^{2}$ sq. units
∴ Side of the equilateral triangle = $[\sqrt{(\frac{4 \times Area}{\sqrt{3}})}]$ cm
= $[\sqrt{(\frac{4 \times 4 \sqrt{3}}{\sqrt{3}})}]$ cm = $(\sqrt{4 \times 4})$ cm = $(\sqrt{16})$ cm = 4 cm

Question 14:

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

(a) 32 cm²
(b) 64 cm²
(c) $16 \sqrt{3}$ cm²
(d) $16 \sqrt{2}$ cm²

Answer 14:

(c) $16 \sqrt{3}$ cm²

It is given that one side of an equilateral triangle is 8 cm.
∴ Area of the equilateral triangle = $\frac{\sqrt{3}}{4} {(Side)}^{2}$ sq. units
= $\frac{\sqrt{3}}{4} {(8)}^{2}$ cm²
= $(\frac{\sqrt{3}}{4} \times 64)$ cm²= $16 \sqrt{3}$ cm²

Question 15:

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 15:

(b) $2 \sqrt{3}$ cm²

Let ΔABC be an equilateral triangle with one side of the length a cm.
Diagonal of an equilateral triangle = $\frac{\sqrt{3}}{2} a$ cm
⇒ $\frac{\sqrt{3}}{2} a = \sqrt{6}$
⇒ a = $\frac{\sqrt{6} \times 2}{\sqrt{3}} = \frac{\sqrt{3} \times \sqrt{2} \times 2}{\sqrt{3}} = 2 \sqrt{2}$ cm
Area of the equilateral triangle = $\frac{\sqrt{3}}{4} a^{2}$
= $\frac{\sqrt{3}}{4} {(2 \sqrt{2})}^{2}$ cm² = $(\frac{\sqrt{3}}{4} \times 8)$ cm² = $2 \sqrt{3}$ cm²

Question 16:

One side of a parallelogram is 16 cm and the distance of this side from the opposite side is 4.5 cm. The area of the parallelogram is

(a) 36 cm²
(b) 72 cm²
(c) 18 cm²
(d) 54 cm²

Answer 16:

(b) 72 cm²

Base of the parallelogram = 16 cm
Height of the parallelogram = 4.5 cm
∴ Area of the parallelogram = Base × Height
= (16 × 4.5) cm²= 72 cm²

Question 17:

The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively. Its area is

(a) 432 cm²
(b) 216 cm²
(c) 108 cm²
(d) 144 cm²

Answer 17:

(b) 216 cm²

Length of one diagonal = 24 cm
Length of the other diagonal = 18 cm
∴ Area of the rhombus = $\frac{1}{2}$ × (Product of the diagonals)
= $(\frac{1}{2} \times 24 \times 18)$ cm² = 216 cm²

Question 18:

The difference between the circumference and radius of a circle is 37 cm. The area of the circle is

(a) 111 cm²
(b) 148 cm²
(c) 154 cm²
(d) 259 cm²

Answer 18:

(c) 154 cm²

Let the radius of the circle be r cm.
Circumference = $2 πr$

(Circumference) - (Radius) = 37
∴ $(2 πr - r) = 37$
⇒ $r (2 π - 1) = 37$
⇒ r = $\frac{37}{(2 π - 1)}$ = $\frac{37}{(2 \times \frac{22}{7} - 1)} = \frac{37}{(\frac{44}{7} - 1)} = \frac{37}{(\frac{44 - 7}{7})} = (\frac{37 \times 7}{37}) = 7$
∴ Radius of the given circle is 7 cm.
∴ Area = ${πr}^{2}$ = $(\frac{22}{7} \times 7 \times 7)$ cm² = 154 cm²

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Question 19:

The perimeter of the floor of a room is 18 m and its height is 3 m. What is the area of 4 walls of the room?

(a) 21 m²
(b) 42 m²
(c) 54 m²
(d) 108 m²

Answer 19:

(c) 54 m²

Given:
Perimeter of the floor = 2(l + b) = 18 m
Height of the room = 3 m

∴ Area of the four walls = {2(l + b) × h}
                          = Perimeter × Height
                                        = 18 m × 3 m = 54 m²

Question 20:

How many metres of carpet 63 cm wide will be required to cover the floor of a room 14 m by 9 m?

(a) 200 m
(b) 210 m
(c) 220 m
(d) 185 m

Answer 20:

(a) 200 m

Area of the floor of a room = 14 m × 9 m = 126 m²

Width of the carpet = 63 cm = 0.63 m (since 100 cm = 1 m)

∴ Required length of the carpet = $\frac{Area of the floor of a room}{Width of the carpet}$
= $(\frac{126}{0.63}) m = 200 m$

Question 21:

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

(a) 100 cm²
(b) 110 cm²
(c) 120 cm²
(d) 150 cm²

Answer 21:

(c) 120 cm²

Let the length of the rectangle be x cm and the breadth be y cm.
Area of the rectangle = xy cm²
Perimeter of the rectangle = 2( x + y) = 46 cm          (given)
⇒ 2( x + y) = 46
⇒ ( x + y) = $(\frac{46}{2})$ cm = 23 cm

Diagonal of the rectangle = $\sqrt{x^{2} + y^{2}}$ = 17 cm
⇒ $\sqrt{x^{2} + y^{2}}$ = 17

Squaring both the sides, we get:
⇒ x² + y² = (17)²
⇒ x² + y² = 289

Now, (x² + y²) = ( x + y)² - 2xy
⇒ 2xy = ( x + y)²- (x² + y²)
           = (23)² - 289
           = 529 - 289 = 240
∴ xy = $(\frac{240}{2})$ cm² = 120 cm²

Question 22:

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 22:

(b) 3:1

Let a side of the first square be a cm and that of the second square be b cm.
Then, their areas will be a² and b², respectively.
Their perimeters will be 4a and 4b, respectively.

According to the question:
$\frac{a^{2}}{b^{2}} = \frac{9}{1}$ ⇒ ${(\frac{a}{b})}^{2} = \frac{9}{1} = {(\frac{3}{1})}^{2}$ ⇒ $\frac{a}{b} = \frac{3}{1}$

∴ Required ratio of the perimeters = $\frac{4 a}{4 b} = \frac{4 \times 3}{4 \times 1} = \frac{3}{1}$ = 3:1

Question 23:

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 23:

(d) 4:1

Let the diagonals be 2d and d.
Area of the square = $[\frac{1}{2} \times {(Diagonal)}^{2}]$ sq. units
Required ratio = $\frac{A_{1}}{A_{2}} = \frac{\frac{1}{2} {(2 d)}^{2}}{\frac{1}{2} {(d)}^{2}} = \frac{4 d^{2}}{d^{2}} = \frac{4}{1} = 4 : 1$

Question 24:

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

(a) 7 m
(b) 14 m
(c) 49 m
(d) none of these

Answer 24:

(c) 49 m

Let the width of the rectangle be x m.

Given:
Area of the rectangle = Area of the square
⇒ Length × Width = Side × Side
⇒ (144 × x) = 84 × 84
∴ Width (x) = $(\frac{84 \times 84}{144})$ m = 49 m
Hence, width of the rectangle is 49 m.

Question 25:

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 25:

(d) $4 : \sqrt{3}$

Let one side of the square and that of an equilateral triangle be the same, i.e. a units.
Then, Area of the square = (Side)² = (a)²
Area of the equilateral triangle = $\frac{\sqrt{3}}{4} {(Side)}^{2}$ = $\frac{\sqrt{3}}{4} {(a)}^{2}$
∴ Required ratio = $\frac{a^{2}}{\frac{\sqrt{3}}{4} a^{2}} = \frac{4}{\sqrt{3}} = 4 : \sqrt{3}$

Question 26:

The area of a square is equal to the area of a circle. What is the ratio between the side of the square and the radius of the circle?

(a) $\sqrt{π} : 1$
(b) $1 : \sqrt{π}$
(c) $1 : π$
(d) $π : 1$

Answer 26:

(a) $\sqrt{π} : 1$

Let the side of the square be x cm and the radius of the circle be r cm.
Area of the square = Area of the circle
⇒ (x)² = ${πr}^{2}$
∴ Side of the square (x) = $\sqrt{π} r$
Required ratio = $\frac{Side of the square}{Radius of the circle}$
= $\frac{x}{r} = \frac{\sqrt{π} r}{r} = \frac{\sqrt{π}}{1} = \sqrt{π} : 1$

Question 27:

Each side of an equilateral triangle is equal to the radius of a circle whose area is 154 cm². The area of the triangle is

(a) $\frac{7 \sqrt{3}}{4} {cm}^{2}$
(b) $\frac{49 \sqrt{3}}{4} {cm}^{2}$
(c) $35 {cm}^{2}$
(d) $49 {cm}^{2}$

Answer 27:

(b) $\frac{49 \sqrt{3}}{4} {cm}^{2}$

Let the radius of the circle be r cm.
Then, its area = ${πr}^{2}$ cm²
∴ ${πr}^{2}$ = 154
⇒ $\frac{22}{7} \times r \times r = 154$
⇒ r² = $(\frac{154 \times 7}{22})$ = 49
⇒ r = $\sqrt{49}$ cm = 7 cm

Side of the equilateral triangle = Radius of the circle
                                                 = 7 cm
∴ Area of the equilateral triangle = $\frac{\sqrt{3}}{4} {(side)}^{2}$ sq. units

                                                     = $\frac{\sqrt{3}}{4} {(7)}^{2}$ cm²

                                                     = $\frac{49 \sqrt{3}}{4}$ cm²

Question 28:

The area of a rhombus is 36 cm² and the length of one of its diagonals is 6 cm. The length of the second diagonal is

(a) 6 cm
(b) $6 \sqrt{2}$ cm
(c) 12 cm
(d) none of these

Answer 28:

(c) 12 cm

Area of the rhombus = $\frac{1}{2}$ × (Product of the diagonals)
Given:
Length of one diagonal = 6 cm
Area of the rhombus = 36 cm²

∴ Length of the other diagonal = $(\frac{36 \times 2}{6})$ cm = 12 cm

Question 29:

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 29:

(d) 24 cm

Let the length of the shorter diagonal of the rhombus be x cm.
∴ Longer diagonal = 2x

Area of the rhombus = $\frac{1}{2}$ × (Product of its diagonals)
⇒ 144 = $(\frac{1}{2} \times x \times 2 x)$

⇒ 144 = $(\frac{2 x^{2}}{2})$ = $(x^{2})$

∴ x = $(\sqrt{144})$ cm = 12 cm

∴ Length of the longer diagonal = 2x
= (2 × 12) cm
= 24 cm

Question 30:

The area of a circle is 24.64 m². The circumference of the circle is

(a) 14.64 m
(b) 16.36 m
(c) 17.60 m
(d) 18.40 m

Answer 30:

(c) 17.60 m

Let the radius of the circle be r m.
Area = ${πr}^{2}$ m²
∴ ${πr}^{2}$ = 24.64
⇒ $(\frac{22}{7} \times r \times r)$ = 24.64
⇒ r² = $(\frac{24.64 \times 7}{22})$ = 7.84
⇒ r = $\sqrt{7.84}$ = 2.8 m
⇒ Circumference of the circle = $(2 πr)$ m
= $(2 \times \frac{22}{7} \times 2.8)$ m = 17.60 m

Page-256

Question 31:

The area of a circle is increased by 22 cm² when its radius is increased by 1 cm. The original radius of the circle is

(a) 6 cm
(b) 3.2 cm
(c) 3 cm
(d) 3.5 cm

Answer 31:

(c) 3 cm

Suppose the radius of the original circle is r cm.
Area of the original circle = ${πr}^{2}$

Radius of the circle = (r +1) cm
According to the question:
$π {(r + 1)}^{2} = {πr}^{2} + 22$
⇒ $π (r^{2} + 1 + 2 r) = {πr}^{2} + 22$
⇒ ${πr}^{2} + π + 2 πr = {πr}^{2} + 22$
⇒ $π + 2 πr = 22$ [cancel ${πr}^{2}$ from both the sides of the equation]
⇒ $π (1 + 2 r) = 22$
⇒ $(1 + 2 r) = \frac{22}{π} = (\frac{22 \times 7}{22}) = 7$
⇒ 2r = 7 -1 = 6
∴ r = $(\frac{6}{2})$ cm = 3 cm
∴ Original radius of the circle = 3 cm

Question 32:

The radius of a circular wheel is 1.75 m. How many revolutions will it make in travelling 11 km?

(a) 10
(b) 100
(c) 1000
(d) 10000

Answer 32:

(c) 1000

Radius of the wheel = 1.75 m
Circumference of the wheel = $2 πr$
= $(2 \times \frac{22}{7} \times 1.75)$ cm = (2 × 22 × 0.25) m = 11 m

Distance covered by the wheel in 1 revolution is 11 m.
Now, 11 m is covered by the car in 1 revolution.
(11 × 1000) m will be covered by the car in $(1 \times \frac{1}{11} \times 11 \times 1000)$ revolutions, i.e. 1000 revolutions.
∴ Required number of revolutions = 1000