Rd Sharma 2020 2021 for Class 7 Maths Chapter 21 - Mensuration Ii

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Rd Sharma 2020 2021 Solutions for Class 7 Maths Chapter 21 Mensuration Ii are provided here with simple step-by-step explanations. These solutions for Mensuration Ii are extremely popular among Class 7 students for Maths Mensuration Ii Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 2021 Book of Class 7 Maths Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 2021 Solutions. All Rd Sharma 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

Page No 21.15:

Question 1:

Find the area of a circle whose radius is
(i) 7 cm
(ii) 2.1 m
(iii) 7 km

Answer:

(i) We know that the area A of a circle of radius r is given by A = $π r^{2}$ .
Here, r = 7 cm
∴ $A = \frac{22}{7} \times {(7)}^{2} {cm}^{2}$ .
$\Rightarrow A = (\frac{22}{7} \times 7 \times 7) {cm}^{2} = (22 \times 7) {cm}^{2} = 154 {cm}^{2}$ .
(ii) We know that the area A of a circle of radius r is given by A = $π r^{2}$ .
Here, r = 2.1 m
∴ $A = \frac{22}{7} \times {(2.1)}^{2} m^{2}$ .
$\Rightarrow A = (\frac{22}{7} \times 2.1 \times 2.1) m^{2} = (22 \times 0.3 \times 2.1) m^{2} = 13.86 m^{2}$ .
(iii) We know that the area A of a circle of radius r is given by A = $π r^{2}$ .
Here, r = 7 km
∴ $A = \frac{22}{7} \times {(7)}^{2} {km}^{2}$ .
$\Rightarrow A = (\frac{22}{7} \times 7 \times 7) {km}^{2} = (22 \times 7) {km}^{2} = 154 {km}^{2}$ .

Page No 21.15:

Question 2:

Find the area of a circle whose diameter is
(i) 8.4 cm
(ii) 5.6 m
(iii) 7 km

Answer:

(i) Let r be the radius of the circle. Then, r = 8.4 $\div$ 2 = 4.2 cm.
∴ Area of the circle = ${πr}^{2}$
   $\Rightarrow A = \frac{22}{7} \times {(4.2)}^{2} {cm}^{2} \Rightarrow A = (\frac{22}{7} \times 4.2 \times 4.2) {cm}^{2} = (22 \times 0.6 \times 4.2) {cm}^{2} = 55.44 {cm}^{2} .$ .

(ii) Let r be the radius of the circle. Then, r = 5.6 $\div$ 2 = 2.8 m.
Area of the circle = ${πr}^{2}$
   $\Rightarrow A = \frac{22}{7} \times {(2.8)}^{2} m^{2} \Rightarrow A = (\frac{22}{7} \times 2.8 \times 2.8) m^{2} = (22 \times 0.4 \times 2.8) m^{2} = 24.64 m^{2} .$
(iii) Let r be the radius of the circle. Then, r = 7 $\div$ 2 = 3.5 km.
Area of the circle = ${πr}^{2}$
   $\Rightarrow A = \frac{22}{7} \times {(3.5)}^{2} {km}^{2} \Rightarrow A = (\frac{22}{7} \times 3.5 \times 3.5) {km}^{2} = (22 \times 0.5 \times 3.5) {km}^{2} = 38.5 {km}^{2} .$

Page No 21.15:

Question 3:

The area of a circle is 154 cm². Find the radius of the circle.

Answer:

Let the radius of the circle be r cm.
Area of the circle (A) = 154 cm²
$\Rightarrow 154 = \frac{22}{7} \times {(r)}^{2} {cm}^{2} \Rightarrow r^{2} = (\frac{154 \times 7}{22}) \Rightarrow r^{2} = (\frac{1078}{22}) \Rightarrow r^{2} = (49) \Rightarrow r = 7 cm .$

Hence, the radius of the circle is 7 cm.

Page No 21.15:

Question 4:

Find the radius of a circle, if its area is
(i) 4 π cm²
(ii) 55.44 m²
(iii) 1.54 km²

Answer:

(i) Let the radius of the circle be r cm.

∴ Area of the circle (A) = 4 $π$ cm²
   $\Rightarrow 4 π = π \times {(r)}^{2} {cm}^{2} \Rightarrow r^{2} = (\frac{4 π}{π}) = 4 \Rightarrow r = 2 cm .$

(ii) Let the radius of the circle be r cm.

∴ Area of the circle (A) = 55.44 m²
   $\Rightarrow 55.44 = \frac{22}{7} \times {(r)}^{2} m^{2} \Rightarrow r^{2} = (\frac{55.44 \times 7}{22}) \Rightarrow r^{2} = (\frac{5.04 \times 7}{2}) \Rightarrow r^{2} = (17.64) \Rightarrow r = 4.2 m$

(iii) Let the radius of the circle be r cm.

∴ Area of the circle (A) = 1.54 km²
   $\Rightarrow 1.54 = \frac{22}{7} \times {(r)}^{2} {km}^{2} \Rightarrow r^{2} = (\frac{1.54 \times 7}{22}) \Rightarrow r^{2} = (\frac{10.78}{22}) \Rightarrow r^{2} = (0.49) \Rightarrow r = 0.7 km = 700 m$

Page No 21.15:

Question 5:

The circumference of a circle is 3.14 m, find its area.

Answer:

We have :

Circumference of the circle = 3.14 m =

2 πr

\Rightarrow 3.14 m = (2 \times \frac{22}{7} x r) m \Rightarrow r = (\frac{3.14 \times 7}{2 \times 22}) m = (\frac{1}{2}) m .

Area of the circle (A) =

{πr}^{2}

\Rightarrow A = \frac{22}{7} \times {(\frac{1}{2})}^{2} m^{2} \Rightarrow A = (\frac{22}{7} \times \frac{1}{2} \times \frac{1}{2}) m^{2} = (\frac{22}{28}) m^{2} = 0.785 m^{2} .

Page No 21.15:

Question 6:

If the area of a circle is 50.24 m², find its circumference.

Answer:

We have :
Area of the circle (A) = ${πr}^{2}$ = 50.24 m²
$\Rightarrow 50.24 m^{2} = \frac{22}{7} \times {(r)}^{2} \Rightarrow r^{2} = (\frac{50.24 \times 7}{22}) m^{2} = (\frac{351.68}{22}) m^{2} = 15.985 m^{2} \Rightarrow r = 3.998 m .$

Circumference of circle (C) =

2 πr

\Rightarrow C = (2 \times \frac{22}{7} x 3.998) m \Rightarrow C = (44 \times 0.571) m = 25.12 m .

Page No 21.15:

Question 7:

A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22/7).

Answer:

We have :
Length of the string = 28 m
The area over which the horse can graze is the same as the area of a circle of radius 28 m.
Hence, required area = πr² = $(\frac{22}{7} \times 28 \times 28) m^{2} = (22 \times 4 \times 28 \times) m^{2} = 2464 m^{2}$ .

Page No 21.15:

Question 8:

A steel wire when bent in the form of a square encloses an area of 121 cm². If the same wire is bent in the form of a circle, find the area of the circle.

Answer:

We have :
Area of the square = 121 cm²
⇒ (side)² = (11)² cm²
⇒ side = 11 cm.
So, the perimeter of the square = 4(side) = (4 x 11) cm = 44 cm.

Let r be the radius of the circle. Then,
Circumference of the circle = Perimeter of the square
⇒ 2πr = 44
⇒ 2 x $\frac{22}{7}$ x r = 44
⇒ r = 7 cm.
∴ Area of the circle = πr² = $(\frac{22}{7} \times 7 \times 7) = (22 \times 7) = 154 {cm}^{2}$ .

Page No 21.15:

Question 9:

A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of of road.

Answer:

We have :
Circumference of the circular park = 2πr = 352 m
⇒ 2 x $\frac{22}{7}$ x r = 352
⇒ r = 56 m.
Radius of the path including the 7m wide road = (r +7) = 56 +7 = 63 m.
∴ Area of the road :
$= [π {(63)}^{2} - π {(56)}^{2}] m^{2} = π [{(63)}^{2} - {(56)}^{2}] m^{2} = π [(63 + 56) (63 - 56)] m^{2} = π [(119) (7)] = (\frac{22}{7} \times 119 \times 7) m^{2} = (22 \times 119) m^{2} = 2618 m^{2} .$

Page No 21.15:

Question 10:

Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is π h (2r + h).

Answer:

Radius of the circular region = r
Radius of the circular path of uniform width h surrounding the circular region of radius r = (r + h).
∴ Area of the path
$= [π {(r + h)}^{2} - {πr}^{2}] = π [{(r + h)}^{2} - r^{2}] = π [r^{2} + 2 rh + h^{2} - r^{2}] = π [2 rh + h^{2}] = πh [2 r + h]$

Page No 21.16:

Question 11:

The perimeter of a circle is 4πr cm. What is the area of the circle?

Answer:

We have :
Given perimeter of the circle = 4 $π$ r cm = 2 $π$ (2r) cm

We know that, the perimeter of a circle = 2 $π$ r

∴ Radius of the circle = 2r cm

Area of the circle=

π

r² =

π

(2r)² = 4

π

r^{2 .}

Page No 21.16:

Question 12:

A wire of 5024 m length is in the form of a square. It is cut and made a circle. Find the ratio of the area of the square to that of the circle.

Answer:

We have:
Perimeter of the square = 5024 m = Circumference of the circle
⇒ 4 x Side of the square = 5024
∴ Side of the square = $\frac{5024}{4} = 1256 m$ .
Let the area of the square be A₁ and the area of the circle be A₂.
Area of the square (A₁)= side x side = $(\frac{5024}{4} \times \frac{5024}{4}) m^{2}$ .
Circumference of the circle = 5024 m
⇒ 2 $π$ r = 5024 m

\Rightarrow (2 \times \frac{22}{7} \times r) = 5024 m \Rightarrow r = \frac{5024 \times 7}{2 \times 22} .

Area of the circle (A₂)=

π

r² =

[\frac{22}{7} \times \frac{5024 \times 7}{2 \times 22} \times \frac{5024 \times 7}{2 \times 22}] = [\frac{5024 \times 5024 \times \times 7}{2 \times 2 \times 22}] m^{2}

∴ A_{1} : A_{2} = [\frac{5024}{4} \times \frac{5024}{4}] : [\frac{5024 \times 5024 \times \times 7}{2 \times 2 \times 22}] \Rightarrow \frac{A_{1}}{A_{2}} = [\frac{5024}{4} \times \frac{5024}{4}] \div [\frac{5024 \times 5024 \times \times 7}{2 \times 2 \times 22}] \Rightarrow \frac{A_{1}}{A_{2}} = [\frac{5024}{4} \times \frac{5024}{4}] \times [\frac{2 \times 2 \times 22}{5024 \times 5024 \times \times 7}] \Rightarrow \frac{A_{1}}{A_{2}} = \frac{11}{14} . \Rightarrow A_{1} : A_{2} = 11 : 14 .

Hence, the ratio of the area of the square to the area of the circle is 11:14.

Page No 21.16:

Question 13:

The radius of a circle is 14 cm. Find the radius of the circle whose area is double of the area of the circle.

Answer:

Let the area of the circle whose radius is 14 cm be A₁_.
Let the radius and area of the circle, whose area is twice the area of the circle A₁_,be r₂ and A₂, respectively.

Thus,
A₁ = $π {(r)}^{2} = π {(14)}^{2} = (\frac{22}{7} \times 14 \times 14) {cm}^{2} = (44 \times 14) {cm}^{2} = 616 {cm}^{2}$ .
A₂ = 2 $\times$ A₁ = 2 $\times$ 616 = 1232 cm²
A₂ = $π {(r_{2})}^{2}$ = 1232 cm²
^{$\Rightarrow (\frac{22}{7} \times {r_{2}}^{2}) = 1232 {cm}^{2} \Rightarrow {r_{2}}^{2} = \frac{1232 \times 7}{22} = (56 \times 7) {cm}^{2} \Rightarrow r_{2} = \sqrt{56 \times 7} = \sqrt{7 \times 8 \times 7} = \sqrt{7 \times 7 \times 4 \times 2} = 14 \sqrt{2} cm$}

Hence, the radius of the circle A₂ is 14 $\sqrt{2}$ cm.

Page No 21.16:

Question 14:

The radius of one circluar field is 20 m and that of another is 48 m. Find the radius of the third circular field whose area is equal to the sum of the areas of two fields.

Answer:

Let the area of the circle whose radius is 20 m be A₁ , and the area of the circle whose radius
is 48 m be A₂. Let A₃ be the area of a circle that is equal to the sum of the areas of the two fields, with the radius of its field being r cm.
∴ A₃= A₁ + A₂

A_{1} = π {(20)}^{2} = (\frac{22}{7} \times 20 \times 20) m^{2} = (400 π) m^{2} A_{2} = π (48)^{2} = (\frac{22}{7} \times 48 \times 48) m^{2} = (2304 π) m^{2} A_{3} = A_{1} + A_{2} = (400 π) + (2304 π) = π (400 + 2304) m^{2} \Rightarrow A_{3} = π {(r)}^{2} = π (400 + 2304) m^{2} \Rightarrow {(r)}^{2} = (400 + 2304) m^{2} \Rightarrow r = \sqrt{2704} m = 52 m

Page No 21.16:

Question 15:

The radius of one circular field is 5 m and that of the other is 13 m. Find the radius of the circular field whose area is the difference of the areas of first and second field.

Answer:

Let the area of the circular field whose radius is 5 m be A₁ , and the area of the circular field whose radius is 13 m be A₂. Let A₃and r cm be the area and radius of the circular field, that is equal to the difference of the areas of the two fields.
∴ A₃= A₂- A₁

A_{1} = π {(5)}^{2} = (25 π) m^{2} A_{2} = π (13)^{2} = (169 π) m^{2} A_{3} = A_{2} - A_{1} = (169 π) - (25 π) = 144 π m^{2} \Rightarrow A_{3} = π {(r)}^{2} = 144 π m^{2} \Rightarrow {(r)}^{2} = 144 m^{2} \Rightarrow r = \sqrt{144} m = 12 m

Hence, the radius of the circular field is 12 m.

Page No 21.16:

Question 16:

Two circles are drawn inside a big circle with diameters $\frac{2}{3} rd$ and $\frac{1}{3} rd$ of the diameter of the big circle as shown in Fig. 18. Find the area of the shaded portion, if the length of the diameter of the circle is 18 cm.

Answer:

Let the left circle be denoted as the 1st circle and the right circle be denoted as the 2nd circle.
Diameter of the big circle = 18 cm
Radius of the big circle = 9 cm
Diameter of the 1st circle = $\frac{2}{3} \times 18 = 12 cm$
Radius of the 1st circle = 6 cm
Diameter of the 2nd circle = $\frac{1}{3} \times 18 = 6 cm$
Radius of the 2nd circle = 3 cm
Area of the 1st circle = $π (6)^{2} = 36 {πcm}^{2}$
Area of the 2nd circle = $π (3)^{2} = π \times 3 \times 3 = 9 π {cm}^{2}$
Area of the big circle = $π (9)^{2} = π \times 9 \times 9 = 81 π {cm}^{2}$
Area of the shaded portion = Area of the big circle - (Area of the I^st circle + Area of the II^nd circle)
Area of the shaded portion = $81 π - (36 π + 9 π) = 36 π$ cm².

Page No 21.16:

Question 17:

In Fig. 19, the radius of quarter circular plot taken is 2 m and radius of the flower bed is 2 m. Find the area of the remaining field.

Answer:

Radius of the quarter circular plot = 2 m
Area of the quarter circular plot = $π (2)^{2} = \frac{22}{7} \times 4 = = 12.57 m^{2}$
Radius of each flower bed = 2 m
Area of four flower beds = $4 \times \frac{1}{4} \times π (2)^{2} = 12.57 m^{2}$
Area of the rectangular region = Length $\times$ Breadth
Area of the rectangular region = 8 $\times$ 6 = 48 m²
Area of the remaining field = Area of the rectangular region - (Area of the quarter circle + Area of the four flower beds)

Area of the remaining field = [48 - (12.57 + 12.57)] m² = 22.86 m²^.

Page No 21.16:

Question 18:

Four equal circles, each of radius 5 cm, touch each other as shown in Fig. 20. Find the area included between them. (Take π = 3.14).

Answer:

Side of the square = 10 cm
Area of the square = Side $\times$ Side
Area of the square =^{$10 \times 10 = 100 {cm}^{2}$}

Area of the four quarter circles = ^{$4 \times \frac{1}{4} \times \frac{22}{7} \times 5^{2} = 78.57 {cm}^{2}$}
Area included in them = Area of the square - Area of the four quarter circles
Area included in them = ( 100 - 78.57 ) cm² = 21.43 cm² .

Page No 21.16:

Question 19:

The area of circle is 100 times the area of another circle. What is the ratio of their circumferences?

Answer:

Let the area of the first circle be A₁, the circumference be C₁and the radius be r₁.
Let the area of the second circle be A₂_,the circumference be C₂and the radius be r₂.

Thus,

$C_{1} : C_{2} = 2 π r_{1} : 2 π r_{2} \Rightarrow \frac{C_{1}}{C_{2}} = \frac{2 π r_{1}}{2 {πr}_{2}} = \frac{r_{1}}{r_{2}}$

We know that :

A₁= 100A₂
_{∴ $\Rightarrow ({πr}_{1}^{2}) = 100 \times ({πr}_{2}^{2}) \Rightarrow {r_{1}}^{2} = 100 \times {r_{2}}^{2} \Rightarrow r_{1} = 10 \times r_{2} \Rightarrow \frac{r_{1}}{r_{2}} = 10$}

Substituting the values, we get:
∴

\frac{C_{1}}{C_{2}} = \frac{r_{1}}{r_{2}} = \frac{10}{1} C_{1} : C_{2} = 10 : 1

Hence, the ratio of their circumferences is 10:1.

Page No 21.17:

Question 1:

The ratio of the perimeter (circumference) and diameter of a circle is

(a) $π$ (b) 2 $π$ (c) $\frac{π}{2}$ (d) $\frac{π}{4}$

Answer:

Let r be the radius of the circle. Then
Perimeter of circle = 2 $π$ r
Diameter of circle = 2r
Now
$\frac{Perimeter of circle}{Diameter of circle} = \frac{2 π r}{2 r} = π$
Thus, the required ratio is $π$ .
Hence, the correct option is (a)

Page No 21.17:

Question 2:

The ratio of the area and circumference of a circle of diameter d is

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

Answer:

Let r and d be respectively the radius and diameter of the circle. Then
d = 2r
Circumference of circle = 2 $π$ r = $2 π \times \frac{d}{2} = π d$
Area of circle = $π$ r² = $π {(\frac{d}{2})}^{2} = \frac{π d^{2}}{4}$
Now
$\frac{Area of circle}{Circumference of circle} = \frac{\frac{π d^{2}}{4}}{π d} = \frac{d}{4}$
Hence, the correct option is (c)

Page No 21.17:

Question 3:

The cost of fencing a circular garden of radius 21 m at ₹10 per metre is

(a) ₹1320 (b) ₹132 (c) ₹1200 (d) ₹660

Answer:

Radius (r) = 21 m
Cost per metre = ₹10
Circumference of circle = $2 π r = 2 \times \frac{22}{7} \times 21 = 132 m$
Cost of fencing = Circumference $×$ Cost per metre
= 132 $×$ ₹10
= ₹1320
Hence, the correct option is (a).

Page No 21.17:

Question 4:

If the diameter of a circle is equal to the diagonal of a square, then the ratio of their areas is

(a) 7 : 1 (b) 1 : 1 (c) 11 : 7 (d) 22 : 7

Answer:

Let r and a be the diameter of the circle and side of the square respectively. Then
Diameter of circle = 2r
Diagonal of square = $a \sqrt{2}$
Now, as per the question
Diameter of circle = Diagonal of square
2r = $a \sqrt{2}$ $\Rightarrow a = \sqrt{2} r$
Therefore
$\frac{Area of circle}{Area of square} = \frac{π r^{2}}{a^{2}} = \frac{\frac{22}{7} \times r^{2}}{{(\sqrt{2} r)}^{2}} = \frac{\frac{22}{7} \times r^{2}}{2 r^{2}} = \frac{11}{7}$
Hence, the correct option is (c).

Page No 21.17:

Question 5:

A circle is inscribed in a square of side 14 m. The ratio of the area of the circle and that of
the square is

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

Answer:

Let a and r be the side of the square and radius of the circle respectively.
Here, the diameter of the circle is equal to the side of the square. So
Diameter of circle = 2r = a
Therefore
$\frac{Area of circle}{Area of square} = \frac{π r^{2}}{a^{2}} = \frac{π \times r^{2}}{{(2 r)}^{2}} = \frac{π}{4}$
Hence, the correct option is (b).

Page No 21.17:

Question 6:

How many times should a wheel of radius 7 m rotate to go around the perimeter
of a rectangular field of length 60 m and breadth 50 m?

(a) 3 (b) 4 (c) 5 (d) 6

Answer:

Here, Radius (r) = 7 m, Length (l) = 60 m and Breadth (b) = 50 m.
Perimeter of circle = $2 π r = 2 \times \frac{22}{7} \times 7 = 44 m$
Perimeter of rectangle = $2 (l + b) = 2 (60 + 50) = 220 m$
Therefore
Number of turns = $\frac{Perimeter of rectangle}{Perimeter of circle} = \frac{220}{44} = 5$
Hence, the correct option is (c).

Page No 21.17:

Question 7:

The minute hand of a clock is 14 cm long. How far does the tip of the minute hand
move in 60 minutes?
(a) 22 cm (b) 44 cm (c) 33 cm (d) 88 cm

Answer:

Length of minute hand = 14 cm
Distance covered by minute hand in one round = $2 π r = 2 \times \frac{22}{7} \times 14 = 88 cm$
Thus, the minute hand move 88 cm in 60 minutes.
Hence, the correct option is (d).

Page No 21.17:

Question 8:

The cost of fencing a semi-circular garden of radius 14 m at ₹10 per metre is

(a) ₹1080 (b) ₹1020 (c) ₹700 (d) ₹720

Answer:

Radius of circle (r) = 14 m
Perimeter of semi-circular garden
$= π r + 2 r = \frac{22}{7} \times 14 + 2 \times 14 = 44 + 28 = 72 m$
Cost of fencing = 72 $×$ ₹10 = ₹720
Hence, the correct option is (d).

Page No 21.17:

Question 9:

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

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

Answer:

Let a and r be respectively the side of the square and radius of the circle.
Here, the area of square is equal to the area of the circle. So
$a^{2} = π r^{2} \Rightarrow \frac{a^{2}}{r^{2}} = π \Rightarrow \frac{a}{r} = \sqrt{π}$
Hence, the correct option is (a).

Page No 21.17:

Question 10:

If A is the area and C be the circumference of a circle, then its radius is
(a) $\frac{A}{C}$ (b) $\frac{2 A}{C}$ (c) $\frac{3 A}{C}$ (d) $\frac{4 A}{C}$

Answer:

Let r be the radius of the circle. Then
$A = π r^{2} a n d C = 2 π r \Rightarrow \frac{A}{C} = \frac{π r^{2}}{2 π r} \Rightarrow \frac{A}{C} = \frac{r}{2} \Rightarrow r = \frac{2 A}{C}$
Hence, the correct option is (b).

Page No 21.17:

Question 11:

The area of a circle of circumference C is

(a) $\frac{C^{2}}{4 π}$ (b) $\frac{C^{2}}{2 π}$ (c) $\frac{C^{2}}{π}$ (d) $\frac{4 C^{2}}{π}$

Answer:

Let r be the radius of the circle. Then
$C = 2 π r \Rightarrow r = \frac{C}{2 π}$
Therefore
Area of circle = $π r^{2} = π {(\frac{C}{2 π})}^{2} = \frac{C^{2}}{4 π}$
Hence, the correct option is (a).

Page No 21.17:

Question 12:

The circumference of a circle is 44 cm. Its area is

(a) 77 cm² (b) 154 cm² (c) 208 cm² (d) 144 cm²

Answer:

Let r be the radius of the circle. Then
$44 = 2 π r \Rightarrow r = \frac{44}{2 \times \frac{22}{7}} = 7 cm$
Therefore
Area of circle = $π r^{2} = \frac{22}{7} \times 7^{2} = 154 {cm}^{2}$
Hence, the correct option is (b).

Page No 21.17:

Question 13:

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) $\frac{7 \sqrt{3}}{4}$ cm² (b) $\frac{49 \sqrt{3}}{2}$ cm² (c) $\frac{49 \sqrt{3}}{4}$ cm² (d) $\frac{7 \sqrt{3}}{2}$ cm²

Answer:

Let r be the radius of the circle and a be the side of the equilateral triangle. Then
Area of circle = 154 cm²
$\Rightarrow 154 = \frac{22}{7} \times r^{2} \Rightarrow r = \sqrt{\frac{154 \times 7}{22}} = 7 cm$
Therefore
Area of equilateral triangle = $\frac{a^{2} \sqrt{3}}{4} = \frac{7^{2} \sqrt{3}}{4} = \frac{49 \sqrt{3}}{4} (∵ a = r = 7 cm)$
Hence, the correct option is (c).

Page No 21.18:

Question 14:

The area of a circle is 9 $π$ cm². Its circumference is

(a) 6 $π$ cm (b) 36 $π$ cm (c) 9 $π$ cm (d) 36 $π^{2}$ cm

Answer:

Let r be the radius of the circle. Then
Area of circle = 9 $π$ cm²
$\Rightarrow π r^{2} = 9 π \Rightarrow r = 3 cm$
Therefore
Circumference of the circle = $2 π r = 2 π \times 3 = 6 π cm$
Hence, the correct option is (a).

Page No 21.18:

Question 15:

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 cm (c) 4 cm (d) 3.5 cm

Answer:

Let r be the radius of the circle. Then
Area of original circle = $π$ r² cm²
Radius of circle after increment = (r + 1) cm
Thus,as per the question
$π {(r + 1)}^{2} - π r^{2} = 22 \Rightarrow {(r + 1)}^{2} - r^{2} = \frac{22}{\frac{22}{7}} = 7 \Rightarrow r^{2} + 2 r + 1 - r^{2} = 7 \Rightarrow r = 3 cm$
Thus, the original radius of the circle is 3 cm.
Hence, the correct option is (b).

Page No 21.18:

Question 16:

The radii of two circles are in the ratio 2 : 3. The ratio of their areas is

(a) 2 : 3 (b) 4 : 9 (c) 3 : 2 (d) 9 : 4

Answer:

Let r₁ and r₂ be the radius of the two circles. So
$\frac{r_{1}}{r_{2}} = \frac{2}{3}$
Now
$Ratio of areas = \frac{π {r_{1}}^{2}}{π {r_{2}}^{2}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9}$
Thus, the required ratio is 4 : 9.
Hence, the correct option is (b).

Page No 21.18:

Question 17:

The areas of two circles are in the ratio 49 : 36. The ratio of their circumferences is

(a) 7 : 6 (b) 6 : 7 (c) 3 : 2 (d) 2 : 3

Answer:

Let r₁ and r₂ be the radius of the two circles. Then
$\frac{π {r_{1}}^{2}}{π {r_{2}}^{2}} = \frac{49}{36} \Rightarrow {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{7}{6})}^{2} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{7}{6}$
Now
$Ratio of circumferences = \frac{2 π r_{1}}{2 π r_{2}} = \frac{r_{1}}{r_{2}} = \frac{7}{6}$
Thus, the required ratio is 7 : 6.
Hence, the correct option is (a).

Page No 21.18:

Question 18:

The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is

(a) 3 : 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9

Answer:

Let r₁ and r₂ be the radius of the two circles. Then
$\frac{2 π r_{1}}{2 π r_{2}} = \frac{3}{4} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{3}{4}$
Now
$Ratio of areas = \frac{π {r_{1}}^{2}}{π {r_{2}}^{2}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{3}{4})}^{2} = \frac{9}{16}$
Thus, the required ratio is 9 : 16.
Hence, the correct option is (c).

Page No 21.18:

Question 19:

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) 258 cm²

Answer:

Let r₁ and r₂ be the radius of the two circles. Then
$2 π r - r = 37 \Rightarrow 2 \times \frac{22}{7} \times r - r = 37 \Rightarrow \frac{44 r - 7 r}{7} = 37 \Rightarrow r = \frac{37 \times 7}{37} = 7 cm$
Now
$Area of circle = π r^{2} = \frac{22}{7} \times 7 \times 7 = 154 {cm}^{2}$
Hence, the correct option is (c).

Page No 21.7:

Question 1:

Find the circumference of a circle whose radius is
(i) 14 cm
(ii) 10 m
(iii) 4 km

Answer:

(i) The circumference C of a circle of radius 'r' is given by:
C = 2 $π$ r
Here, r = 14 cm
∴ C = $2 \times \frac{22}{7} \times 14 = 88 cm$ .
(ii) The circumference C of a circle of radius 'r' is given by:
C = 2 $π$ r
Here, r = 10 m
∴ C = $2 \times \frac{22}{7} \times 10 = 62.86 m$ .
(iii) The circumference C of a circle of radius 'r' is given by:
C = 2 $π$ r
Here, r = 4 km
∴ C = $2 \times \frac{22}{7} \times 4 = 25.142 km$ .

Page No 21.7:

Question 2:

Find the circumference of a circle whose diameter is
(i) 7 cm
(ii) 4.2 cm
(iii) 11.2 km

Answer:

(i) We have:
Diameter = 7 cm
∴ Radius = $\frac{7}{2}$ cm
Let C be the circumference of a circle. Then,
C = 2 $π$ r
∴ C = $2 \times \frac{22}{7} \times \frac{7}{2} = 22 cm$ .
(ii) We have:
Diameter = 4.2 cm
∴ Radius = $\frac{4.2}{2} cm = 2.1 cm$
Let C be the circumference of a circle. Then,
C = 2 $π$ r
∴ C = $2 \times \frac{22}{7} \times 2.1 = 13.2 cm$ .
(iii) We have:
Diameter = 11.2 km
∴ Radius = $\frac{11.2}{2} km = 5.6 km$
Let C be the circumference of a circle. Then,
C = 2 $π$ r
∴ C = $2 \times \frac{22}{7} \times 5.6 = 35.2 km$ .

Page No 21.7:

Question 3:

Find the radius of a circle whose circumference is
(i) 52.8 cm
(ii) 42 cm
(iii) 6.6 km

Answer:

(i) We know that the circumference C of a circle of radius 'r' is given by:
  C = 2 $π$ r
   $\Rightarrow r = \frac{C}{2 π} Here, C = 52.8 cm \Rightarrow r = \frac{52.8 \times 7}{2 \times 22} cm = \frac{369.6}{44} cm = 8.4 cm .$ .

(ii) We know that the circumference C of a circle of radius 'r' is given by:
C = 2 $π$ r
   $\Rightarrow r = \frac{C}{2 π} Here, C = 42 cm \Rightarrow r = \frac{42 \times 7}{2 \times 22} cm = \frac{294}{44} cm = 6.68 cm .$

(iii) We know that the circumference C of a circle of radius 'r' is given by:
  C = 2 $π$ r
   $\Rightarrow r = \frac{C}{2 π} Here, C = 6.6 km \Rightarrow r = \frac{6.6 \times 7}{2 \times 22} km = \frac{2.1}{2} km = 1.05 km .$

Page No 21.7:

Question 4:

Find the diameter of a circle whose circumference is
(i) 12.56 cm
(ii) 88 m
(iii) 11.0 km

Answer:

(i) We know that the circumference C of a circle of diameter 'd' is given by:
C = $π$ d
$\Rightarrow d = \frac{C}{π} Here, C = 12.56 cm \Rightarrow d = \frac{12.56 \times 7}{22} cm = \frac{87.92}{22} cm = 3.99 cm .$

(ii) We know that the circumference C of a circle of diameter 'd' is given by:
C = $π$ d
$\Rightarrow d = \frac{C}{π} Here, C = 88 m \Rightarrow d = \frac{88 \times 7}{22} m = 28 m .$

(iii) We know that the circumference C of a circle of diameter 'd' is given by:
C = $π$ d
$\Rightarrow d = \frac{C}{π} Here, C = 11 km \Rightarrow d = \frac{11 \times 7}{22} km = \frac{7}{2} km = 3.5 km .$

Page No 21.7:

Question 5:

The ratio of the radii of two circles is 3 : 2. What is the ratio of their circumferences?

Answer:

We have, the ratio of the radii = 3:2
So, let the radii of the two circles be 3r and 2r, respectively.
Let C₁ and C₂ be the circumferences of the two circles of radii 3r and 2r, respectively. Then,
C₁= 2 $π$ x 3r = 6 $π$ r, and C₂ = 2 $π$ x2r = 4 $π$ r
∴
$\frac{C_{1}}{C_{2}} = \frac{6 π r}{4 π r} = \frac{6}{4} = \frac{3}{2}$ , or

C₁: C₂ = 3:2.

Page No 21.7:

Question 6:

A wire in the form of a rectangle 18.7 cm long and 14.3 cm wide is reshaped and bent into the form of a circle. Find the radius of the circle so formed.

Answer:

Length of the wire = Perimeter of the rectangle
= 2 ( l +b ) = 2 x (18.7 + 14.3) = 66 cm

Let the wire be bent in the form of a circle of radius r cm. Then,

circumference = 66 cm

$\Rightarrow 2 πr = 66 cm \Rightarrow 2 \times \frac{22}{7} \times r = 66 cm \Rightarrow r = \frac{66 \times 7}{22 \times 2} = \frac{21}{2} cm = 10.5 cm .$

Page No 21.7:

Question 7:

A piece of wire is bent in the shape of an equilateral triangle of each side 6.6 cm. it is re-bent to form a circular ring. What is the diameter of the ring?

Answer:

We have:
Length of the wire = The perimeter of the equilateral triangle

= 3 x side = 3 x 6.6 = 19.8 cm.

Let the wire be bent to form a circular ring of radius r cm.Then,

Circumference = 19.8 cm

\Rightarrow 2 πr = 19.8 cm \Rightarrow 2 \times \frac{22}{7} \times r = 19.8 cm \Rightarrow r = \frac{19.8 \times 7}{22 \times 2} = \frac{138.6}{44} cm = 3.15 cm .

So, the diameter of the ring = 2 x 3.15 = 6.30 cm.

Page No 21.7:

Question 8:

The diameter of a wheel of a car is 63 cm. Find the distance travelled by the car during the period, the wheel makes 1000 revolutions.

Answer:

It may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.
Now, the diameter of the wheel = 63 cm
∴ Circumference of the wheel = $πd = (\frac{22}{7} \times 63) cm = 198 cm$ .
Thus, the cycle covers 198 cm in one revolution.
∴ The distance covered by the cycle in 1000 revolutions = (198 x 1000 ) cm = 198000 cm = 1980 m.

Page No 21.7:

Question 9:

The diameter of a wheel of a car is 98 cm. How many revolutions will it make to travel 6160 metres.

Answer:

We have:
Diameter of the wheel of the car = 98 cm
∴ Circumference of the wheel of the car = $πd = (\frac{22}{7} \times 98) cm = 308 cm$ .
Note that, in one revolution of the wheel, the car travels a distance equal to the circumference of the wheel.

∴ The distance travelled by the car in one revolution of the wheel = 308 cm.
Total distance travelled by the car = 6160 m = 616000 cm.
∴ Number of revolutions = $\frac{616000}{308} = 2000$ .

Page No 21.7:

Question 10:

The moon is about 384400 km from the earth and its path around the earth is nearly circular. Find the circumference of the path described by the moon in lunar month.

Answer:

We have:
The radius of the path described by the moon around the earth = 384400 km
∴ The circumference of the path described by the moon C = $2 π r = 2 \times (\frac{22}{7} \times 384400) km = 2416228.57 km$ .

Page No 21.7:

Question 11:

How long will John take to make a round of a circular field of radius 21 m cycling at the speed of 8 km/hr?

Answer:

We have:
The radius of the circular field = 21 m
∴ Circumference of the circular field = $2 π r = 2 \times (\frac{22}{7} \times 21) m = 132 m$ .
If John cycles at the speed of 8 km/hr (In 1 hour John covers 8 km = 8000 m ), then,
John covers 8000 m in 1 hour.

∴ Time required to cover 132 m = $\frac{132}{8000} = 0.0165 hours$
1 hour = 3600 seconds
∴ 0.0615 hours = 0.0615 x 3600 = 59.4 seconds.

Page No 21.7:

Question 12:

The hour and minute hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled by their tips in 2 days.

Answer:

The radius of the path inscribed by the hour hand = Length of the hour hand = 4 cm
The radius of the path inscribed by the minute hand = Length of the minute hand = 6 cm

The circumference of the path inscribed by the hour hand = $2 π r = 2 \times \frac{22}{7} x 4 = \frac{176}{7} cm$ .
The hour hand makes 2 revolutions in one day.
∴ The distance covered by the hour hand in 2 days = $\frac{176}{7} \times 2 \times 2 = 100.57 cm$ .
The distance covered by the minute hand in 1 revolution = $2 πr = 2 \times \frac{22}{7} x 6 = \frac{264}{7} cm$
The minute hand makes 1 revolution in one hour.
∴ In 1 day, it makes 24 revolutions.
In 2 days, it makes 2 x 24 revolutions.
∴ The distance covered by the minute hand in 2 days = $(2 \times 24 \times \frac{264}{7}) cm = (\frac{12672}{7}) cm = 1810.28 cm$
The sum of the distances travelled by the hour and minute hands in 2 days = 1810.28 + 100.57 = 1910.85 cm.

Page No 21.7:

Question 13:

A rhombus has the same perimeter as the circumference of a circle. If the side of the rhombus is 2.2 m, find the radius of the circle.

Answer:

We have:

The side of a rhombus = 2.2 m
Let C be the circumference of a circle having a radius r cm. Then,

the perimeter of the rhombus = 4

\times side

= 4 x 2.2 = 8.8 m.

We know:

Perimeter of the rhombus = Circumference of the circle

\Rightarrow 8.8 m = 2 πr \Rightarrow r = (\frac{8.8}{2 π}) m = (\frac{8.8 \times 7}{2 \times 22}) m = (\frac{88 \times 7}{2 \times 22 \times 10}) m \Rightarrow r = (\frac{2 \times 7}{10}) m = 1.4 m .

The radius of the circle is 1.4 m.

Page No 21.7:

Question 14:

A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of the side of the square.

Answer:

We have:

The radius of the circle = 28 cm
∴ Circumference of the circle =

2 π r = 2 \times \frac{22}{7} \times 28 = 176 cm

.
Let a cm be the side of the square. Then,

the circumference of the circle = the perimeter of the square

\Rightarrow 176 = 4 \times a \Rightarrow a = (\frac{176}{4}) cm = 44 cm .

The side of the square is 44 cm.

Page No 21.7:

Question 15:

A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

Answer:

We have:

Total distance covered in 5000 revolutions = 11 km = 11,000 m
∴ Distance covered in 1 revolution = $\frac{11, 000}{5000} = (\frac{11}{5}) m$ .
Distance covered in 1 revolution = Circumference of the wheel
$(\frac{11}{5}) = πd \Rightarrow d = (\frac{11}{5 \times π}) = (\frac{11 \times 7}{5 \times 22}) m \Rightarrow d = (\frac{7}{10}) = 0.7 m .$
Thus, the diameter of the wheel is 0.7 m = 70 cm.

Page No 21.7:

Question 16:

A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, calculate the speed per hour with which the boy is cycling.

Answer:

We have:
The diameter of the wheel = 60 cm
Distance covered by the wheel in 1 revolution = Circumference of the wheel
∴ Distance covered by the wheel in 1 revolution = $πd = (\frac{22}{7} \times 60) cm .$
∴ Distance covered in 140 revolutions = $(\frac{22}{7} \times 60 \times 140) cm = (\frac{184800}{7}) cm = 26400 cm$ .

Thus, the wheel covers 26400 cm in 1 minute. Then,

Speed = (\frac{26400}{100} \times 60) m / hr = (264 \times 60) m / hr \Rightarrow Speed = (\frac{264 \times 60}{1000}) km / hr = 15.84 km / hr .

The speed with which the boy is cycling is 15.84 km/hr.

Page No 21.7:

Question 17:

The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km per hour?

Answer:

We have:
Diameter of the wheel = 140 cm, desired speed of the bus = 66 km/hr
∴ Distance covered in 1 revolution = The circumference of the wheel = $πd$ = $(\frac{22}{7} \times 140) cm = 440 cm$ .
Now, the desired speed of the bus = 66 km/hr = $\frac{66 \times 1000 \times 100}{60} = 1, 10, 000 cm / \min$ .
∴ Number of revolutions per minute = $\frac{110000}{440} = \frac{1000}{4} = 250$ .

Thus, the bus must make 250 revolutions per minute to keep the speed at 66 km/hr.

Page No 21.8:

Question 18:

A water sprinkler in a lawn sprays water as far as 7 m in all directions. Find the length of the outer edge of wet grass.

Answer:

The wet grass forms a circular region of radius 7 m.
∴ The length of the outer edge of the wet grass is $2 πr = (2 \times \frac{22}{7} \times 7) m = 44 m$ .

Page No 21.8:

Question 19:

A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of the parapet is 660 cm, then find the width of the parapet.

Answer:

We have :
Diameter of the well = 150 cm
Length of the outer edge of the parapet = 660 cm
Width of the parapet = ?
Radius of well = 150 $\div$ 2 = 75 cm.
Let the width of the stone parapet be x cm.Clearly, the outer edge of the parapet forms a circular region of radius (x +75 cm). Therefore,

$660 cm = 2 \times \frac{22}{7} \times (x + 75) \Rightarrow \frac{660 \times 7}{2 \times 22} = (x + 75) \Rightarrow 15 \times 7 = (x + 75) \Rightarrow x = 105 - 75 \Rightarrow x = 30 cm .$
Thus, the width of the parapet is 30 cm.

Page No 21.8:

Question 20:

An ox in a kolhu (an oil processing apparatus) is tethered to a rope 3 m long. How much distance does it cover in 14 rounds?

Answer:

We have :
Radius of the circular path traced by the ox in a kolhu = 3 m
Distance covered by the ox in 1 round = Circumference of the circular path = $2 πr = (2 \times \frac{22}{7} \times 3) m$ .
∴ Distance covered in 14 rounds = $(2 \times \frac{22}{7} \times 3) m \times 14 = 22 \times 12 = 264 m$ .

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