The PDF of RD Sharma Solutions for Class 7 Exercise 21.2 of Chapter 21 Mensuration – II (Area of Circle) is given below. Learners can access and download the PDF for free. The solutions to this exercise come with detailed explanations structured by our expert teachers that further makes learning and understanding of concepts an easy task. RD Sharma Solutions for Class 7 textbook includes the concepts in a detailed and step by step way to help the students understand effectively. This exercise includes the area of the circle.

## Download the PDF of RD Sharma Solutions For Class 7 Chapter 21 – Mensuration – II (Area of Circle) Exercise 21.2

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 21 – Mensuration – II (Area of Circle) Exercise 21.2

**1. Find the area of a circle whose radius is**

**(i) 7 cm**

**(ii) 2.1 m**

**(iii) 7 km**

**Solution:**

(i) Given radius = 7 cm

We know that area of circle = Ï€ r^{2}

A = (22/7) Ã— 7^{2}

A = 22 Ã— 7

A = 154 cm^{2}

(ii) Given radius = 2.1 m

We know that area of circle = Ï€ r^{2}

A = (22/7) Ã— (2.1)^{2}

A = 22/7 Ã— 4.41

A = 13.86 m^{2}

(iii) Given radius = 7 km

We know that area of circle = Ï€ r^{2}

A = (22/7) Ã— 7^{2}

A = 22 Ã— 7

A = 154 km^{2}

**2. Find the area of a circle whose diameter is**

**(i) 8.4 cm**

**(ii) 5.6 m**

**(iii) 7 km**

**Solution:**

(i) Given diameter = 8.4 cm

Therefore radius, r = d/2 = 8.4/2 = 4.2

We know that area of circle = Ï€ r^{2}

A = (22/7) Ã— (4.2)^{2}

A = 55.44 cm^{2}

(ii) Given diameter = 5.6 cm

Therefore radius, r = d/2 = 5.6/2 = 2.8

We know that area of circle = Ï€ r^{2}

A = (22/7) Ã— (2.8)^{2}

A = 24.64 cm^{2}

(iii) Given diameter = 7 km

Therefore radius, r = d/2 = 7/2 = 3.5

We know that area of circle = Ï€ r^{2}

A = (22/7) Ã— (3.5)^{2}

A = 38.5 km^{2}

**3. The area of a circle is 154 cm ^{2}. Find the radius of the circle.**

**Solution:**

Given area of the circle = 154 cm^{2}

A = Ï€ r^{2}

154 = 22/7 r^{2}

r^{2} = (154 Ã— 7)/22

r^{2} = 49

r = 7 cm

**4. Find the radius of a circle, if its area is**

**(i) 4 Ï€ cm ^{2}**

**(ii) 55.44 m ^{2}**

**(iii) 1.54 km ^{2}**

**Solution:**

(i) We know that area of circle = Ï€ r^{2}

Given area = 4 Ï€ cm^{2}

A = Ï€ r^{2}

4 Ï€ = Ï€ r^{2}

r^{2 }= 4

Therefore r = 2 cm

(ii) We know that area of circle = Ï€ r^{2}

Given area = 55.44 cm^{2}

A = Ï€ r^{2}

55.44 = Ï€ r^{2}

r^{2} = (55.44 Ã— 7)/22

r^{2} = 17.64 m

r = 4.2 m

(iii) We know that area of circle = Ï€ r^{2}

Given area = 1.54 km^{2}

A = Ï€ r^{2}

1.54 = Ï€ r^{2}

r^{2} = (1.54 Ã— 7)/22

r^{2} = 0.49 km

r = 0.7 km

r = 700m

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

**Solution:**

Given circumference = 3.14m

We know that circumference of circle = 2 Ï€ r

3.14 = 2 Ã— 3.14 Ã— r

r = 3.14/ (2 Ã— 3.14)

r = 0.5

We know that area of circle = Ï€ r^{2}

A = (22/7) Ã— (0.5)^{2}

A = 0.785 m^{2}

**6. If the area of a circle is 50.24 m ^{2}, find its circumference.**

**Solution:**

Given area of a circle is 50.24 m^{2}

We know that area of circle = Ï€ r^{2}

50.24 = (22/7) Ã— r^{2}

r^{2} = (50.24 Ã— 7)/22

r^{2} = 15.985

r = 3.998 m

We know that circumference of circle = 2 Ï€ r

C = 2 Ã— (22/7) Ã— 3.998

C = 25.12 m

**7. A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (TakeÂ Ï€Â = 22 / 7).**

**Solution:**

Given the length of the string = 28m

The area over which the horse can graze is the same as the area of circle of radius 28 m

Hence required area = Ï€ r^{2}

A = (22/7) Ã— (28)^{2}

A = 2464 m^{2}

**8. A steel wire when bent in the form of a square encloses an area of 121 cm ^{2}. If the same wire is bent in the form of a circle, find the area of the circle.**

**Solution:**

Given area of the square = 121 cm^{2}

(Side) ^{2} = 121

Therefore side = 11 cm

We know that the perimeter of the square = 4 (side)

= 4 (11)

= 44 cm

According to the question circumference of the circle = perimeter of the square

So let r be the radius of the circle

2 Ï€ r = 44

2 Ã— (22/7) Ã— r = 44

Therefore r = 7 cm

We know that area of the circle = Ï€ r^{2}

A = (22/7) Ã— (7)^{2}

A = 154 cm^{2}

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

**Solution:**

Given circumference of park = 352 m

But we know that circumference of circle = 2 Ï€ r = 352 m

2 Ã— (22/7) Ã— r = 352

r = (352 Ã— 7)/ 44

r = 56 m

Radius of the path including the 7m wide road = (r + 7) = 56 + 7 = 63m

Therefore area of the road = Ï€ Ã— (63)^{2} – Ï€ Ã— (56)^{2}

A = 22/7 Ã— 63 Ã— 63 â€“ (22/7) Ã— 56 Ã— 56

A = 22 (9 Ã— 63 â€“ 8 Ã— 56)

A = 22 (567 â€“ 448)

A = 2618 m^{2}

**10. Prove that the area of a circular path of uniform width h surrounding a circular region of radius r isÂ Ï€h (2r + h).**

**Solution:**

Let radius of circular region = r

Radius of circular path of uniform width h surrounding the circular region of radius,

r = r + h

Therefore area of path = Ï€ (r + h)^{2} â€“ Ï€r^{2}

= Ï€ r^{2} + Ï€ h^{2} + 2 Ï€ r h â€“ Ï€ r^{2}

= Ï€ h (2r + h)

Hence the proof.

**11. The perimeter of a circle isÂ 4Ï€rÂ cm. What is the area of the circle?**

**Solution:**

Given perimeter of circle = 4Ï€rÂ cm

Which can be written as = 2 Ï€ (2r)

We know that the perimeter of a circle = 2 Ï€ r

Therefore we have radius = 2r

We also know that area of circle = Ï€ r^{2}

= Ï€ (2r)^{2}

= 4 Ï€ r^{2} cm^{2}

**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.**

**Solution:**

It is given that,Â Perimeter of the square = 5024 m

â‡’ 4Â Ã— side = 5024

â‡’ Side = 5024/4

â‡’ Side = 1256 m

The same wire is converted into the form of a circle. Therefore,

Circumference of the circle = Perimeter of the square

â‡’ 2Ï€r = 5024

â‡’ 2Â Ã— Ï€ Ã— r = 5024

â‡’ r = 2512/Ï€

We know that area of the square: Area of the circle = (side)^{2}Â :Â Ï€r^{2}

Area of square/ area of circle = (side)^{2}/ Ï€r^{2}

Area of square/ area of circle = (1256 Ã— 1256)/ [Ï€ Ã— (2512/ Ï€) Ã— (2512/ Ï€)]

= (1256 Ã— 1256 Ã— 22)/ (2512 Ã— 2512 Ã— 7)

= 11/14

Area of the square: Area of the circle = 11: 14

**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.**

**Solution:**

Let the area of the circle whose radius is 14 cm be A_{1}.

We know that area of the circle = Ï€r^{2}

Therefore,

A_{1}Â =Â Ï€ (14)^{2}

Let A_{2}Â and r_{2}Â be the area and radius of the second circle respectively whose area is double the area of circle A_{1}.

A_{2}Â = 2 A_{1}

â‡’Â Ï€ (r_{2})^{2}Â = 2 Ã— Ï€ (14)^{2}

â‡’Â (r_{2})^{2}Â =Â 2 Ã— (14)^{2}

â‡’ r_{2}Â = 14âˆš2 cm

Hence the radius of the circle A_{2}Â isÂ 14âˆš2 cm.

**14. The radius of one circular 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.**

**Solution:**

Let A_{1Â }= theÂ area of the circle whose radius is 20 m [given]

A_{2Â }= theÂ area of the circle whose radius is 48 m [given]

Now we have to find the radius of third circle such that whose area is equal to the sum of areas of two fields.

Hence,

A_{3Â }=Â A_{1}Â + A_{2}

â‡’ Ï€ r^{2}Â =Â Ï€ (20)^{2}Â +Â Ï€ (48)^{2}

â‡’ Ï€ r^{2}Â =Â Ï€ [(20)^{2}Â + (48)^{2}]

â‡’ r^{2}Â = 400Â + 2304

â‡’ r = 52 m

Therefore radius = 52 m

**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.**

**Solution:**

Let A_{1Â }= theÂ area of the circular field whose radius is 5 m [given]

A_{2Â }= theÂ area of the circular field whose radius is 13Â m [given]

Now we have to find the area of circular field such that area is the difference of the areas of first and second field

A_{3Â }=Â A_{2}Â â€“ A_{1}

â‡’ Ï€ r^{2}Â =Â Ï€ (13)^{2}Â â€“ Ï€ (5)^{2}

â‡’ Ï€ r^{2}Â =Â Ï€ (13)^{2}Â â€“ Ï€ (5)^{2}

â‡’ Ï€ r^{2}Â =Â Ï€ [(13)^{2}Â â€“ (5)^{2}]

â‡’ r^{2}Â = 169Â â€“ 25

â‡’ r^{2}Â = 144

â‡’ r = 12 m

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

**16. Two circles are drawn inside a big circle with diametersÂ 2/3rd andÂ 1/3rd 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.**

**Solution:**

It is given that, diameter of the big circle = 18 cm

Radius of the big circle = 9 cm

Area of the big circle, A =Â Ï€ r^{2}Â =Â Ï€ (8)^{2}Â = 81Ï€ cm^{2}

Let d_{1}Â = (2/3)Â Ã— 18 = 12 cm

r_{1}Â = 6 cm

Area of the circle, A_{1}Â =Â Ï€ r^{2}Â =Â Ï€ (6)^{2}Â = 36Ï€ cm^{2}

d_{2}Â = (1/3)Â Ã— 18 = 6 cm

r_{2}Â = 3 cm

Area of the circle, A_{2}Â =Â Ï€ r^{2}Â =Â Ï€ (3)^{2}Â = 9Ï€Â cm^{2}

Area of the shaded portion = A â€“ (A_{1}Â + A_{2})

Area of the shaded portion = 81Ï€ â€“ (36Ï€ â€“ 9Ï€) = 36Ï€Â cm^{2}

**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.**

**Solution:**

Given that Radius of flower bed = 2 m

Area of flower bed = Ï€ r^{2}Â = Ï€ (2)^{2}Â = 4Ï€

Radius of the quarter circular plot = 2 m

Area of the quarter circular plot = (Ï€ r^{2})/4

Area of 4 quarter circular plots = 4 Ã— (Ï€ r^{2})/4

=Â Ï€ r^{2Â }

= Ï€ (2)^{2}Â

= 4Ï€

We know that area of the rectangular region = Length x Breadth

Area of the rectangular region = 8 x 6 = 48 m^{2}

Area of the remaining field = Area of the rectangular region â€“ (Area of 4 quarter circular plots + Area of the flower bed)

Area of the remaining field = 48 â€“ (4Ï€ + 4Ï€)

= 48 â€“ 25.14

= 22.86 m^{2}

**18. Four equal circles, each of radius 5 cm, touch each other as shown in Fig. 20. Find the area included between them. (TakeÂ Ï€Â = 3.14).**

**Solution:**

From the figure we can see that,

Side of the square = 10 cm

We know that area of the square = side x side =Â 10 x 10 = 100 cm^{2}

Radius of theÂ quarter circle = 5 cm

Area of the quarter circle = (Ï€ r^{2})/4

Area of 4 quarter circle = 4 Ã— (Ï€ r^{2})/4

=Â Ï€ r^{2Â }

= 3.14Â Ã— (5)^{2}Â

= 78.5Â cm^{2}

Area included in the quarter circle = Area of the square â€“ Area of the four quarter circles

Area included in them = (100 â€“ 78.5) cm^{2Â }

= 21.5 cm^{2}

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

**Solution:**

Let the area of the circles be A_{1}Â and A_{2}Â and their circumference be c_{1}Â and c_{2Â }respectively.

According to the question it is clear that A_{1}Â = 100 A_{2}

â‡’ Ï€ (r_{1})^{2}Â = 100Â Ã—Â Ï€ (r_{2})^{2}

â‡’Â r_{1}Â = 10 r_{2}

â‡’ r_{1}/r_{2}Â = 10/1 â€¦ (i)

Finding the ratios of the circumference;

C_{1}: C_{2}Â =Â 2Ï€r_{1}:Â Â 2Ï€r_{2}

C_{1}/C_{2}Â = (2Ï€r_{1})/ (2Ï€r_{2})

C_{1}/C_{2}Â = r_{1}/r_{2}

Putting the value of r_{1}/r_{2Â }from equation (i)

C_{1}/C_{2}Â = 10/1

C_{1}: C_{2}Â = 10: 1

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