# Rd Sharma Solutions Class 7 Chapter 21

## Rd Sharma Solutions Class 7 Chapter 21 Mensuration – II Area of circle

Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Rd Sharma Class 7 Mathematics, Chapter 21, Mensuration II (Area of circle). Here students can easily find Exercise wise solution for chapter 21, Mensuration II (Area of circle). Students will find proper solutions for Exercise 21.1 and 21.2. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.

Mensuration – II Area of circle Exercise 21.1 Solution

Question no – (1)

Solution :

(i) 14 cm

In the given question,

We know that,

(Circumference= 2πr)

Circumference,

= 2 × 22/7 × 14

= 88 cm

(ii) 10 m

In the given question,

We know that,

(Circumference = 2πr)

∴ Circumference,

= 2 × 22/7 × 10

=  440/7 m

=  62.86 m

(iii) 4 km

In the given question,

We know that,

(Circumference= 2πr)

∴ Circumference,

= 2 × 22/7 × 4

=  176/7 km

= 25.142 km

Question no – (2)

Solution :

(i) 7 cm

In the given question,

Diameter = 7 cm,

Circumference,

= 2 π r = 2 × 22/7 × 7/2

= 22 cm

(ii) 4.2 cm

Given in the question,

Diameter = 4.2 then

Circumference,

= 2 × 22/7 × 4.2/2

= 2 × 22/7 × 42/2 × 10

= 132/10

= 13.2 cm

(iii) 11.2 km

From the question we get,

Diameter = 11.2 km,

Circumference

= 2 × 22/7 × 11.2/2

= 2 × 22/7 × 112/2 × 10

= 352/10

= 35.2 km

Question no – (3)

Solution :

(i) 52.8 cm

We know, circumference = 2 π r

So, Let radius be = r

2 π r = 52.8 …(given in question)

= r = 528/10 × 7/22 × 1/2

= 84/10

= 8.4 cm

Therefore, the radius of a circle is 8.4 cm.

(ii) 42 cm

We know circumference = 2 π r

Let radius be = r cm

So, 2 π r = 42 …(From the question)

= r = 42/2 × 7/22

= 147/22 cm

= 6.68 cm

Thus, the radius of circle is 6.68 cm.

(ii) 42 cm

We know, circumference = 2 π r

So, 2 π r = 6.6

= r = 66/10 × 7/22 × 1/2

= 1.05

Hence, the radius of a circle is 1.05 km

Question no – (4)

Solution :

(i) 12.56 cm

Given in the question,

Circumference = 12.56 cm

So, 2 π r = 12.56

= r = 1256/100 × 7/22 × 1/2

= 4396/2200

Diameter,

= 4396/200 × 2

= 3.99 cm

Hence, the diameter of a circle is 3.99 cm.

(ii) 88 m

Given in the question,

Circumference = 88 m

So, 2 π r = 88

= r = 88/2 × 7/22

= 14

Diameter,

= 2 × 14

= 28 m

Thus, the diameter of a circle 28 m.

Question no – (5)

Solution :

Let, radii of two circle be 3r and 2r

Circumference of the first circle,

= 2 π (3r)

= 6 π r

Circumference second circle,

= 2 π (2r)

= 4 π r

Ratio of the circumference of the two circle,

= 6 π r/4 π r

= 6/4

= 3 : 2

Question no – (6)

Solution :

Total length of the wire,

= 2 (18.7 + 14.3)

= 2 × 330

= 66 cm

Let radii of the reshaped wire r

So, 2 π r = 66

= r = 66/2 × 7/22

= 21/2

= 10.5 cm

Therefore, the radius of the circle so formed is 10.5 cm.

Question no – (7)

Solution :

Total length of the wire,

= 3 × 6.6

= 20.8 cm

Let the radius of reshaped circle be r

So, 2 π r = 20.8

= r = 208/10 × 7/22 × 1/2

= 33.09/10

= 3.30

Diameter of the circular ring,

= 2 × 3.30

= 6.6

Therefore, the diameter of the ring is 6.6 cm.

Question no – (8)

Solution :

As per the given question,

Diameter = 63

So, radius of the wheel = 63/2

Circumference,

= 2 × 7/22 × 63/2

= 198 cm

The wheel travelled,

= 198 × 1000

= 198000 cm

= 1.98 km

Therefore, the car will cover 1.98 km.

Question no – (9)

Solution :

Given, diameter of the wheel = 98

= 49

Circumference = 2 × π × r

= 2 × 22/7 × 49

= 308 cm

The wheel revolute,

= 6160 m/308 cm

= 6160/3.08

= 2000 times

Therefore, it will make 2000 revolutions to travel 6160 metres.

Question no – (10)

Solution :

According to the given question, ,

Circumference,

= 2 × 22/7 × 384400

= 2416228.57 km

Therefore, the circumference of the path is 2416228.57 km.

Question no – (11)

Solution :

As per the given question,

Circular field of radius 21 m

Circumference

= 2 × 22/7 × 21

= 132 m

Cycling speed = 8 km/hr

= 8000/3600 m/s

Then time take by John,

= 132 800/3600

= 132 × 3600/8000

= 66 × 9/10

= 594/10

= 59.4 s

Therefore, the John will take 59.4m/s to make a round of a circular field.

Question no – (12)

Solution :

According to the given question,

Radii of hour hand = 4 cm

and radii of minute hand = 6 cm

Circumference of hour hand,

= 2 × 22/7 × 4

Circumference of minute hand,

= 2 × 22/7 × 6

Total distance coves by two hands in 2 days,

= 2 × (× 22/7 × 4 + 2 × 22/7 × 6)

= 2 × 22/7 (4 + 6)

= 4 × 22/7 × 10

= 1910.8 cm

Therefore, the sum of the distances travelled 1910.8 cm.

Question no – (13)

Solution :

According to the given question,

Side of rhombus = 2.2 m

Perimeter of a rhombus,

= 2.2 + 2.2 + 2.2 + 2.2

= 8.8 m

Let, radius of the circle = r m

2 π r = 8.8

= r = 8.8/2 × 7/22

= 88 × 7/10 × 2 × 22

= 14/10

= 1.4

Therefore, the radius of the circle will be 1.4 m

Question no – (14)

Solution :

Circumference of the circle,

= 2 × 22/7 × 28

= 176 cm

So, Let side be x

Circumference of square,

= 4 × x

= 4x = 176

= x = 44

Hence, the length of the side of the square is 44 cm.

Question no – (15)

Solution :

As per the given question,

Moving = 11 km

Bicycle wheel makes = 5000 revolutions

Distance travelled in one revolution,

= 11 km/5000

= 11000/5000

= 11/5

Circumference is equal to distance travelled in one revolution

So, let radius be = r

2 π r = 11/5

= r = 11/5 × 7/22 × 1/2

= r = 7/

Diameter,

= 7/22 × 2

= 0.7 m

Therefore, the diameter of the wheel is 0.7 m.

Question no – (16)

Solution :

According to given question,

Diameter of wheel = 60 cm

= 30

Circumference of wheel,

= 2 × 7/22 × 30

= 1320/7 cm

Wheel travel’s by in one minute,

= 1320/7 × 140

= 26400 cm

Wheel travel’s in one hour,

= 26400 × 60

= 1584000 cm/hr

= 15.84 km/hour

Therefore, the speed per hour the boy is cycling is 15.84 km/hour.

Question no – (17)

Solution :

Diameter of wheel = 140

= 70

Circumference of the wheel,

= 2 × 22/7 × 70

= 440 cm

Wheel speed = 66 km/hr

= 6600000/60 cm/min

Revolution per minute,

= 6600000/60/440

= 6600000/60 × 440

= 1500/6

= 250

Therefore, the wheel need to make 250 revaluation per minute.

Question no – (18)

Solution :

Given in the question,

Radius of the water sprinkler = 7 m

Circumference,

= 2 × 22/7 × 7

= 44 m

Therefore, the length of outer edge of wet grass is 44 m.

Question no – (19)

Solution :

From the question we get,

Diameter of the well = 150 cm

= 75 cm

Let, radius of parapet = r

Circumference = 2 × 22/7 r

= 2 × 22/7 r = 660

= r = 660/22 × 7/2

= r = 105

Width of the parapet,

= 105 – 75

= 30 cm

Therefore, the width of the parapet is 30 cm.

Next Chapter Solution :

👉 Chapter 22 👈

Updated: June 9, 2023 — 3:00 pm