RS Aggarwal Solutions for Class 7 Maths Exercise 20E Chapter 20 Mensuration in simple PDF are given here. This exercise of RS Aggarwal Solutions for Class 7 Chapter 20 contains questions related to the circumference of a circle. The perimeter of a circle is called its circumference. The length of a thread that winds around the circle exactly gives the circumference of the circle. Students can practice RS Aggarwal Solutions for Class 7 Maths Chapter 20 Mensuration Exercise 20E with the help of solutions provided here.

## Download the PDF of RS Aggarwal Solutions For Class 7 Maths Chapter 20 Mensuration – Exercise 20E

### Access answers to Maths RS Aggarwal Solutions for Class 7 Chapter 20 – Mensuration Exercise 20E

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

**(i) 28 cm, (ii) 1.4 m**

**Solution:-**

(i) 28 cm

Here, radius r = 28 cm

Circumference of a circle = (2Ï€r) units

Where, Ï€ = (22/7)

= 2 Ã— (22/7) Ã— 28

= 2 Ã— 22 Ã— 4

= 176 cm

(ii) 1.4 m

Here, radius r = 1.4 m = (14/10) m

Circumference of a circle = (2Ï€r) units

Where, Ï€ = (22/7)

= 2 Ã— (22/7) Ã— (14/10)

= 1 Ã— 22 Ã— (2/5)

= 8.8 m

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

**(i) 35 cm, (ii) 4.9 m**

**Solution:-**

(i) 35 cm

Here, diameter of the circle is given d = 35 cm

Circumference of a circle when diameter is given = Ï€d units

Where, Ï€ = (22/7)

= (22/7) Ã— 35

= 22 Ã— 5

= 110 cm

(ii) 4.9 m

Here, diameter of the circle is given d = 4.9 m = (49/10)

Circumference of a circle when diameter is given = Ï€d units

Where, Ï€ = (22/7)

= (22/7) Ã— (49/10)

= 11 Ã— (7/5)

= 15.4 m

**3. Find the circumference of a circle of radius 15 cm. [Take Ï€ = 3.14]**

**Solution:-**

Here, radius r = 15 cm

Circumference of a circle = (2Ï€r) units

Where, Ï€ = 3.14 â€¦ [given]

= 2 Ã— 3.14 Ã— 15

= 94.2 cm

**4. Find the radius of a circle whose circumference is 57.2 cm**

**Solution:-**

Circumference of a given circle is 57.2 cm

âˆ´C = 57.2 cm

Let the radius of the given circle be r cm

Then,

= C = (2Ï€r) units

= r = (C/2Ï€)

= r = (57.2/ (2 Ã— (22/7)))

= r = (57.2/2) Ã— (7/22)

= r = (28.6 Ã— 0.3182)

= r = 9.1 cm

Hence, the radius of a circle is 9.1 cm

**5. Find the diameter of a circle whose circumference is 63.8 m**

**Solution:-**

Circumference of a given circle is 63.8 m

âˆ´C = 63.8 m

Let the radius of the given circle be r m

Then,

= C = (2Ï€r) units

= r = (C/2Ï€)

= r = (63.8/ (2 Ã— (22/7)))

= r = (63.8/2) Ã— (7/22)

= r = (31.9 Ã— 0.3182)

= r = 10.15 m

âˆ´Diameter of the circle = 2r = (2 Ã— 10.15) = 20.3 cm

**6. The circumference of a circle exceeds its diameter by 30 cm. Find the radius of the circle.**

**Solution:-**

Let the radius of the circle be r cm.

Then, its circumference = (2Ï€r) cm

Now, (circumference) â€“ (diameter) = 30 cm â€¦ [given]

âˆ´ (2Ï€r â€“ 2r) = 30

By taking 2r common we get,

= 2r (Ï€ â€“ 1) = 30

= 2r Ã— ((22/7) â€“ 1) = 30

= 2r Ã— ((22 â€“ 7)/7) = 30

= 2r Ã— (15/7) = 30

= (30/7) r = 30

= r = (30) Ã— (7/30)

= r = 7 cm

Hence, the radius of the circle is 7 cm.

**7. The ratio of the radii of two circle is 5: 3. Find the ratio of their circumferences.**

**Solution:-**

Let the radii of the given circles be 5x and 3x respectively and let their circumferences be C_{1} and C_{2 }respectively.

Circumference of a circle = 2Ï€r

Then, C_{1} = 2 Ã— Ï€ Ã— 5x = 10Ï€x and C_{2} = 2 Ã— Ï€ Ã— 3x = 6Ï€x

âˆ´ (C_{1}/C_{2}) = (10Ï€x/ 6Ï€x)

= (C_{1}/C_{2}) = (5/ 3)

= (C_{1}: C_{2}) = 5: 3

Hence, the ratio of the circumference of the given circles is 5: 3

**8. How long will a man take to make a round of a circular field of radius 21 m, cycling at the speed of 8 km/h?**

**Solution:-**

From the question,

Radius of the circular field, r = 21 m

Cycling at the speed of = 8 km/h

= ((8 Ã— 1000)/ (3600)) m/s â€¦ [Because 1 km = 1000 m, 1hr = 3600 sec]

= (8000/3600)

= (80/36) â€¦ [divide by 4]

= (20/9) m/s

Distance covered by the cyclist = Circumference of the circular field

= 2Ï€r

= (2 Ã— (22/7) Ã— 21)

= (2 Ã— 22 Ã— 3)

= 132 m

Time taken by the cyclist to cover the field = (Distance covered by the cyclist/Speed of the cyclist)

= (132/ (20/9))

= 132 Ã— (9/20)

= 33 Ã— (9/5)

= 59.4 sec

**9. A racetrack is in the form of a ring whose inner circumference is 528 m and the outer circumference is 616 m. Find the width of the track.**

**Solution:-**

Let the inner and outer radii of the track be r meters and R meters respectively.

Then,

= 2Ï€r = 528 m

= 2 Ã— (22/7) Ã— r = 528

= r = (528/ (2 Ã— (22/7)))

= r = (528/ 2) Ã— (7/22)

= r = (264/2) Ã— (7/11)

= r = 132 Ã— 0.6364

= r = 84 m

And,

= 2Ï€R = 616 m

= 2 Ã— (22/7) Ã— R = 616

= R = (616/ (2 Ã— (22/7)))

= R = (616/ 2) Ã— (7/22)

= R = (308/2) Ã— (7/11)

= R = 154 Ã— 0.6364

= R = 98 m

Now,

(R â€“ r) = (98 â€“ 84)

= 14 m

Hence, the width of the track is 14 m

**10. The inner circumference of a circular track is 330 m. The track is 10.5 m wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of â‚¹ 20 per meter.**

**Solution:-**

Let the inner and outer radii of the track be r meters and (r + 10.5) meters respectively.

Inner circumference = 330 m

âˆ´ 2Ï€r = 330

= 2 Ã— (22/7) Ã— r = 330

= (44/7) Ã— r = 330

= r = 330 Ã— (7/44)

= r = 30 Ã— (7/4)

= r = 52.5 m

Because, Inner radius of the track = 52.5 m

âˆ´Outer radius of the track = (r + 10.5) = (52.5 + 10.5) = 63 m

Then,

Circumference of the outer circle = 2Ï€r

= 2 Ã— (22/7) Ã— 63

= (2 Ã— 22 Ã— 9)

= 396 m

Rate of fencing = â‚¹ 20 per meter.

Total cost of fencing the outer circle = â‚¹ (396 Ã— 20)

= â‚¹ 7920