## RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

**Question 1.**

**Solution:**

Radius of the circle (r) = 28 cm

**Question 2.**

**Solution:**

(i) Diameter of circle (d) = 35 cm

**Question 3.**

**Solution:**

Radius of a circle = 15 cm

Circumference = 2πr = 2 x 3.14 x 15 = 94.20 cm = 94.2 cm

**Question 4.**

**Solution:**

Circumference of a circle (c) = 57.2 cm

**Question 5.**

**Solution:**

Circumference (c) = 63.8 m

**Question 6.**

**Solution:**

Let c be the circumference and d be the diameter of the circle.

c = d + 30

⇒ dπ = d + 30

⇒ dπ – d = 30

⇒ d (π – 1) = 30

**Question 7.**

**Solution:**

The ratio of the radii of the circles = 5 : 3

Let radius of first circle = 5x

and radius of second circle = 3x

Circumference of first circle = 2πr = 2π x 5x = 10πx

and circumference of second circle = 2π x 3x = 6πx

Ratio = 10πx : 6πx = 10 : 6 = 5 : 3

**Question 8.**

**Solution:**

Radius of circular field (r) = 21 m

Circumference = 2πr

**Question 9.**

**Solution:**

Outer circumference = 616 m

Width of track = R – r = 98 – 84 = 14m

**Question 10.**

**Solution:**

Inner circumference of the circular track = 330 m

Rate of fencing = Rs. 20 per metre

Total cost = Rs. 20 x 396 = Rs. 7920

**Question 11.**

**Solution:**

Radius of inner circle (r) = 98 cm

Inner circumference = 2πr = 2 x \(\frac { 22 }{ 7 }\) x 98 cm = 616 cm

Radius of the outer circle (R) = 1 m 26cm = 126 cm

Outer circumference = 2πR = 2 x \(\frac { 22 }{ 7 }\) x 126 cm = 792 cm

**Question 12.**

**Solution:**

Side of equilateral triangle = 8.8 cm

Its perimeter = 8.8 x 3 = 26.4 cm

By bending their wire into a circular shape,

the circumference = 26.4 cm

Let d be the diameter,

Then C = dπ

**Question 13.**

**Solution:**

Each side of rhombus = 33 cm

Perimeter = 4 x 33 = 132 cm

Perimeter of circle = 132 cm

Let r be the radius

**Question 14.**

**Solution:**

Length of rectangle (l) = 18.7 cm

Breadth (b) = 14.3 cm

Perimeter = 2 (l + b) = 2 (18.7 + 14.3) cm = 2 x 33 = 66 cm

Circumference of the so formed circle = 66 cm

**Question 15.**

**Solution:**

Radius of the circle (r) = 35 cm

Its circumference (c) = 2πr

**Question 16.**

**Solution:**

Diameter of well (d) = 140 cm

Outer circumference of parapet = 616 cm

Let D be the diameter, then

**Question 17.**

**Solution:**

Diameter of wheel (d) = 98 cm

**Question 18.**

**Solution:**

Diameter of cycle wheel (d) = 70 cm

**Question 19.**

**Solution:**

Diameter of car wheel (d) = 77 cm

Circumference = πd = \(\frac { 22 }{ 7 }\) x 77 cm

**Question 20.**

**Solution:**

No. of revolutions = 5000

Distance covered = 11 km = 11 x 1000 = 11000 m

**Question 21.**

**Solution:**

Length of hour hand (r) = 4.2 cm

and length of minutes hand (R) = 7cm

Distance covered by hour hand in 24 hours = 2πR

= 2 x \(\frac { 22 }{ 7 }\) x 4.2

= 26.4 cm

But distance covered by minute hand in one hour = 2πR = 2 x \(\frac { 22 }{ 7 }\) x 7 = 44 cm

and distance covered by minute hand in 24 hours = 44 x 24 cm = 1056 cm

Sum of distance covered by these hands = 26.4 + 1056 = 1082.4 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E are helpful to complete your math homework.

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