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

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

**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:**

(i) Radius of the circle (r) = 21 cm

**Question 2.**

**Solution:**

(i) Diameter of the circle = 28 cm

**Question 3.**

**Solution:**

Circumference of a circle = 264 cm

**Question 4.**

**Solution:**

Circumference of the circle (c) = 35.2 m

**Question 5.**

**Solution:**

Area of the circle = 616 cm²

**Question 6.**

**Solution:**

Area of a circle = 1386 m²

**Question 7.**

**Solution:**

Ratio in the radii of two circles = 4 : 5

Let radius of first circle (r1) = 4x

and radius of the second circle (r2) = 5x

**Question 8.**

**Solution:**

Length of rope (r) = 21 m

Area of the circle = πr² = \(\frac { 22 }{ 7 }\) x 21 x 21 m² = 1386 m²

The horse will graze on 1386 m² area

**Question 9.**

**Solution:**

Area of a square made of a wire = 121 cm²

Side = √Area = √121 = 11 m

Perimeter of wire = 4 x side = 4 x 11 = 44 cm

Circumference of circular wire = 44 cm

**Question 10.**

**Solution:**

Radius of circular wire = 28 cm

Circumference = 2πr = 2 x \(\frac { 22 }{ 7 }\) x 28 cm = 176 cm

Perimeter of the square formed by this wire = 176 cm

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

Area of square so formed = a² = (44)² cm² = 1936 cm²

**Question 11.**

**Solution:**

Length of rectangular sheet (l) = 34 cm

and breadth (b) = 24 cm

Area = l x b = 34 x 24 cm² = 816 cm²

Diameter of one button = 3.5 cm

Radius (r) = \(\frac { 3.5 }{ 2 }\) cm

and area of one button = πr²

Area of 64 buttons = 9.625 x 64 cm² =616 cm²

Area of remaining sheet = 816 – 616 = 200 cm²

**Question 12.**

**Solution:**

Length of ground (l) = 90 m

and breadth (b) = 32 m

Area = l x b = 90 x 32 m² = 2880 m²

Radius of circular tank (r) = 14 m

= 616 m²

Area of remaining portion = 2880 – 616 = 2264 m²

Rate of turfing = Rs. 50 per sq.m.

Total cost = Rs. 50 x 2264 = Rs. 113200

**Question 13.**

**Solution:**

Each side of square = 14 cm

Area of square = a² = 14 x 14 = 196 cm²

Radius of each circle at each corner of square = \(\frac { 14 }{ 2 }\) = 7 cm

Area of shaded portion = Area of square – area of 4 quadrants

= 196 – 154 = 42 cm²

**Question 14.**

**Solution:**

Length of field = 60 m

and breadth = 40 m

Length of rope = 14 m

Area covered by the horse

**Question 15.**

**Solution:**

Diameter of largest circle (outer circle) = 21 cm

Area of shaded portion = 346.5 – (154 + 38.5) = (346.5 – 192.5) = 154 cm²

**Question 16.**

**Solution:**

Length of plot (l) = 8m

and breadth (b) = 6m

Area of plot = l x b = 8 x 6 = 48 m²

Radius of each quadrant at the corner = 2 m

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

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