## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

**Ex 11.3 Class 7 Maths Question 1.**

**Find the circumference of the circles with the following radius. (Take π = 22/7)**

**(a) 14 cm**

**(b) 28 mm**

**(c) 21 cm**

**Solution:**

(a) Given: Radius (r) = 14 cm

∴ Circumference = 2πr = 2 × 22/7 × 14

= 88 cm

(b) Given: Radius (r) = 28 mm

∴ Circumference = 2πr = 2 × 22/7 × 28

= 176 mm

(c) Given: Radius (r) = 21 cm

∴ Circumference = 2πr = 2 × 22/7 × 21

= 132 cm

**Ex 11.3 Class 7 Maths Question 2.**

**Find the area of the following circles, given that (Take π = 22/7)**

**(a) radius = 14 mm**

**(b) diameter = 49 m**

**(c) radius = 5 cm**

**Solution:**

(a) Here, r = 14 mm

∴ Area of the circle = πr2

= π × 14 × 14 = 22/7 × 14 × 14

(b) Here, diameter = 49 m 49

(c) Here, radius = 5 cm

**Ex 11.3 Class 7 Maths Question 3.**

**If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22/7)**

**Solution:**

Given: Circumference = 154 m

∴ 2πr = 154

**Ex 11.3 Class 7 Maths Question 4.**

**A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds offence. Also find the cost of the rope, if it costs ₹ 4 per metre. (Take π = 22/7)**

**Solution:**

Diameter of the circular garden = 21 m

∴ Radius = 21/2 m

∴ Circumference = 2πr = 2 × 22/7 × 21/2

= 66 m

Length of rope needed for 2 rounds

= 2 × 66 m = 132 m

Cost of the rope = ₹4 × 132 = ₹ 528

**Ex 11.3 Class 7 Maths Question 5.**

**From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)**

**Solution:**

Radius of the circular sheet = 4 cm

∴ Area = πr² = π × 4 × 4 = 16π cm²

Radius of the circle to be removed = 3 cm

∴ Area of sheet removed = πr² = 9π cm²

Area of the remaining sheet

= (16π – 9π) cm² = 7π cm²

= 7 × 3.14 cm² = 21.98 cm²

Hence, the required area = 21.98 cm².

**Ex 11.3 Class 7 Maths Question 6.**

**Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one metre of the lace costs ₹ 15. (Take π = 3.14)**

**Solution:**

Diameter of the table cover = 1.5 m

∴ Radius = 1.5/2 = 0.75 m

∴ Length of the lace = 2πr = 2 × 3.14 × 0.75

= 4.710 m

Cost of the lace = ₹ 15 × 4.710 = ₹ 70.65

**Ex 11.3 Class 7 Maths Question 7.**

**Find the perimeter of the given figure, which is a semicircle including its diameter.**

**Solution:**

Given: Diameter = 10 cm

Hence, the required perimeter

= 25.7 cm. (approx.)

**Ex 11.3 Class 7 Maths Question 8.**

**Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15 m². (Take π = 3.14)**

**Solution:**

Given:

Diameter = 1.6 m

∴ Radius = 1.6/2 = 0.8 m

Area of the table-top = πr²

= 3.14 × 0.8 × 0.8 m²

= 2.0096 m²

∴ Cost of polishing = ₹ 15 × 2.0096

= ₹ 30.14 (approx.)

**Ex 11.3 Class 7 Maths Question 9.**

**Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π = 22/7)**

**Solution:**

Length of the wire to be bent into a circle = 44 cm

2πr = 44

Now, the length of the wire is bent into a square.

Here perimeter of square

= Circumference of line k

Length of each side of the square

= Perimeter/4 = 44/4 =11 cm

Area of the square = (Side)² = (11)² = 121 cm²

Since, 154 cm² >121 cm²

Thus, the circle encloses more area.

**Ex 11.3 Class 7 Maths Question 10.**

**From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the given figure below). Find the area of the remaining sheet. (Take π = 22/7)**

**Solution:**

Radius of the circular sheet = 14 cm

∴ Area = πr² =22/7 × 14 × 14 cm²

= 616 cm²

Area of 2 small circles = 2 × πr²

= 2 × 22/7 × 3.5 × 3.5 cm²

= 77.0 cm²

Area of the rectangle = l × b

= 3 × 1 cm² = 3 cm²

Area of the remaining sheet after removing the 2 circles and 1 rectangle

= 616 cm² – (77 + 3) cm²

= 616 cm² – 80 cm² = 536 cm²

**Ex 11.3 Class 7 Maths Question 11.**

**A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)**

**Solution:**

Side os the square sheet = 6 m

∴ Area of the sheet = (Side)² = (6)² = 36 cm²

Radius of the circle = 2 cm

∴ Area of the circle to be cut out = πr2

= 22/7 × 2 × 2 = 88/7 cm²

Area of the left over sheet

**Ex 11.3 Class 7 Maths Question 12.**

**The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take π = 3.14)**

**Solution:**

Circumference of the circle = 31.4 cm

2πr = 31.4

∴ r = 31.4/2 × 3.14 = 5cm

Area of the circle = 7πr² = 3.14 × 5 × 5 = 78.5 cm²

Hence, the required radius = 5 cm and area = 78.5 cm².

**Ex 11.3 Class 7 Maths Question 13.**

**A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (Take π = 3.14)**

**Solution:**

Diameter of the flower bed = 66 m .

∴ Radius = 66/2 = 33 m

Let r1 = 33 m

Width of the path = 4 m

Radius of the flower bed included path

= 33 m + 4 m = 37m

Let r2 = 37m

Area of the circular path

=

= 3.14 (37²– 33²)

= 3.14 × (37 + 33) (37 – 33) [Y a² – b² = (a + b)(a-b)]

= 3.14 × 70 × 4 = 879.20 m²

Hence, the required area = 879.20 m²

**Ex 11.3 Class 7 Maths Question 14.**

**A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler can water the entire garden?**

**[Take π = 3.14]**

**Solution:**

Area of the flower garden = 314 m²

Radius of the circular portion covered by the sprinkler = 12 m

∴ Area = 7πr² = 3.14 × 12 × 12

= 3.14 × 144 m² = 452.16 m²

Since 452.16 m² > 314 m²

Yes, the sprinkler will water the entire garden.

**Ex 11.3 Class 7 Maths Question 15.**

**Find the circumference of the inner and the outer circles, shown in the given figure. (Take π = 3.14)**

**Solution:**

Radius of the outer circle = 19 m

∴ Circumference of the outer circle = 2πr

= 2 × 3.14 × 19 = 3.14 × 38 m

= 119.32 m

Radius of the inner circle

= 19m – 10m = 9m

∴ Circumference = 2πr = 2 × 3.14 × 9

= 56.52 m

Here the required circumferences are 56.52 m and 119.32 m.

**Ex 11.3 Class 7 Maths Question 16.**

**How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)**

**Solution:**

Radius of the wheel = 28 cm

∴ Circumference = 2πr = 2 × 22/7 × 28 = 176 cm

Number of rotations made by the wheel in going 352 m or 35200 cm

= 35200/176 = 200

Hence, the required number of rotation = 200.

**Ex 11.3 Class 7 Maths Question 17.**

**The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)**

**Solution:**

Length of minute hand = 15 cm

∴ Radius = 15 cm

Circumference = 2πr

= 2 × 3.14 × 15 cm = 94.2 cm

Since the minute hand covers the distance in 1 hour equal to the circumference of the circle. Here the required distance covered by the minute hand = 94.2 cm.