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

- Class 7 Maths Perimeter and Area Exercise 11.1
- Class 7 Maths Perimeter and Area Exercise 11.2
- Class 7 Maths Perimeter and Area Exercise 11.3
- Class 7 Maths Perimeter and Area Exercise 11.4

**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3**

Ex 11.3 Class 7 Maths Question 1.

Find the circumference of the circles with the following radius. (Take π =(frac{22}{7}))

(a) 14 cm

(b) 28 mm

(c) 21 cm

Solution:

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

∴ Circumference = 2πr = 2 × (frac{22}{7}) × 14

= 88 cm

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

∴ Circumference = 2πr = 2 × (frac{22}{7}) × 28

= 176 mm

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

∴ Circumference = 2πr = 2 × (frac{22}{7}) × 21

= 132 cm

Ex 11.3 Class 7 Maths Question 2.

Find the area of the following circles, given that (Take π =(frac{22}{7}))

(a) radius = 14 mm

(b) diameter = 49 m

(c) radius = 5 cm

Solution:

(a) Here, r = 14 mm

∴ Area of the circle = πr^{2}

= π × 14 × 14 = (frac{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 =(frac{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 =(frac{22}{7}))

Solution:

Diameter of the circular garden = 21 m

∴ Radius = (frac{21}{2}) m

∴ Circumference = 2πr = (2 times frac{22}{7} times frac{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 = πr^{2} = π × 4 × 4 = 16π cm^{2}

Radius of the circle to be removed = 3 cm

∴ Area of sheet removed = πr^{2} = 9π cm^{2}

Area of the remaining sheet

= (16π – 9π) cm2 = 7π cm^{2}

= 7 × 3.14 cm^{2} = 21.98 cm^{2}

Hence, the required area = 21.98 cm^{2}.

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 = (frac{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 m^{2}. (Take π = 3.14)

Solution:

Given:

Diameter = 1.6 m

∴ Radius = (frac{1.6}{2}) = 0.8 m

Area of the table-top = πr^{2}

= 3.14 × 0.8 × 0.8 m^{2}

= 2.0096 m^{2}

∴ 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 =(frac{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

(=frac{text { Perimeter }}{4}=frac{44}{4}=11 mathrm{cm})

Area of the square = (Side)^{2} = (11)^{2} = 121 cm^{2}

Since, 154 cm^{2} >121 cm^{2}

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 = (frac{22}{7}))

Solution:

Radius of the circular sheet = 14 cm

∴ Area = πr^{2} = (frac{22}{7}) × 14 × 14 cm^{2}

= 616 cm^{2}

Area of 2 small circles = 2 × πr^{2}

= 2 × (frac{22}{7}) × 3.5 × 3.5 cm^{2}

= 77.0 cm^{2}

Area of the rectangle = l × b

= 3 × 1 cm^{2} = 3 cm^{2}

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

= 616 cm^{2} – (77 + 3) cm^{2}

= 616 cm^{2} – 80 cm^{2} = 536 cm^{2}

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 cm^{2}

Radius of the circle = 2 cm

∴ Area of the circle to be cut out = πr^{2
}= (=frac{22}{7} times 2 times 2=frac{88}{7} mathrm{cm}^{2})

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=frac{31.4}{2 times 3.14}) = 5cm

Area of the circle = 7πr^{2} = 3.14 × 5 × 5 = 78.5 cm^{2}

Hence, the required radius = 5 cm and area = 78.5 cm^{2}.

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 = (frac{66}{2}) = 33 m

Let r_{1} = 33 m

Width of the path = 4 m

Radius of the flower bed included path

= 33 m + 4 m = 37m

Let r_{2} = 37m

Area of the circular path = (pileft(r_{2}^{2}-r_{1}^{2}right))

= 3.14 (37^{2}– 33^{2})

= 3.14 × (37 + 33) (37 – 33) [Y a^{2} – b^{2} = (a + b)(a-b)]

= 3.14 × 70 × 4 = 879.20 m^{2}

Hence, the required area = 879.20 m^{2}

Ex 11.3 Class 7 Maths Question 14.

A circular flower garden has an area of 314 m^{2}. 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 m^{2}

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

∴ Area = 7πr^{2} = 3.14 × 12 × 12

= 3.14 × 144 m^{2} = 452.16 m^{2}

Since 452.16 m^{2} > 314 m^{2}

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 = (frac{22}{7}))

Solution:

Radius of the wheel = 28 cm

∴ Circumference = 2πr = 2 × (frac{22}{7}) × 28 = 176 cm

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

(=frac{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.

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