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

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area
Ex 11.3 are the part of NCERT Solutions for Class 7 Maths. Here you can find
the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3.

**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1****NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2****NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3****NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4**

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

Find the circumference of the circles with the
following radius. (Take Ï€ = 22/7)(a) 14 cm

(b) 28 mm

(c) 21 cm

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

∴ Circumference of a circle = 2Ï€r = 2 × 22/7 × 14

= 88 cm

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

∴ Circumference of a circle = 2Ï€r = 2 × 22/7 × 28

= 176 mm

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

∴ Circumference of a circle = 2Ï€r = 2 × 22/7 × 21

= 132 cm

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

Find the area of the following circles, given that
(Take Ï€ = 22/7)(a) radius = 14 mm

(b) diameter = 49 m

(c) radius = 5 cm

**Solution:
**(a) Given, r = 14 mm

∴ Area of a circle = Ï€r

^{2}

= Ï€ × 14 × 14 = 22/7 × 14 × 14

= 616 mm

^{2}

(b)
Given, diameter = 49 m

**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 Ï€ = 22/7)**Solution:****
**Given, circumference
= 154 m

∴ 2Ï€r = 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 Ï€ = 22/7)

**Solution:
**Given, diameter of the circular garden = 21
m

∴ Radius = 21/2 m

∴ Circumference = 2Ï€r = 2 × 22/7 × 21/2

= 66 m

Length of the 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 Ï€ = 3.14)**Solution:
**Radius of the circular sheet = 4 cm

∴ Area of the circular sheet = Ï€r

^{2}= Ï€ × 4 × 4 = 16Ï€ cm

^{2 }

Radius of the circular sheet to be removed = 3 cm

∴ Area of the circular sheet removed = Ï€r

^{2}= 9Ï€ cm

^{2}

Area of the remaining sheet

= (16Ï€ – 9Ï€) cm

^{2}= 7Ï€ cm

^{2}

= 7 × 3.14 cm

^{2}= 21.98 cm

^{2}

Hence, the area of the remaining sheet is 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 Ï€ = 3.14)**Solution:
**Given, diameter of the table cover = 1.5 m

∴ Radius = 1.5/2 = 0.75 m

∴ Length of the lace = 2Ï€r = 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

**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 per m^{2}. (Take Ï€ = 3.14)

**Solution:
**Given, diameter = 1.6 m

∴ Radius = 1.6/2 = 0.8 m

Area of the table-top = Ï€r

^{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 Ï€ = 22/7)**Solution:
**Length of the wire to be bent into a circle = 44 cm

2Ï€r = 44

Here, perimeter of square = Length of the wire

4 × Side = 44

Side
= 44/4 = 11 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 than the square.

**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 Ï€ = 22/7)

**Solution:
**Radius of the circular sheet = 14 cm

∴ Area of the circular sheet = Ï€r

^{2}= 22/7 × 14 × 14 cm

^{2}

= 616 cm

^{2}

Area of 2 small circles = 2 × Ï€r

^{2}

= 2 × 22/7 × 3.5 × 3.5 cm

^{2}

= 77 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 Ï€ = 3.14)**Solution:
**Given, side of 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 = Ï€r

^{2 }= 3.14 × 2 × 2 = 12.56 cm

^{2}

Area of the left over sheet = 36 cm

^{2 }– 12.56 cm

^{2}

= 23.44 cm

^{2}

**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 Ï€ = 3.14)**Solution:
**Circumference of the circle = 31.4 cm

2Ï€r = 31.4

∴ r = 31.4/(2 × 3.14) = 5 cm

Area of the circle = Ï€r

^{2}= 3.14 × 5 × 5 = 78.5 cm

^{2}

Hence, the radius of the circle is 5 cm and area is 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 Ï€ = 3.14)**Solution:
**Given, diameter of the flower bed = 66 m

∴ Radius = 66/2 = 33 m

Let r

_{1}= 33 m

Width of the path = 4 m

Radius of the flower bed including path

= 33 m + 4 m = 37 m

Let r

_{2}= 37 m

Area of the circular path = Ï€(r

_{2}

^{2}– r

_{1}

^{2})

= 3.14 (37

^{2 }– 33

^{2})

= 3.14 × (37 + 33) (37 – 33) [Using a

^{2}– b

^{2}= (a + b)(a – b)]

= 3.14 × 70 × 4 = 879.20 m

^{2}

Hence, the area of the path is 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 Ï€ = 3.14]

**Solution:
**Area of the flower garden = 314 m

^{2}

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

∴ Area of the circular portion covered by the sprinkler

= Ï€r^{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 can 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 Ï€ = 3.14)**Solution:
**Radius of the outer circle = 19 m

∴ Circumference of the outer circle = 2Ï€r

= 2 × 3.14 × 19 = 3.14 × 38 m

= 119.32 m

Radius of the inner circle = 19 m – 10 m = 9 m

∴ Circumference of the inner circle = 2Ï€r = 2 × 3.14 × 9

= 56.52 m

Hence, 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 Ï€ = 22/7)**Solution:
**Given, radius of the wheel = 28 cm

∴ Circumference = 2Ï€r = 2 × 22/7 × 28 = 176 cm

Number of rotations made by the wheel in 1 rotation = 176 cm

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

= 35200/176 = 200

Hence, the number of rotations made by the wheel is 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 Ï€ = 3.14)**Solution:
**Given, length of the minute hand = 15 cm

∴ Radius = 15 cm

Circumference = 2Ï€r

= 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. Hence, the required distance covered by the minute hand is 94.2 cm.

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