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

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.



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 of a circle = 2πr = 2 × 22/7 × 14
= 88 cm
(b) Given: Radius (r) = 28 mm
Circumference of a circle = 2πr = 2 × 22/7 × 28
= 176 mm
(c) Given: Radius (r) = 21 cm
Circumference of a circle = 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) Given, r = 14 mm
Area of a circle = πr2
= π × 14 × 14 = 22/7 × 14 × 14
= 616 mm2

(b) Given, diameter = 49 m

(c) Given, 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:
Given, diameter of the circular garden = 21 m
Radius = 21/2 m
Circumference = 2πr = 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 π = 3.14)

Solution:
Radius of the circular sheet = 4 cm
Area of the circular sheet = πr2 = π × 4 × 4 = 16π cm2
Radius of the circular sheet to be removed = 3 cm
Area of the circular sheet removed = πr2 = 9π cm2
Area of the remaining sheet
= (16π – 9π) cm2 = 7π cm2
= 7 × 3.14 cm2 = 21.98 cm2
Hence, the area of the remaining sheet is 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:
Given, 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 is 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 per 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 = Length of the wire
4 × Side = 44

Side = 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 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 π = 22/7)

Solution:
Radius of the circular sheet = 14 cm
Area of the circular sheet = πr2 = 22/7 × 14 × 14 cm2
= 616 cm2
Area of 2 small circles = 2 × πr2
= 2 × 22/7 × 3.5 × 3.5 cm2
= 77 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:
Given, side of 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
= 3.14 × 2 × 2 = 12.56 cm2
Area of the left over sheet = 36 cm2 12.56 cm2
= 23.44 cm2

 

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) = 5 cm
Area of the circle = πr2 = 3.14 × 5 × 5 = 78.5 cm2
Hence, the radius of the circle is 5 cm and area is 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:
Given, 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 including path
= 33 m + 4 m = 37 m
Let r2 = 37 m
Area of the circular path = π(r22r12)
= 3.14 (372 – 332)
= 3.14 × (37 + 33) (37 – 33)     [Using a2 – b2 = (a + b)(a – b)]
= 3.14 × 70 × 4 = 879.20 m2
Hence, the area of the path is 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 of the circular portion covered by the sprinkler

= πr2 = 3.14 × 12 × 12

= 3.14 × 144 m2 = 452.16 m2
Since 452.16 m2 > 314 m2
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 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 = 19 m – 10 m = 9 m
Circumference of the inner circle = 2πr = 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 π = 22/7)

Solution:
Given, radius of the wheel = 28 cm
Circumference = 2πr = 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 π = 3.14)

Solution:
Given, length of the 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. Hence, the required distance covered by the minute hand is 94.2 cm.


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