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

NCERT Solutions Class 7 Mathematics 
Chapter – 11 (Perimeter and Area)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 11 Perimeter and Area Exercise 11.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 11 Perimeter and Area

• NCERT Solution Class 7 Maths Exercise – 11.1
• NCERT Solution Class 7 Maths Exercise – 11.2
• NCERT Solution Class 7 Maths Exercise – 11.4

Exercise – 11.3 

1. Find the circumference of the circle with the following radius: (Take π = )
(a) 14 cm          (b) 28 cm                   (c) 21 cm

Solution –
(a) 14 cm
Given, radius of circle = 14 cm
Circumference of the circle = 2πr
= 2 × × 14
= 2 × 22 × 2
= 88 cm

(b) 28 cm
Given, radius of circle = 28 cm
Circumference of the circle = 2πr
= 2 × × 28
= 2 × 22 × 4
= 176 cm

(c) 21 cm
Given, radius of circle = 21 cm
Circumference of the circle = 2πr
= 2 × × 21
= 2 × 22 × 3
= 132 cm

2. Find the area of the following circles, given that:
(a) Radius = 14 mm (Take π = )
(b) Diameter = 49 m
(c) Radius = 5 cm

Solution –
(a) Radius = 14 mm (Take π = )
Given, radius of circle = 14 mm
Then,
Area of the circle = πr²
= × 14²

= × 196
= 22 × 28
= 616 mm²

(b) Diameter = 49 m
Given, diameter of circle (d) = 49 m
We know that, radius (r) =
= = 24.5 m

Then,
Area of the circle = πr²
= × (24.5)²

= × 600.25
= 22 × 85.75
= 1886.5 m²

(c) Radius = 5 cm
Given, radius of circle = 5 cm
Then,
Area of the circle = πr²
= × 5²

= × 25

=
= 78.57 cm²

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

Solution –
Given that,
Circumference of the circle = 154 m
Then,
We know that, Circumference of the circle = 2πr
154 = 2 × × r
154 = × r

r =

r =

r =

r =

r = 24.5 m
Now,
Area of the circle = πr²
= × (24.5)²

= × 600.25
= 22 × 85.75
= 1886.5 m²
So, the radius of circle is 24.5 and area of circle is 1886.5 

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

Solution –
Given that,
Diameter of the circular garden = 21 m
We know that, radius (r) = d/2
= 21/2
= 10.5 m
Then,
Circumference of the circle = 2πr
= 2 × × 10.5

=
= 66 m
So, the length of rope required = 2 × 66 = 132 m
Cost of 1 m rope = ₹ 4
Cost of 132 m rope = ₹ 4 × 132
= ₹ 528

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 –
Give that,
Radius of circular sheet R = 4 cm
A circle of radius to be removed r = 3 cm
Then,
The area of the remaining sheet = πR² – πr²
= π (R² – r²)
= 3.14 (4² – 3²)
= 3.14 (16 – 9)
= 3.14 × 7
= 21.98 cm²
So, the area of the remaining sheet is 21.98 cm².

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 meter of the lace costs ₹ 15. (Take π = 3.14)

Solution –
Given that,
Diameter of the circular table cover = 1.5 m
We know that, radius (r) = d/2
= 1.5/2
= 0.75 m
Then,
Circumference of the circular table cover = 2πr
= 2 × 3.14 × 0.75
= 4.71 m
So, the length of lace = 4.71 m
Cost of 1 m lace = ₹ 15
Cost of 4.71 m lace = ₹ 15 × 4.71
= ₹ 70.65

7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Solution –
Given that,
Diameter of semi-circle = 10 cm
We know that, radius (r) = d/2
= 10/2
= 5 cm
Then,
Circumference of the semi-circle = πr
= × 5

=
= 15.71 cm

Now,
Perimeter of the given figure = Circumference of the semi-circle + semi-circle diameter
= 15.71 + 10
= 25.71 cm

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 that,
Diameter of the circular table-top = 1.6 m
We know that, radius (r) = d/2
= 1.6/2
= 0.8 m
Then,
Area of the circular table-top = πr²
= 3.14 × 0.8²
= 3.14 × 0.8 ×0.8
= 2.0096 m²
Cost of polishing 1 m² area = ₹ 15
Cost of polishing 2.0096 m² area = ₹ 15 × 2.0096
= ₹ 30.144
Hence, the cost of polishing 2.0096 m² area is ₹ 30.144.

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 = )

Solution –
Given that,
Length of wire that Shazli took =44 cm
Then,
If the wire is bent into a circle,
We know that, circumference of the circle = 2πr
44 = 2 × × r

44 = × r

  = r

r = 7 cm

Area of the circle = πr²
= × 7²

= × 7 ×7

= 22 × 7
= 154 cm²

Now,
If the wire is bent into a square,
Length of wire = perimeter of square
44 = 4 x side
44 = 4s
s =

s = 11cm
Area of square = (side)² =  11²
= 121 cm²
Therefore circle has more area than square

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 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = )

Solution –
Given that,
Radius of the circular card sheet = 14 cm
Radius of the two small circles = 3.5 cm
Length of the rectangle = 3 cm
Breadth of the rectangle = 1 cm
First we have to find out the area of circular card sheet, two circles and rectangle to find out the remaining area.

Now,
Area of the circular card sheet = πr²
= × 14²

= × 14 × 14
= 22 × 2 × 14
= 616 cm²
Area of the 2 small circles = 2 × πr²
= 2 × ( × 3.5²)

= 2 × ( × 3.5 × 3.5)

= 2 × (22 × 0.5 × 3.5)
= 2 × 38.5
= 77 cm²

Area of the rectangle = Length × Breadth
= 3 × 1
= 3 cm²
Now,
Area of the remaining sheet = Area of circular card sheet – (Area of two small circles + Area of the rectangle)
= 616 – (77 + 3)
= 616 – 80
= 536 cm²
Hence, the area of the remaining sheet is 536 cm²

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 that,
Radius of circle = 2 cm
Square sheet side = 6 cm
First we have to find out the area of square aluminium sheet and circle to find out the remaining area.
Now,
Area of the square = side²
Hence, the area of the square aluminium sheet = 6² = 36 cm²
Area of the circle = πr²
= 3.14 × 2²
= 3.14 × 2 × 2

= 3.14 × 4
= 12.56 cm²
Now,
Area of aluminium sheet left = Area of the square aluminum sheet – Area of the circle
= 36 – 12.56
= 23.44 cm²
Hence, the area of the aluminium sheet left is 23.44 cm²

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

Solution –
Given that,
Circumference of a circle = 31.4 cm
Circumference of a circle = 2πr
31.4 = 2 × 3.14 × r
31.4 = 6.28 × r
= r
r = 5 cm
Then,
Area of the circle = πr²
= 3.14 × (5cm)²
= 3. 14 × 25 cm²
= 78.5 cm²
Therefore, radius of the circle is 5 cm and area of the circle is 78.5 cm²

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? (π = 3.14)

Solution –
Given that,
Diameter of the flower bed = 66 m
Then,
Radius of the flower bed = d/2
= 66/2
= 33 m
Area of flower bed = πr²
= 3.14 × 33²
= 3.14 × 1089
= 3419.46 m
Now we have to find area of the flower bed and path together
So, radius of flower bed and path together = 33 + 4 = 37 m
Area of the flower bed and path together = πr²
= 3.14 × 37²
= 3.14 × 1369
= 4298.66 m
Area of the path = Area of the flower bed and path together – Area of flower bed
= 4298.66 – 3419.46
= 879.2 m²
Hence, the area of the path is 879.2 m²

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 water the entire garden? (Take π = 3.14)

Solution –
Given that,
Area of the circular flower garden = 314 m²
Sprinkler at the centre of the garden can cover an area that has a radius = 12 m
Area of the circular flower garden = πr²
314 = 3.14 × r²

= r²

r² = 100
r = √100
r = 10 m
∴ Radius of the circular flower garden is 10 m.
Since, the sprinkler can cover an area of radius 12 m
Hence, the sprinkler will water the whole garden.

15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Solution –
Radius of inner circle = outer circle radius – 10
= 19 – 10
= 9 m
Circumference of the inner circle = 2πr
= 2 × 3.14 × 9
= 56.52 m
Then,
Radius of outer circle = 19 m
Circumference of the outer circle = 2πr
= 2 × 3.14 × 19
= 119.32 m
Therefore, the circumference of the inner circle is 56.52 m and the circumference of the outer circle is 119.32 m

16. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = )

Solution –
Given that,
Radius of the wheel = 28 cm
Total distance = 352 m = 35200 cm
Circumference of the wheel = 2πr
= 2 × × 28
= 2 × 22 × 4
= 176 cm
Now we have to find the number of rotations of the wheel,
Number of times the wheel should rotate = Total distance covered by wheel / Circumference of the wheel
=

=

= 200
Hence, wheel rotates 200 times

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 that,
Length of the minute hand of the circular clock = 15 cm
Then,
Distance travelled by the tip of minute hand in 1 hour = circumference of the clock
= 2πr
= 2 × 3.14 × 15
= 94.2 cm
Therefore, the minute hand moves 94.2 cm in 1 hour