Home Class 7th NCERT Solutions Class 7th Math Chapter – 9 Perimeter and Area Exercise 9.2

NCERT Solutions Class 7th Math Chapter – 9 Perimeter and Area Exercise 9.2

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NCERT Solutions Class 7th Math Chapter 9 Perimeter and Area

Textbook NCERT
Class 7th
Subject Mathematics
Chapter 9th
Chapter Name Perimeter and Area
Category Class 7th Mathematics
Medium  English
Source Last Doubt

NCERT Solutions Class 7th Math Chapter 9 Perimeter and Area

Chapter – 9

Perimeter and Area

Exercise 9.2

1. Find the circumference of the circle with the following radius: (Take π = 22/7)

(a) 14 cm

Solution:Given, radius of circle = 14 cm
Circumference of the circle = 2πr
= 2 × (22/7) × 14
= 2 × 22 × 2
= 88 cm

(b) 28 cm

Solution: Given, radius of circle = 28 cm
Circumference of the circle = 2πr
= 2 × (22/7) × 28
= 2 × 22 × 4
= 176 cm

(c) 21 cm

Solution:Given, radius of circle = 21 cm
Circumference of the circle = 2πr
= 2 × (22/7) × 21
= 2 × 22 × 3
= 132 cm

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

(a) Radius = 14 mm 

Solution: Given, radius of circle = 14 mm
Then,
Area of the circle = πr2
= 22/7 × 142
= 22/7 × 196
= 22 × 28
= 616 mm2

(b) Diameter = 49 m

Solution: Given, diameter of circle (d) = 49 m
We know that, radius (r) = d/2
= 49/2
= 24.5 m
Then,
Area of the circle = πr2
= 22/7 × (24.5)2
= 22/7 × 600.25
= 22 × 85.75
= 1886.5 m2

(c) Radius = 5 cm

Solution: Given, radius of circle = 5 cm
Then,
Area of the circle = πr2
= 22/7 × 52
= 22/7 × 25
= 550/7
= 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 = 22/7)

Solution: From the question it is given that,Circumference of the circle = 154 m Then, We know that, Circumference of the circle = 2πr
154 = 2 × (22/7) × r
154 = 44/7 × r
r = (154 × 7)/44
r = (14 × 7)/4
r = (7 × 7)/2
r = 49/2
r = 24.5 m
Now,
Area of the circle = πr2
= 22/7 × (24.5)2
= 22/7 × 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 = 22/7)

Ch 11 11.3

Solution: From the question it is 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 × (22/7) × 10.5
= 462/7
= 66 m
So, the length of rope required = 2 × 66 = 132 m
Cost of 1 m rope = ₹ 4 [given]
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: From the question it is 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: From the question it is 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 [given]
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.Ch 11 11.3

Solution: From the question it is 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
= (22/7) × 5
= 110/7
= 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: From the question it is 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 [given]
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 = 22/7)

Solution: From the question it is 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 × (22/7) × r
44 = 44/7 × r
(44 × 7)/44 = r
r = 7 cm
Area of the circle = πr2
= 22/7 × 72
= 22/7 × 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 = 44/4
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 = 22/7)

Ch 11 11.3

Solution: From the question it is 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
= 22/7 × 142
= 22/7 × 14 × 14
= 22 × 2 × 14
= 616 cm2
Area of the 2 small circles = 2 × πr2
= 2 × (22/7 × 3.52)
= 2 × (22/7 × 3.5 × 3.5)
= 2 × ((22/7) × 12.25)
= 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: From the question it is 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: From the question it is given that, Circumference of a circle = 31.4 cm We know that,
Circumference of a circle = 2πr
31.4 = 2 × 3.14 × r
31.4 = 6.28 × r
31.4/6.28 = 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)

Ch 11 11.3

Solution: From the question it is 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
Finally,
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: From the question it is 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
314/3.14 = 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)

Ch 11 11.3

Solution: From the figure, 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 = 22/7)

Solution: From the question it is given that, Radius of the wheel = 28 cm
Total distance = 352 m = 35200 cm
Circumference of the wheel = 2πr
= 2 × 22/7 × 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
= 352 m/176 cm
= 35200 cm/ 176 cm
= 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: From the question it is 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