MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3

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) r = 14 cm
∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 14 =88 cm
(b) r = 28 mm
∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 28 = 176 mm
(c) r = 21 cm
∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 21 = 132 cm

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) r = 14 mm
MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 1

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:
Circumference = 2πr = 154 m
MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 2

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 of fence. Also find the cost of the rope, if it costs ₹ 4 per meter. (Take π = \(\frac{22}{7}\))
Solution:
Diameter (d) = 21 m 21
∴ Radius (r) = \(\frac{21}{2}\)m
Circumference = 2πr = 2 × \(\frac{22}{7} \times \frac{21}{2}\) = 66 m
Length of rope required for fencing = 2 × 66 m = 132 m
Cost of 1 m rope = ₹ 4
Cost of 132 m rope = 4 × 132 = ₹ 528

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3

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:
Outer radius of circular sheet (R) = 4 cm
Inner radius of circular sheet (r) = 3 cm
Remaining area = πR2 – πr2
= 3.14 × 4 × 4 – 3.14 × 3 × 3
= 50.24 – 28.26 = 21.98 cm2

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 meter of the lace costs ₹ 15. (Take π = 3.14)
Solution:
The length of the lace required = circumference of circular table
Circumference = 2πr = 2 × 3.14 × \(\frac{d}{2}\)
= 2 × 3.14 × \(\frac{1.5}{2}\) = 4.71 m
Cost of 1 m lace = ₹ 15
Cost of 4.71 m lace = 4.71 × 15 = ₹ 70.65

Question 7.
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 3
Solution:
Diameter = 10 cm
Radius = \(\frac{10}{2}\) = 5 cm
Circumference of semicircle = \(\frac{2 \pi r}{2}\)
= 2 × \(\frac{1}{2} \times \frac{22}{7}\) × 5 = 15.71 cm
Total perimeter = Circumference of semicircle + Length of diameter
= 15.71 + 10 = 25.71 cm

Question 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:
Diameter = 1.6 m
∴ Radius = \(\frac{1.6}{2}\) = 0.8 m
Area = πr2 = 3.14 × 0.8 × 0.8 = 2.0096 m2
Cost for polishing 1 m2 area = ₹ 15
Cost for polishing 2.0096 m2 area
= 15 × 2.0096 = ₹ 30.14
Therefore, it will cost ₹ 30.14 for polishing circular table.

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3

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:
If the wire is bent into a circle, then the length of wire = circumference of the circle
⇒ 2πr = 44 cm
⇒ 2 × \(\frac{22}{7}\) × r = 44
⇒ r = 7 cm
Area = πr2= \(\frac{22}{7}\) × 7 × 7 = 154 cm2
If the wire is bent into a square, then the length of the wire = perimeter of the square
⇒ 4 × side = 44cm ⇒ side = \(\frac{44}{4}\) = 11 cm
Area of square = (11)2 = 121 cm2
As 154 > 121,
Therefore, circle encloses more area.

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 following figure). Find the area of the remaining sheet. (Take π = \(\frac{22}{7}\))
MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 4
Solution:
Area of bigger circle = \(\frac{22}{7}\) × 14 × 14 = 616 cm2
Area of 2 small circles = 2 × πr2
= 2 × \(\frac{22}{7}\) × 3.5 × 3.5 = 77 cm2
Area of rectangle = Length × Breadth = 3 × 1
= 3 cm2
Area of remaining sheet = Area of bigger circle – (Area of 2 small circles + Area of rectangle)
= 616 – (77 + 3) = 536 cm2

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:
Area of square-shaped sheet = (Side)2
= (6)2 = 36 cm2
Area of circle = 3.14 × 2 × 2= 12.56 cm2
Area of remaining sheet = Area of square sheet – area of circle
= 36 – 12.56 = 23.44 cm2

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:
Let r be the radius of circle. Circumference = 2πr = 31.4 cm
⇒ 2 × 3.14 × r = 31.4 cm
⇒ r = 5 cm
Area = 3.14 × 5 × 5 = 78.50 cm2

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? (π = 3.14)
MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 5
Solution:
Radius of flower bed = \(\frac{66}{2}\) = 33 m
Width of the path = 4 m
Radius of flower bed and path together = 33 + 4 = 37 m
Area of flower bed and path together
= 3.14 × 37 × 37 = 4298.66 m2
Area of flower bed = 3.14 × 33 × 33 = 3419.46 m2
Area of path = Area of flower bed and path together – Area of flower bed
= 4298.66 – 3419.46 = 879.20 m2

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 water the entire garden? (Take π = 3.14)
Solution:
Area = πr2 = 314 m2
3.14 × r2 = 314 ⇒ r2 = 100 ⇒ r = 10 m
Yes, the sprinkler will water the whole garden.

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3

Question 15.
Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)
MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 6
Solution:
Radius of outer circle = 19 m
Circumference = 2πr =2 × 3.14 × 19 = 119.32 m
Radius of inner circle = 19 – 10 = 9 m
Circumference = 2πr = 2 × 3.14 × 9 = 56.52 m

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = \(\frac{22}{7}\))
Solution:
r = 78 cm
MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 7
Therefore, it will rotate 200 times.

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:
Distance travelled by the tip of minute hand = Circumference of the clock
= 2πr = 2 × 3.14 × 15 = 94.2 cm

MP Board Class 7th Maths Solutions

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