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 part of NCERT Solutions for Class 7 Maths. Here we have given 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

Question 1.
Find the circumference of the circles with the following radus ( Take π =  )
(a) 14 cm
(b) 28 mm
(c) 21 cm.
Solution:

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

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet ( Take π =  )
Solution:
Circumference of the circular sheet = 154 m
Let the radius of the circular sheet be r cm
Then, its circumference = 2nr m According to the question,
Circumference = 2πr = 154

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 meter. ( Take π =  )
Solution:
Diameter of the circular garden (r) = 21 m
= Radius of the circular garden (r) =  m
∴ Circumference of the circular garden = 2πr
= 2 ×  ×  m = 66m
⇒ Length of the rope needed to make 1 round of fence = 66 m
⇒ Length of the rope needed to make 2 rounds of fence
= 66 × 2 m = 132 m
Cost of rope per meter = ₹ 4
∴ Cost of the rope = ₹ 132 × 4 = ₹ 528.

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 outer circle (R) = 4 cm

Area of the outer circle
= πR2 = 3.14 (4)² cm²
= 3.14 × 16 cm² = 50.24 cm².
Radius of inner circle (r) = 3 cm
∴ Area of the inner circle
= πr² = 3.14 × (3)² cm² = 28.26 cm²
∴ Area of the remaining sheet = Area of the outer circle – Area of the inner circle
= 50.24 cm² – 28.26 cm² = 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:
Diameter of the table cover = 1.5 m
⇒ Radius of the table cover (r) =  m
⇒ Circumference of the table cover = 2πr
= 2 × 3.14 ×  m = 4.71 m
⇒ Length of the lace required = 4.71 m
∵ Cost of lace per meter = ₹ 15
∴ Cost of the lace = ₹ 4.71 × 15 = ₹ 70.65

Question 7.
Find the perimeter of the following figure, which is a semicircle including its diameter.
Solution:

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 of the table-top = 1.6 m
⇒ Radius of the table-top (r) =  m = 0.8 m
∴ Area of the table-top
= πr2 = 3.14 × (0.8)2 m2
= 3.14 × 0.64 m2
= 2.0096 m2
∵ Rate of polishing = ₹ 15 per m2
∴ Cost of polishing the table-top
= ₹ 2.0096 × 15 = ₹ 30.144 = ₹ 30.14 (approx.).

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 side? Which figure encloses more area, the circle or the square? ( Take π =  )
Solution:

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 following figure). Find the area of the remaining sheet. ( Take π = )

Solution:

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:

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:

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)

Solution:

Question 14.
A circular flower garden has an area of about 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? (Taken π = 3.14)
Solution:
Radius of the area covered by the sprinkler at the centre of the garden (r) = 12 cm
∴ Area covered by the sprinkler at the centre of the garden
= πr2 = 3.14(12)² m²
= 452.16 m² (> 314 m²)
Hence, the sprinkler will water the entire garden.

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

Solution:
Radius of inner circle = 19 – 10 = 9 m
∴ Circumference of the inner circle = 2 πr = 2 × 3.14 × 9 m = 56.52 cm
Radius of the outer circle = 19 m
∴ Circumference of the outer circle = 2πr = 2 × 3.14 × 19 m = 119.32 m.

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

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:
Length of the minute hand of the circular clock = 15 cm
⇒ Radius of the circular clock (r) = 15 cm
∴ Circumference of the circular clock
= 2πr = 2 × 3.14 × 15 cm = 94.2 cm
⇒ Length moved by the tip of the minute hand in 1 hour = 94.2 cm
[As the minute hand makes one complete revolution round the clock in 1 hour].

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