RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 7 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 7. Students can also read RBSE Class 7 Maths Important Questions for exam preparation. Students can also go through RBSE Class 7 Maths Notes to understand and remember the concepts easily. Students can access the data handling class 7 extra questions with answers and get deep explanations provided by our experts.

RBSE Class 7 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
Answer:
Circumference = 2πr
(a) r = 14 cm,
circumference = 2 × \(\frac{22}{7}\) × 14 cm = 88 cm

(b) r = 28 mm,
circumference = 2 × \(\frac{22}{7}\) × 28 cm = 176 mm

(c) r = 21 cm,
circumference = 2 × \(\frac{22}{7}\) × 21 cm = 132 cm

RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

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
Answer:
∵ Area of circle = πr2

(a) radius (r) = 14 cm
A = \(\frac{22}{7}\) × 14 × 14 = 616 cm2

(b) diameter (d)= 49 m, r = \(\frac{49}{2}\) m
A = \(\frac{22}{7} \times \frac{49}{2} \times \frac{49}{2}\) = 1886.5 m2

(c) radius = 5 cm
(A) = \(\frac{22}{7}\) × 5 × 5 = 78.57 m2.

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet.
(Take π = \(\frac{22}{7}\))
Answer:
Circumference = 154 m
r = \(\frac{\text { Circumference }}{2 \pi}\)
= \(\frac{154 \times 7}{2 \times 22}\) = \(\frac{49}{2}\)
= 24.5 m
A = πr2 = \(\frac{22}{7}\) × 24.5 × 24.5 = 1886.5 m2
So, the radius of circular sheet is 24.5 m and its area is 1886.5 m2.

RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

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 costs of the rope, if it cost ₹ 4 per metre.
(Take π = \(\frac{22}{7}\))
Answer:
Diameter of circular garden = 21 m
Length of rope required to fence
= 2πr = 2 × \(\frac{22}{7}\) × \(\frac{21}{2}\) = 66m
Length of rope to make 2 round of fence
= 2 × 66 m = 132 m
∵ Cost of 132 m 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)
Answer:
Radius of outer circular sheet = 4 cm
Radius of inner circular sheet = 3 cm
Area of remaining sheet = π(42 - 32)
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 1
= π(16 - 9) cm2 = 3.14 × 7 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 metre of the lace costs ₹ 15. (Take π = 3.14)
Answer:
Diameter of the table cover = 1.5 m
⇒ Radius of the table cover (r) = \(\frac{1.5}{2}\)
Circumference of the table cover = 2πr
= 2 × 3.14 × m = 4.71 m.
Hence, Length of the lace required = 4.71 m
∵ Cost of lace per metre = ₹ 15
∴ Cost of the lace = ₹ 4.71 × 15
= ₹ 70.65.

RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

Question 7.
Find the perimeter of the adjoining figure, which is a semi-circle including its diameter.
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 2
Answer:
Radius = \(\frac{10 \mathrm{~cm}}{2}\) = 5 cm
Perimeter of figure = πr + d
= 3.14 × 5 + 10
= 15.7 + 10 cm
= 25.7 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)
Answer:
Diameter = 1.6 m
Radius = \(\frac{1.6}{2}\)m = 0.8 m
Area of circular table-top = πr2
= 3.14 × 0.8 × 0.8
= 2.0096 m2
Cost of polishing 1 m2 = ₹ 15
Cost of polishing 2.0096 m2 = ₹ 15 × 2.0096 = ₹ 30.144
So, the cost of polishing circular table-top is ₹ 30.14.

RBSE Solutions for Class 7 Maths 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 = \(\frac{22}{7}\))
Answer:
Length of wire =44 cm
Since, the wire is bent in the shape of a circle of circumference = 44 cm
r = \( \frac{\text { Circumference }}{2 \pi}\) = \(\frac{44 \times 7}{2 \times 22}\) = 7cm
Area = πr2 = \(\frac{22}{7}\)) × 7 × 7 = 154 cm2
Since, the same wire is rebent into a squre,
Side of square = \(\frac{\text { Perimeter }}{4}\) = \(\frac{44}{4}\) cm = 11 cm
Area of square = (Side)2 = (11 cm)2 = 121 cm2
∴ The circle encloses more area.

Question 10.
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and rectangle of length 3 cm and breadth 1 cm are removed (as shown in the adjoning figure). Find the area of the remaining sheet.
(Take π = \(\frac{22}{7}\))
Answer:
Radius of circular card sheet = 14 cm
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 3
Area = πr2 = \(\frac{22}{7}\) × 14 × 14cm2
= 616 cm2
Radius of 1 circle = 3.5 cm
Area = πr2 = \(\frac{22}{7}\) × 3.5 × 3.5
= 38.8 cm2
Area of 2 circle = 2 × 38.5 cm2 = 77 cm2
Area of rectangle = l × b
= 3 cm × 1 cm = 3 cm2
Area to be removed = Area of 2 circles + Area of rectangle = 77 cm2 + 3 cm2 = 80 cm2
Area of remaining sheet = Area of circle - Area to be removed
= 616 cm2 - 80 cm2 = 536 cm2
So, the area of remaining sheet is 536 cm2.

RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheetofside6cm. What is the area of the leftover aluminium sheet? (Take π = 3.14)
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 4
Answer:
Area of square = (Side)2
= (6 cm)2 = 36 cm2
Area of circle = πr2 = 3.14 × 2 × 2
= 12.56 cm2
Area leftover = Area of square - Area of circular cut-out
= 36 cm2 - 12.56 cm2
= 23.44 cm2
So, the area of the leftover aluminium sheet is 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)
Answer:
Circumference of circle = 31.4
Radius = \(\frac{\text { Circumference }}{2 \pi}\) = \(\frac{31.4}{2 \times 3.14}\) = 5cm
Area = πr2 = 3.14 × 5 × 5 = 78.5 cm2.

Question 13.
A circular flower bed is surrounded by a path 4m wide. The diameter of the flower bed is 66 m. What is the area of this path? (Take π = 3.14)
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 5
Answer:
Diameter of flower bed = 66 m
Radius of flower bed = \(\frac{\text { Diameter }}{2}\)
= \(\frac{66 \mathrm{~m}}{2}\) = 33 m
Width of path = 4 m
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 6
Radius of path = 33m + 4m = 37m
Area of path = Area of flower bed with path - Area of flower bed
= π(37)2 - π(33)2
= π(372 - 332) = π(1369 - 1089)
= 3.14 × 280 = 879.2 m2.
So, the area of path is 879.2 m2.

RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

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)
Answer:
Area of circular flower garden = 314 m2
Radius of sprinkler = 12 m
Area covered by sprinkler = πr2
= 3.14 × (12)2 m2
= 3.14 × 144 m2
= 452.16 m2
Since, 452.16 > 314
∴ The sprinkler will water the entire garden.

Question 15.
Find the circumference of the inner and the outer circles, shown in the adjoining figure. (Take π = 3.14)
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 7
Answer:
Outer radius = 19 m
Outer circumference = 2πr
= 2 × 3.14 × 19 m
= 119.32 m
Inner radius = 19m - 10m = 9m
Inner circumference = 2πr
= 2 × 3.14 × 9 m
= 56.52 m.

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m?
(Taken π = \(\frac{22}{7}\))
Answer:
Radius of wheel = 28 cm
Distance travelled by wheel in 1 rotation = 2πr
= 2 × \(\frac{22}{7}\) × 28 = 176 cm
Total distance covered by wheel
= 352 m = 352 × 100 cm = 35200 cm
No. of rounds
= \(\frac{\text { Total distance to be covered }}{\text { Distance covered in } 1 \text { round }}\)
= \(\frac{35200}{176}\) = 200 times
So, the wheel must rotate 200 times to go 352 m.

RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

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?
Answer:
Radius of circular clock =15 cm
Distance covered by minute hand in 1 hr
= 2πr = 2 × 3.14 × 15 cm = 94.2 cm
So, the tip of the minute hand will move 94.2 cm in 1 hour.

Bhagya
Last Updated on June 15, 2022, 3:28 p.m.
Published June 15, 2022