RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area Ex 16.3

RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area Ex 16.3 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Exercise 16.3.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 16
Chapter Name	Perimeter and Area
Exercise	Ex 16.3
Number of Questions	16
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Ex 16.3

Question 1
Find the circumference of circle having (RBSESolutions.com) following radius (π = \(\frac { 22 }{ 7 }\))
(i) 21 cm
(ii) 28 cm
(iii) 10.5 cm
Solution:
(i) Radius of circle = 21 cm
Circumference of Circle = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 21 = 2 x 22 x 3 = 44 x 3
= 132 cm

(ii) Radius of circle = 28 mm
circumference = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 28 = 2 x 22 x 4 = 2 x 88
= 176cm

(iii) Radius of circle = 10.5 cm
Circumference of circle = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 10.5
= 2 x 22 x 1.5
= 44 x 1.5 = 66 cm.

Question 2
Find the area of the (RBSESolutions.com) following circles. Given
(i) Radius = 5 cm
(ii) Diameter = 42 metre
(iii) Radius = 5.6 cm
Solution:
(i) Radius (r)= 5 cm
Area of circle = πr²
= 3.14 x (5)² = 3.14 x 25 = 78.57 sq. cm

(ii) Diameter (d) = 42 m
Radius (r) = \(\frac { 42 }{ 2 }\) = 21 m
Area of circle = πr²
= \(\frac { 22 }{ 7 }\) x 21 x 21 = 22 x 3 x 21
= 66 x 21
= 1386 sq. cm

(iii) Radius (r) = 5.6 cm
Area of circle = πr²
= \(\frac { 22 }{ 7 }\) x 5.6 x 5.6 = 22 x 0.8 x 5.6 = 98.56 sq. cm.

Question 3
Find the radius of a circular sheet (RBSESolutions.com) whose perimeter is 132 metre. Also find its area (π = \(\frac { 22 }{ 7 }\))
Solution:
Circumference of circle = 132 m
Let the radius of circle is r metre
Circumference = 132 ⇒ 2πr = 132
⇒ 2 x \(\frac { 22 }{ 7 }\) x r = 132
⇒ r = \(\frac { 132\times 7 }{ 22\times 2 }\) = 21 cm
Area of sheet = πr²
= \(\frac { 22 }{ 7 }\) x 21 x 21 = 22 x 3 x 21
= 66 x 21
= 1386 sq. m

Question 4
Circumference of a circle is 44 cm. Find the (RBSESolutions.com) radius and area of triangle. (π = \(\frac { 22 }{ 7 }\))
Solution:
Let radius of cirlce r cm
Circumference of circle = 44
⇒ 2πr = 44
⇒ 2 x \(\frac { 22 }{ 7 }\) x r = 44
r = \(\frac { 44\times 7 }{ 22\times 7 }\) = 7 cm
∴ Area of circle = πr²
= \(\frac { 22 }{ 7 }\) x 7 x 7 = 22 x 7 = 154sq. cm

Question 5
The given figure is a semi circle with 12 cm. diameter. Find out (RBSESolutions.com) its circumference.

Solution:
Diameter (d)
Radius r = \(\frac { 12 }{ 2 }\) = 6 cm
Perimeter of half circle = πr + diameter
= 3.14 x 6 + 12 (∵ π = 3.14)
= 30.84 cm.

Question 6
The radius of a circular pond is 28 meter. 4 path of 1.4 meter width is (RBSESolutions.com) present around it Find the area of path.
Solution:
Radius of circular pond (r) = 28 m
Radius of circular pond including path
(R) = 28 + 1.4
= 29.4 m

Area of path
= πR² – 4²
= π(R² – r²)
= π(29.4² – 28²)
= \(\frac { 22 }{ 7 }\)(864.36 – 784)
= \(\frac { 22 }{ 7 }\) x 80.36
= 252.56 sq. m

Question 7
Area of a circle is 616 square cm. The circle is (RBSESolutions.com) surrounded by a path 2 meter wide. What is the area of this path?
Solution:
Area of path = 616 sq. cm
Let radius of circle is r cm
Area of circle = 616 sq. cm
⇒ πr²= 616
⇒ \(\frac { 22 }{ 7 }\) x r² = 616
⇒ r² = \(\frac { 616\times 7 }{ 22 }\)
⇒ r² = 28 x 7
⇒ r² = 196
⇒ r = 14 cm

After 2 cm wide road let the radius is
R = r + 2
R = 14 + 2 = 16 cm
Area of path = = πR² – πr²
= π(R² – r²) = \(\frac { 22 }{ 7 }\) x (16² – 14²)
= \(\frac { 22 }{ 7 }\) x (256 – 196) = \(\frac { 22 }{ 7 }\) x 60
= 188.57 sq. cm.

Question 8
From a circular card sheet of radius 5 cm, a circular sheet (RBSESolutions.com) of radius of 4 cm is removed. Find the area of the remaining sheet. (π = 3.14)
Solution:
Area of sheet (R) = 5 cm
Radius of circle (r) = 4 cm
Area of remaining part = = πR² – πr²
= π(R² – r²)
= 3.14(5² – 4²)
= 3.14(25 – 16)
= 3.14 x 9
= 28.26 sq. cm

Question 9
From a circular card sheet of radius 14 cm, a square of 4 cm is (RBSESolutions.com) removed as shown in the adjoining figure. Find the area of the remaining sheet
(π = \(\frac { 22 }{ 7 }\))

Solution:
Area of remining sheet = Area of sheet – Area of square
= πr² – 4² = \(\frac { 22 }{ 7 }\) x 14 x 14 – 16
[∵ r = 14 cm]
= 22 x 2 x 14 – 16
= 616 – 16
= 600 sq. cm

Question 10
The ratio of diameter of two circles is 4 : 5 then (RBSESolutions.com) Find out ratio of their areas.
Solution:
Let the diameter are : d₁ and d₂
d₁: d = 4 : 5 or \(\frac { { d }_{ 1 } }{ { d }_{ 2 } }\) = \(\frac { 4 }{ 5 }\)
Let the radius of circles are r₁ and r₂ and their areas are A₁ AND A₂ then,
r₁ = \(\frac { { d }_{ 1 } }{ 2 }\), r = \(\frac { { d }_{ 2 } }{ 2 }\)

Question 11
Durga want to polish a circular table top of diameter 2.8 metre. Find (RBSESolutions.com) the cost of polishing if the rate of polishing is ₹25 per square meter.
Solution:
Diameter of table = 2.8 m
Radius of table (r) = \(\frac { 2.8 }{ 7 }\) = 1.4 m
Area of table = πr²
= \(\frac { 22 }{ 7 }\) x 1.4 x 1.4 = 22 x 0.2 x 1.4
= 4.4 x 1.4 = 6.16 sq. m
cost of polishing per sq.m = ₹25
6.16 वर्ग मीटर टेबल पर पॉलिश कराने का खर्चा = 25 x 6.16 = ₹154

Question 12
Gopi ties his horse with a rope of length 12 m. What (RBSESolutions.com) area of grass can the horse graze upon ?
Solution:
Radius of rope (r) = 12 m
Area of land grazed by horse = πr²
= 3.14 x (12)² = 3.14 x 144
= 452.16 sq.m

Question 13
In the given figure, two semicircular parts of diameter 12 cm, are add on (RBSESolutions.com) both the ends of a rectangular part Length of the part is 15 cm. Find out the area ?

Solution:
Diameter of table = 12 cm
Radius (r) = \(\frac { 12 }{ 2 }\) = 6 cm
Area of ABCD
= length x breadth = 15 x 12 = 180 sq. cm
Area of two semi circles = 2 x \(\frac { 1 }{ 2 }\) πr²
= 3.14 x 6 x 6 = 3.14 x 36 = 113.04sq. cm
Area of whole shape = Area of rectangle + Area of two semi circle
= 180+113.04 = 293.04 sq. cm.

Question 14
How many rounds does a wheel of radius 35 take in order to cover (RBSESolutions.com) a distance of 880 metre? (π = \(\frac { 22 }{ 7 }\))
Solution:
Radius of wheel (r) = 35 m
Circumference of wheel = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 35 = 2 x 22 x 5 = 44 x 5
= 220 m
Wheel covers 220 metre distance revolution.
So to cover a distance of 880 metre number of revolution = \(\frac { 880 }{ 220 }\) = 4 revolution.

Question 15
Parvat has spend how much money to pour the soilalla (RBSESolutions.com) round it circular shape garden in 7 m wide road. It the diameter of garden is 56 metre? (π = \(\frac { 22 }{ 7 }\))

Solution:
Diameter of park = 56 m 56
Radius (r) = \(\frac { 56 }{ 2 }\) = 28 m
The radius of (RBSESolutions.com) park including path (R) = 28 + 7
= 35 m
Area of path = = πR² – πr²
= π(35² – 28²) = \(\frac { 22 }{ 7 }\) x (35 + 28) (35 – 28)
= \(\frac { 22 }{ 7 }\) x 63 x 7 = 22 x 63 = 1386 sq.m
Expenditure of putting soil = Area of road x Expenditure per sq. m
= 1386 x 11 = ₹15,246

Question 16
Length of minute arm of circular shape clock is 20 cm. How (RBSESolutions.com) much distance does tip of minute hand cover in one hour = π = 3.14.
Solution:
Length of minute hand = 20 cm
r = 20 cm
So the distance covered by minute hand is one completer revolution
= 2πr = 2 x 3.14 x 20 = 125.6 cm

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