RBSE Solutions for Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.3

RBSE Solutions for Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.3 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 15 Circumference and Area of a Circle Exercise 15.3.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 15
Chapter Name	Circumference and Area of a Circle
Exercise	Exercise 15.3
Number of Questions Solved	12
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.3

Question 1.
Find the circumference of circle (RBSESolutions.com) Inscribed in a square of sides 14 cm.
Solution :
Given
Side of square = 14 cm
Diameter of circle inscribed in square = side of square = 14 cm

Thus, radius of circle is r = \(\frac { 14 }{ 2 }\) = 7 cm
The circumference of circle = 2πr
= \(\frac { 2\times 22 }{ 7 } \times 7\) = 44 cm
Thus circumference of inscribed circle = 44 cm

Question 2.
The difference between radius and (RBSESolutions.com) circumference of a circle is 74 cm. Find area of this circle.
Solution :
Let r be the radius of circle
According to question
Circumference of circle – radius = 74

Area of circle = πr²
= \(\frac { 22 }{ 7 }\) × (14)² = \(\frac { 22\times 14\times 14 }{ 7 } \)
= 616 sq cm

Question 3.
In given figure 0 is center of circle, ∠AOB = 90° and OA = 3 cm area (RBSESolutions.com) of shaded part.

Solution :
Given :
In given figure 0 is center of circle and ∠AOB = 90°, radius of circle OA = 3 cm
Area of shaded part


Area of shaded part
= Area of sector OARB – Area of ΔOAB
= 7.07 – 4.5
= 2.57 sq cm
Thus, area of shaded part = 2.57 sq cm

Question 4.
If perimeter of a circle is equal to perimeter of a (RBSESolutions.com) square, then find ratio of their areas.
Solution :
Let radius of circle r cm
and side of square = a cm
According to question
Perimeter of circle = Perimeter of square
⇒ 2 πr = 4a
⇒ \(\frac { 7 }{ 5 }\) = \(\frac { 4 }{ 2\pi } \) = \(\frac { 2 }{ \pi } \)

Thus, required ratio = 14 : 11

Question 5.
The radius of a circular park is 3.5 m and it is (RBSESolutions.com) surrounded by 1.4 m broad footpath. Find the area of footpath.
Solution :
Let O is center of circle with center O and radius 3.5 m. Along the outsides of this circle 1.4 m broad path is made.

Thus, r₁ = 3.5 m
Radius of park with footpath r₂ = 3.5 + 1.4 = 4.9 m

Question 6.
Find the area of square inscribed in a (RBSESolutions.com) circle of radius 8 cm.
Solution :

Given:
Radius of circle r = 8 cm
So diameter of circle = 2 × r
= 2 × 8 = 16 cm
Square is inside the circle
Diameter of circle and diagonals of square will be same.
Thus, diagonal of square = 16 cm
But diagonal of square side \(\sqrt { 2 }\)
⇒ side × \(\sqrt { 2 }\) = 16

Question 7.
In given fig. ABMC is a quadrant of a circle of (RBSESolutions.com) radius 14 cm and a semicircle is drawn assuming BC is diameter. Find the area of shaded region.

Solution :
Radius (r) of quadrant ABMC = 14 cm


∴ Area of shaded part = area of semicircle – [area of quadrant ABMC – area of right angled triangle]
= [154 – (154 – 98)]
= 154 – 154 + 98
= 98 sq cm
Thus, area of shaded part 98 sq. cm

Question 8.
In given figure, AB is diameter of circle AC = 6 cm and BC = 8 cm, then (RBSESolutions.com) find area of shaded part.

Solution :
Given a circle with center O and AB its diameter
∵ Angle is semicircle is 90°
Thus, ∠ACB = 90°
By Pythagoras theorem

Area of right angled ∆ACB
= \(\frac { 1 }{ 2 }\) × base × height
= \(\frac { 1 }{ 2 }\) × 6 × 8
= 24 sq cm
Area of shaded part = Area of circle – area of triangle
= 78.57 – 24 = 54.57 sq cm
Thus area of shaded part = 54.57 sq cm

Question 9.
In given fig., find the area of shaded part (RBSESolutions.com) where ABCD is a square of side 10 cm and taking each side as diameter, semicircles and drawn. (π = 3.14)

Solution :
Given
Side of square = 10 cm.
in fig. let unshaped part are I, II, III and IV all these (RBSESolutions.com) part meet at point O in same manner.

= Area of square ABCD – [area of semicircle AOD] + [area of semicircle BOC]

Similarly area of part II + area of part IV
= 21.5 cm²
Thus area of shaded part = area of (RBSESolutions.com) square ABCD of – area of [I + II + III + IV] part
= (10 × 10) – (21.5 + 21.5)
= 100 – (2 × 21.5)
= 100 – 43 = 57 sq cm
Thus area of shaded part = 57 sq cm

Question 10.
In given figure, radius of semi-circle is 7 cm. Find the area of circle formed in semicircle.

Solution :
Radius of semicircle = 7 cm
Diameter of circle formed (RBSESolutions.com) in semicircle
= Radius of semicircle = 7 cm
Radius of circle formed is semicircle (r)
= \(\frac { diameter }{ 5 }\) = \(\frac { 7 }{ 2 }\)
= 3.5 cm
∴ Area of circle formed in semicircle = πr²
= \(\frac { 22 }{ 7 }\) × (3.5)²
= \(\frac { 22 }{ 7 }\) × (3.5) × (3.5)
= 11 × 3.5 = 38.5 sq cm
Thus area of circle formed in semicircle = 38.5 sq cm

Question 11.
The sum of circumference of two circles of radius R₁ and R₂ is equal to the (RBSESolutions.com) circumference of circle of radius R, then correct option is :
(A) R₁ + R₂ = R
(B)R₁ + R₂ > R
(C)R₁ + R₂ < R
(D) nothing is definite
Solution :
Correct option is (A).

Question 12.
The circumference of circle inscribed In a (RBSESolutions.com) square of side 14cm will be:
(A)22 cm
(B)44 cm
(C) 33 cm
(D) 55 cm
Solution :
Side of square ABCD = 14 cm

Radius of circle inscribed (RBSESolutions.com) in a square = \(\frac { 14 }{ 2 }\) cm
circumference = 2πr
= \(\frac { 22 }{ 7 }\) × 2 × 7 = 44 cm
Thus, option (B) is correct

We hope the given RBSE Solutions for Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 15 Circumference and Area of a Circle Exercise 15.3, drop a comment below and we will get back to you at the earliest.