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
Ex 15.3 Class 10 RBSE 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
Exercise 15.3 Class 10 RBSE 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
RBSE Class 10 Maths Chapter 15.3 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
Class 10 Maths RBSE Solution Chapter 15.3 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
RBSE Solutions For Class 10 Maths Chapter 15.3 In Hindi 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
15.3 Class 10 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
15.3 Class 10 RBSE 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
Exercise 15.3 Class 10 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
RBSE 10th Maths Solutions 15.3 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
RBSE Class 10 Maths Chapter 15 Solutions 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
RBSE Class 10 Maths Chapter 15 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).
RBSE Solutions For Class 10 Maths Chapter 15.3 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.