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 = πr^{2}

= \(\frac { 22 }{ 7 }\) × (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, r_{1} = 3.5 m

Radius of park with footpath r_{2} = 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 cm^{2}

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 = πr^{2}

= \(\frac { 22 }{ 7 }\) × (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 R_{1} and R_{2} is equal to the (RBSESolutions.com) circumference of circle of radius R, then correct option is :

(A) R_{1} + R_{2} = R

(B)R_{1} + R_{2} > R

(C)R_{1} + R_{2} < 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.

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