RBSE Solutions for Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.2 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 15.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 15 |

Chapter Name |
Circumference and Area of a Circle |

Exercise |
Exercise 15.2 |

Number of Questions Solved |
9 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.2

Question 1.

Radius of a circle is 7 cm and angle (RBSESolutions.com) subtended at the center is 600. Find the length of arc.

Solution :

Given :

Radius of circle (r) = 7 cm

Angle at center = 60°

We know that length of sector

Thus, length of an arc = 7.3 cm

Question 2.

The radius of a circle is 10.5 cm and angle of sector is 45°. Find the area (RBSESolutions.com) of minor sector. \(\left( \pi =\frac { 22 }{ 7 } \right) \)

Solution :

Radius of circle (r) = 10.5 cm

And angle of section = 45°

We know that area of minor sector

Thus, area of minor sector = 43.31 sq cm

Question 3.

The length of an are of circle ¡s 12 cm and (RBSESolutions.com) radius is 7 cm. Find the area of minor sector.

Solution :

Length of arc of circle = 12 cm

and Radius of circle (r) = 7 cm

We know that if length of arc (L) and radius of circle (r) then arc of minor sector

A = \(\frac { 1 }{ 2 }\) × L × r

= \(\frac { 1 }{ 2 }\) × 12 × 7

= 42 sq cm

Thus, area of minor sector = 42 sq cm

Question 4.

A chord of circle of radius 21 cm. subtends (RBSESolutions.com) an angle of 600 at center. Find

(i) Length of arc

(ii) Area of sector formed by arc

(iii) Area of segment formed by corresponding chord.

Solution :

Given :

Radius of circle (r) = 21 cm

Angle subtended by an arc at center (θ)

= 60°

(i) Length of arc (L)

Thus length of arc = 22 cm

(ii) Area of sector (RBSESolutions.com) formed by arc

Thus required arc of sector = 231 sq cm

(iii) Area of segment formed by corresponding chord

Question 5.

A minute hand of a clock is 10.5 m long. Find the (RBSESolutions.com) area of sector formed by minute hand in 10 minutes.

Solution :

Length of minute hand (r) = 10.5 cm

∵ minute hand subtends an angle of 360° in 60 minutes

∴ angle subtended by min hand in 10 minutes = 6 × 10 = 60°

Thus r = 10.5 cm, θ = 60°

Area of sector

= \(\frac { { \pi r }^{ 2 }\theta }{ { 360 }^{ \circ } }\)

Thus required area of sector = 57.75 sq cm

Question 6.

A chord of circle of radius 3.5 cm subtends an angle of 90° at center. Find the (RBSESolutions.com) area of minor segment formed by chord.

Solution :

Radius of circle (r) = 3.5 cm.

Angle subtended by chord at center = 90°

∴ Area of corresponding minor segment

Thus area of minor segment by chord = 3.5 sq cm

Question 7.

Find the area of fourth quadrant of a (RBSESolutions.com) circle whose circumference is 22 cm.

Solution :

Circumference of circle = 22 cm

⇒ 2πr = 22

r = \(\frac { 22 }{ 2\pi } \) = \(\frac { 22\times 7 }{ 2\times 22 }\)

= \(\frac { 7 }{ 2 }\) cm

Area of fourth quadrant of circle

Thus area of fourth quadrant of circle = 9.625 sq cm

Question 8.

An hour hand of a clock is 5 cm long. Find the (RBSESolutions.com) area formed by this hand in 7 minutes.

Solution :

Length of hour hand (r) = 5 cm

So. hour hand formed a sector of radius 5 cm.

Angle formed by hour hand in 12 hours = 360°

Angle formed by hour hand in 1 hours = \(\frac { { 360 }^{ \circ } }{ 12 } \) = 30°

Angle formed by hour hand in 1 min = \(\frac { { 30 }^{ \circ } }{ { 60 }^{ \circ } } \) = \({ \left( \frac { 1 }{ 2 } \right) }^{ \circ }\)

Thus angle formed by hour hand in 7 min.

= \({ \left( \frac { 1 }{ 2 } \right) }^{ \circ }\) × 7 = \({ \left( \frac { 7 }{ 2 } \right) }^{ \circ }\)

Area of sector formed by hour hand

Question 9.

In given figure, ABCD is a rectangle. Side AB = 10 cm and BC = 7 cm. Circle of (RBSESolutions.com) radius 3.5 cm are drawn at the vertices of rectangle. Find the area of the shaded part. \(\left( \pi =\frac { 22 }{ 7 } \right) \)

Solution:

Length of rectangle = AB = 10 cm

And width of rectangle = BC = 7 cm

And circles are drawn at the vertices of rectangle of radius 3.5 cm

∵ each angle of rectangle is 90°

Thus area of sector of one vertex

Thus area of four vertices = \(\frac { 38.5 }{ 4 }\) × 4 = 38.50 sq cm

Thus area of shaded part = Area of rectangle – Area of circles at (RBSESolutions.com) vertices rectangle

= 10 × 7 – 38.50

= 70 – 38.50

= 31.50 sq cm

Thus area of shaded part = 31.50 sq cm

We hope the given RBSE Solutions for Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 15 Circumference and Area of a Circle Exercise 15.2, drop a comment below and we will get back to you at the earliest.

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