1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
8104911739

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

022-62211530

Mon to Sat - 11 AM to 8 PM

# R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 12 - Circles

Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 12 - Circles.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Exercise/Page

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 12 - Circles Page/Excercise MCQ

Solution 1

Correct option: (b)

Solution 2

Correct option: (c)

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 12 - Circles Page/Excercise 12A

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Let O be the centre of a circle with radius r.

OB = OC = r

Let AC = x

Then, AB = 2x

Let OM AB

OM = p

Let ON AC

ON = q

In ΔOMB, by Pythagoras theorem,

OB2 = OM2 + BM2

In ΔONC, by Pythagoras theorem,

OC2 = ON2 + CN2

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Draw OM PQ and O'N PQ

OM AP

AM = PM (perpendicular from the centre of a circle bisects the chord)

AP = 2AM ….(i)

And, O'N PQ

O'N AQ

AN = QN (perpendicular from the centre of a circle bisects the chord)

AQ = 2AN ….(ii)

Now,

PQ = AP + PQ

PQ = 2AM + 2AN

PQ = 2(AM + AN)

PQ = 2MN

PQ = 2OO'  (since MNO'O is a rectangle)

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 12 - Circles Page/Excercise 12B

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Given, AOD = 90° and OEC = 90°

AOD = OEC

But AOD and OEC are corresponding angles.

OD || BC and OC is the transversal.

DOC = OCE (alternate angles)

DOC = 30° (since OCE = 30°)

We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

DOC = 2DBC

Now, ABE = ABC = ABD + DBC = 45° + 15° = 60°

In ΔABE,

BAE + AEB + ABE = 180°

x + 90° + 60° = 180°

x + 150° = 180°

x = 30°

Solution 14

Construction: Join AC

Given, BD = OD

Now, OD = OB (radii of same circle)

BD = OD = OB

ΔODB is an equilateral triangle.

ODB = 60°

We know that the altitude of an equilateral triangle bisects the vertical angle.

Now, CAB = BDC (angles in the same segment)

CAB = BDE = 30°

Solution 15

Solution 16

We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

APB = 2ACB

Now, ACD is a straight line.

ACB + DCB = 180°

75° + DCB = 180°

DCB = 105°

Again,

Solution 17

Solution 18

Join AC and BC

We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

AOC = 2ABC ….(i)

Similarly, BOD = 2BCD ….(ii)

AOC + BOD = 2ABC + 2BCD

AOC + BOD = 2(ABC + BCD)

AOC + BOD = 2(EBC + BCE)

AOC + BOD = 2(180° - CEB)

AOC + BOD = 2(180° - [180° - AEC])

AOC + BOD = 2AEC

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17