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# RD Sharma Solution for Class 9 Mathematics Chapter 15 - Circles

Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 15 - 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 RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics 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

## RD Sharma Solution for Class 9 Mathematics Chapter 15 - Circles Page/Excercise 15.1

Solution 1

(i) interior/exterior

(ii) concentric

(iii) the exterior

(iv) arc

(v) diameter

(vi) semi-circle

(vii) centre

(viii) three

Solution 2

(i) T

(ii) T

(iii) T

(iv) F

(v) T

(vi) T

(vii) F

(viii) T

## RD Sharma Solution for Class 9 Mathematics Chapter 15 - Circles Page/Excercise 15.2

Solution 1

Solution 2

Solution 3

Solution 4

Steps of construction:

(1) Take three point A, B and C on the given circle.

(2) Join AB and BC.

(3) Draw the perpendicular bisectors of chord AB and BC which interesect each other at O.

(4) Point O will be the required circle because we know that the perpendicular bisector of a chord always passes through the centre.

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Distance of smaller chord AB from centre of circle = 4 cm.
OM = 4 cm

In OMB In OND OD=OB=5cm             (radii of same circle)   So, distance of bigger chord from centre is 3 cm.

Solution 12

Solution 13

Solution 14

Suppose two different circles can intersect each other at three points then they will pass through the three common points but we know that there is one and only one circle with passes through three non-collinear points, which contradicts our supposition.

Hence, two different circles cannot intersect each other at more than two points.

Solution 15

Draw OM  AB and ON  CD. Join OB and OD                                              (Perpendicular from centre bisects the chord) Let ON be x, so OM will be 6 - x
In MOB In NOD   We have OB = OD             (radii of same circle)
So, from equation (1) and (2)   From equation (2)   So, radius of circle is found to be  cm.

Solution 1

Solution 2

Solution 1

Solution 2

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 3(v)

Solution 3(vi)

Solution 3(vii)

Solution 3(viii)

Solution 3(ix)

Solution 3(x)

Solution 3(xi)

Solution 3(xii)

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

## RD Sharma Solution for Class 9 Mathematics Chapter 15 - Circles Page/Excercise 15.5

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

Solution 18(i)

Solution 18(ii)

Solution 18(iii)

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Let O be the centre of the circle circumscribing the cyclic rectangle ABCD. Since ABC = 90o and AC is a chord of the circle, so, AC is a diameter of the circle. Similarly, BD is a diameter.

Hence, point of intersection of AC and BD is the centre of the circle.

Solution 26

Solution 27

Solution 28

Solution 29

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

## RD Sharma Solution for Class 9 Mathematics Chapter 15 - Circles Page/Excercise 15.111

Solution 16

Solution 17

Solution 18

The greatest chord of the circle is diameter of the circle.

Hence, correct option is (c).

Solution 19

Angle formed in a minor segment is always a obtuse angle.

Hence, correct option is (b).

Solution 20

Three non-collinear points make a triangle and there is only one circle that can pass through all three points,

i.e. circumcircle of that triangle.

Hence, correct option is (a).

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

## Browse Study Material

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 15 - Circles for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 15 - Circles.

# Text Book Solutions

CBSE IX - Mathematics

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