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

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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.

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Exercise/Page

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

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Correct option: (b)

  

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Correct option: (c)

 

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

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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

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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

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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° 

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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,

  

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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)

Adding (i) and (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

  

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

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