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Class 9 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 12 - Circles

Circles Exercise MCQ

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Correct option: (b)

  

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Correct option: (c)

 

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Circles Exercise Ex. 12C

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AB is the common hypotenuse of ΔACB and ΔADB.

ACB = 90° and BDC = 90° 

ACB + BDC = 180° 

The opposite angles of quadrilateral ACBD are supplementary.

Thus, ACBD is a cyclic quadrilateral.

This means that a circle passes through the points A, C, B and D.

BAC = BDC (angles in the same segment) 

Solution 26

   

Construction: Take a point E on the circle. Join BE, DE and BD.

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.

BAD = 2BED

  

Now, EBCD is a cyclic quadrilateral.

BED + BCD = 180° 

BCD = 180° - BED

  

In ΔBCD, by angle sum property

CBD + CDB + BCD = 180° 

  

Circles Exercise Ex. 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)

Circles Exercise Ex. 12B

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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