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

## Circles Exercise MCQ

### Solution 1

Correct option: (b)

### Solution 2

Correct option: (c)

## Circles Exercise Ex. 12C

### Solution 25

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.

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

### 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 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

### 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 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,