# 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 = 2∠BED

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,

OB^{2}
= OM^{2} + BM^{2}

In ΔONC, by Pythagoras theorem,

OC^{2}
= ON^{2} + CN^{2}

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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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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 = 2∠DBC

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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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 = 2∠ACB

Now, ACD is a straight line.

⇒ ∠ACB + ∠DCB = 180°

⇒ 75° + ∠DCB = 180°

⇒ ∠DCB = 105°

Again,

### Solution 17