# NCERT solution class 9 chapter 10 Circles exercise 10.5 mathematics

## EXERCISE 10.5

#### Question 1:

In the given figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

It can be observed that

∠AOC = ∠AOB + ∠BOC

= 60° + 30°

= 90°

We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.

#### Question 2:

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

In ΔOAB,

AB = OA = OB = radius

∴ ΔOAB is an equilateral triangle.

Therefore, each interior angle of this triangle will be of 60°.

∴ ∠AOB = 60°

⇒ ∠ADB = 180° − 30° = 150°

Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.

#### Question 3:

In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Consider PR as a chord of the circle.

Take any point S on the major arc of the circle.

∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)

⇒ ∠PSR = 180° − 100° = 80°

We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠POR = 2∠PSR = 2 (80°) = 160°

In ΔPOR,

OP = OR (Radii of the same circle)

∴ ∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)

∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of a triangle)

2 ∠OPR + 160° = 180°

2 ∠OPR = 180° − 160° = 20º

∠OPR = 10°

#### Question 4:

In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

In ΔABC,

∠BAC + ∠ABC + ∠ACB = 180° (Angle sum property of a triangle)

⇒ ∠BAC + 69° + 31° = 180°

⇒ ∠BAC = 180° − 100º

⇒ ∠BAC = 80°

∠BDC = ∠BAC = 80° (Angles in the same segment of a circle are equal)

#### Question 5:

In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

In ΔCDE,

∠CDE + ∠DCE = ∠CEB (Exterior angle)

⇒ ∠CDE + 20° = 130°

⇒ ∠CDE = 110°

However, ∠BAC = ∠CDE (Angles in the same segment of a circle)

⇒ ∠BAC = 110°

#### Question 6:

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

For chord CD,

∠CBD = ∠CAD (Angles in the same segment)

∠BCD + 100° = 180°

∠BCD = 80°

In ΔABC,

AB = BC (Given)

∴ ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)

⇒ ∠BCA = 30°

We have, ∠BCD = 80°

⇒ ∠BCA + ∠ACD = 80°

30° + ∠ACD = 80°

⇒ ∠ACD = 50°

⇒ ∠ECD = 50°

#### Question 7:

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.

(Consider BD as a chord)

∠BCD = 180° − 90° = 90°

(Considering AC as a chord)

90° + ∠ABC = 180°

∠ABC = 90°

Each interior angle of a cyclic quadrilateral is of 90°. Hence, it is a rectangle.

#### Question 8:

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Consider a trapezium ABCD with AB | |CD and BC = AD.

Draw AM ⊥ CD and BN ⊥ CD.

In ΔAMD and ΔBNC,

∠AMD = ∠BNC (By construction, each is 90°)

AM = BN (Perpendicular distance between two parallel lines is same)

∴ ΔAMD ≅ ΔBNC (RHS congruence rule)

∴ ∠ADC = ∠BCD (CPCT) … (1)

∠BAD + ∠BCD = 180° [Using equation (1)]

This equation shows that the opposite angles are supplementary.

Therefore, ABCD is a cyclic quadrilateral.

#### Question 9:

Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.

Join chords AP and DQ.

For chord AP,

∠PBA = ∠ACP (Angles in the same segment) … (1)

For chord DQ,

∠DBQ = ∠QCD (Angles in the same segment) … (2)

ABD and PBQ are line segments intersecting at B.

∴ ∠PBA = ∠DBQ (Vertically opposite angles) … (3)

From equations (1), (2), and (3), we obtain

∠ACP = ∠QCD

#### Question 10:

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Consider a ΔABC.

Two circles are drawn while taking AB and AC as the diameter.

Let they intersect each other at D and let D not lie on BC.

∠ADB = 90° (Angle subtended by semi-circle)

∠ADC = 90° (Angle subtended by semi-circle)

Therefore, BDC is a straight line and hence, our assumption was wrong.

Thus, Point D lies on third side BC of ΔABC.

#### Question 11:

ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

In ΔABC,

∠ABC + ∠BCA + ∠CAB = 180° (Angle sum property of a triangle)

⇒ 90° + ∠BCA + ∠CAB = 180°

⇒ ∠BCA + ∠CAB = 90° … (1)

∠CDA + ∠ACD + ∠DAC = 180° (Angle sum property of a triangle)

⇒ 90° + ∠ACD + ∠DAC = 180°

⇒ ∠ACD + ∠DAC = 90° … (2)

Adding equations (1) and (2), we obtain

∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°

⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°

∠BCD + ∠DAB = 180° … (3)

However, it is given that

∠B + ∠D = 90° + 90° = 180° … (4)

From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.

Consider chord CD.

∠CAD = ∠CBD (Angles in the same segment)

#### Question 12:

Prove that a cyclic parallelogram is a rectangle.

Let ABCD be a cyclic parallelogram.

∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral) … (1)

We know that opposite angles of a parallelogram are equal.

∴ ∠A = ∠C and ∠B = ∠D

From equation (1),

∠A + ∠C = 180°

⇒ ∠A + ∠A = 180°

⇒ 2 ∠A = 180°

⇒ ∠A = 90°

Parallelogram ABCD has one of its interior angles as 90°. Therefore, it is a rectangle