Haryana State Board HBSE 9th Class Maths Solutions Chapter 10 Circles Ex 10.5 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 10 Circles Exercise 10.5

Question 1.

In Fig. 10.36, 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.

Solution:

∠AOC = ∠AOB + ∠BOC

∠AOC = 60° + 30°

∠AOC = 90°

∠AOC is an angle subtended by an arc AC on centre and ∠ADC on the remaining part of the circle.

∴ ∠AOC = 2 × ∠ADC [By theorem 10.8]

⇒ 90° = 2 × ∠ADC

⇒ ∠ADC = \(\frac{90^{\circ}}{2}\) = 45°3

Hence, ∠ADC = 45°.

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.

Solution:

Let AB be a chord which is equal to radius. Join OA and OB, where O is the centre of the circle.

∵ AO = OB = AB (given)

∴ AOB is an equilateral triangle.

⇒ ∠AOB = 60°

Reflex ∠AOB = 360° – 60° = 300°

∠AOB = 2 × ∠ACB [By theorem 10.8]

⇒ 60° = 2 × ∠ACB

⇒ ∠ACB = \(\frac{60^{\circ}}{2}\) = 30°

and Reflex (∠AOB) = 2 × ∠APB [By theorem 10.8]

⇒ 300° = 2 × ∠APB

⇒ ∠APB = \(\frac{300^{\circ}}{2}\) = 150°

Hence, angle subtended by the chord at a point on the minor arc is 150° and major arc is 30°.

Question 3.

In Fig. 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:

Since, reflex ∠POR is an angle subtended by an arc PR at the centre and ∠PQR at the remaining part of the circle.

∴ Reflex ∠POR = 2 × ∠PQR

⇒ Reflex ∠POR = 2 × 100° = 200°

∠POR = 360° – Reflex ∠POR

⇒ ∠POR = 360° – 200° = 160°

In ΔPOR, OP = OR [Radii of the same circle]

⇒ ∠OPR = ∠ORP [Angles opposite to equal sides are equal]

Let ∠OPR = ∠ORP = x°

In ΔPOR, we have

∠OPR + ∠ORP + ∠POR = 180° [∵ Sum of angles of a triangle is 180°]

⇒ x° + x° + 160° = 180°

⇒ 2x° = 180° – 160° = 20°

⇒ x° = \(\frac{20^{\circ}}{2}\) = 10°

Hence, ∠OPR = 10°.

Question 4.

In Fig. 10.38, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Solution:

In ΔBAC, we have

∠ABC + ∠BCA + ∠BAC = 180° [∵ Sum of angles of a triangle = 180°]

⇒ 69° + 31° + ∠BAC = 180°

⇒ 100° + ∠BAC = 180°

⇒ ∠BAC = 180° – 100°

⇒ ∠BAC = 80°

∠BDC = ∠BAC

[∵ Angles in the same segment are equal]

Hence ∠BDC = 80°.

Question 5.

In Fig. 10.39, 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.

Solution:

In ΔECD, we have

∠BEC = ∠EDC + ∠ECD [∵ Exterior angle is equal to sum of its opposite two interior angles]

⇒ 130° = ∠EDC + 20°

⇒ ∠EDC = 130° – 20°

⇒ ∠EDC = 110°

⇒ ∠BDC = 110°

∠BAC = ∠BDC [∵ Angles in the same segment are equal]

Hence, ∠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.

Solution:

In ΔABC,

AB = BC (given)

⇒ ∠BAC = ∠BCA [∵ Angles opp. to equal sides are equal]

30° = ∠BCA

⇒ ∠BCA = 30°

In ΔABC, we have

∠BAC + ∠BCA + ∠ABC = 180°

⇒ 30° +30° +70° + ∠ABE = 180°

⇒ 130° + ∠ABE = 180°

⇒ ∠ABE = 180° – 130° = 50°

⇒ ∠ABD = 50°

∠ACD = ∠ABD [∵ Angles in the same segment of a circle are equal]

⇒ ∠ACD = 50°

⇒ ∠ECD = 50°

⇒ ∠BCD = ∠BCA + ∠ACD

⇒ ∠BCD = 30° +50°

⇒ ∠BCD = 80°

Hence, ∠BCD = 80° and ∠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.

Solution:

Given: Diagonals AC and BD of cyclic quadrilateral are thediameters of the circle through the vertices of the quadrilateral ABCD.

To prove: Quadrilateral ABCD is a rectangle.

Proof: Since BD is a diameter.

∠C = ∠A = 90°

[Angle in a semicircle is 90°]

Similarly, AC is a diameter.

∠B = ∠D = 90°

[Angle in a semicircle is 90°]

Thus, in a quadrilateral ABCD

∠A = ∠B = ∠C = ∠D = 90°

Hence, quadrilateral ABCD is a rectangle

Proved.

Question 8.

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

Solution:

A trapezium ABCD in which AB || CD and AD = BC.

To prove: Trapezium ABCD is cyclic.

Construction: Draw AE || BC.

Proof: Since, AB || CD (given)

and AE || BC (By construction)

∴ AE = BC …..(i) (Opposite sides of a parallelogram)

AD = BC (given) … (ii)

From (i) and (ii), we get

AE = AD

⇒ ∠2 = ∠D …..(iii)

[Angles opposite to equal sides are equal]

∠B = ∠1 …..(iv)

[Opposite angles of a parallelogram]

But, ∠1 + ∠2 = 180°

[By linear pair axiom]

∠B + ∠D = 180° [Using (iii) and (iv)]

⇒ ABCD is a cyclic.

Hence, trapezium ABCD is cyclic.

Proved

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 Fig. 10.40). Prove that ∠ACP = ∠QCD.

Solution:

∠ABP = ∠ACP ……(i)

[Angles in a same segment are equal]

∠QBD = ∠QCD …..(ii)

[Angles in a same segment are equal]

∠ABP = ∠QBD …..(iii)

(Vertically opposite angles)

From (i), (ii) and (iii), we get

∠ACP = ∠QCD. Hence proved

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.

Solution:

Given: Two circles are drawn with sides AB and AC of ΔABC as diameters. The circles intersect each other at D.

To prove: Point D lies on BC i.e., BDC is a straight line.

Construction: Join AD.

Proof: Since AB and AC are diameters of two circles.

∴ ∠ADB = 90° …..(i)

[Angle in semicircle is 90°]

and ∠ADC = 90° …..(ii)

[Angle in a semicircle is 90°]

Adding (i) and (ii), we get

∠ADB + ∠ADC = 90° + 90°

⇒ ∠ADB + ∠ADC = 180°

⇒ BDC is a straight line.

Hence, D lies on BC. Proved.

Question 11.

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

Solution:

∠ABC = ∠ADC = 90° (given)

In a quadrilateral ABCD,

∠ABC + ∠ADC = 90° + 90°

= 180°

Therefore, ABCD is a cyclic quadrilateral.

∠CAD = ∠CBD [Angles in a same segment are equal]

Hence proved

Question 12.

Prove that a cyclic parallelogram is a rectangle.

Solution:

Given: A cyclic parallelogram ABCD.

To prove: ABCD is a rectangle.

Proof : Since ABCD is a parallelogram. (given)

∴ ∠A = ∠C

(Opposite angles of a || gm) …(i)

But, ∠A + ∠C = 180° [By theorem]

⇒ ∠A + ∠A = 180° [Using (i)]

⇒ 2∠A = 180°

⇒ ∠A = \(\frac{180^{\circ}}{2}\)

So, ∠A = ∠C = 90°

Thus, ABCD is a parallelogram in which

∠A = ∠C = 90°.

Hence, ABCD is a rectangle. Proved