## MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.5

Question 1.

In the 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.

Solution:

We have a circle with centre O, such that ∠AOB = 60° and ∠BOC = 30°

∴ ∠AOB + ∠BOC = ∠AOC

∠AOB = 60° + 30° = 90°

Now, tne arc ABC subtends ∠AOC = 90° at the centre and ∠ADC at a point D on the circle other than the arc ABC.

∴ ∠ADC = \(\frac{1}{2}\) [∠AOC]

∠ADC = \(\frac{1}{2}\) (90°) = 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:

We have a circle having a chord AB equal to radius of the circle.

∴ AO = BO = AB

∆AOB is an equilateral triangle.

Since, each angle of an equilateral triangle = 60

∠AOB = 60°

Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ABC at a point on the minor arc of the circle.

∠ACB = \(\frac{1}{2}\) [reflex ∠AOB]

= \(\frac{1}{2}\) [300°] = 150°

Similarly, ∠ADB = \(\frac{1}{2}\) [∠AOB]

= \(\frac{1}{2}\) x [60°] = 30°

Thus, the angle subtended by the chord on the minor arc = 150° and on the major arc = 30°.

Question 3.

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

Solution:

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.

∴ reflex ∠POR = 2∠PQR,

But ∠PQR = 100°

reflex ∠POR = 2 x 100° = 200°

Since, ∠POR + reflex ∠POR = 360°

∠POR + 200° = 360°

∠POR = 360° – 200°

∠POR = 160°

Since, OP = OR [Radii of the same circle]

∴ In ∆POR, ∠OPR = ∠ORP [Angles opposite to equal sides of a triangle are equal]

Also, ∠OPR + ∠ORP + ∠POR = 180° [Sum of the angles of a triangle = 180°]

∠OPR + ∠OPR + 160° = 180° [∴ ∠OPR = ∠ORP]

2∠OPR = 180° – 160° = 20°

∠OPR = \(\frac{20^{\circ}}{2}\) =10°

Question 4.

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

Solution:

We have, in ∆ABC,

∠ABC = 69° and

∠ACB = 31°

But ∠ABC + ∠ACB + ∠BAC = 180°

∴ 69° +31° + ∠BAC = 180°

∠BAC = 180° – 69° – 31°

= 80°

Since, angles in the same segment are equal.

∴ ∠BDC = ∠BAC

∠BDC =80°

Question 5.

In the 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.

Solution:

In ∆CDE,

Exterior ∠BEC – ∠AED = 130°

130° = ∠EDC + ∠ECD

130° = ∠EDC + 20°

∠EDC = 130° – 20° = 110°

∠BDC = 110°

Since, angles in the same segment are equal.

∠BAC = ∠BDC

∠BAC = 110°

Question 6.

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

Solution:

Given:

∠BAC = 30°, ∠DBC = 70°

To find:

∠BCD and ∠ECD.

∠BDC = ∠BAC = 30° (∠s on the same segment of a circle are equal)

In ∆BCD, 70° + 30° + ∠C = 180° (ASP)

100° + ∠C = 180°

∠C = 80°

AB = BC (Given)

∠BAC = ∠BCA = 30°

∠BCD = ∠BCA + ∠ECD

80° = 30° + ∠ECD

∠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

ABCD is a cyclic quadrilateral in which AC and BD are diagonals.

To prove:

ABCD is a rectangle.

Proof:

BD is the diameter of the circle.

∠BAD = 90°(angle in a semicircle)

Similarly ∠BCD = 90°

AC is the diameter of the circle

∠ABC = 90° (angle in a semicircle)

Similarly ∠ADC = 90°

In quadrilateral ABCD, ∠A = ∠B = ∠C = ∠D = 90°

ABCD is a rectangle.

Question 8.

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

Solution:

Given:

AD = BC, AB ∥ DC.

To prove:

ABCD is cyclic quadrilateral.

Construction:

Draw AE and BF As on DC.

Proof:

In ∆ADE and ∆BCF

∴ AE = BF

(∴ Distance between two parallel lines are equal)

∠E = ∠F (Each 90°)

AD = BC (Given)

∆ADE = ∆BCF (By RHS)

and so ∠D = ∠C (By CPCT)

AB ∥ DC and AD is the transversal.

∴ ∠BAD + ∠ADC =180° (CIA’s)

⇒ ∠BAD + ∠BCD = 180° (∠ADC = ∠BCD)

∴ ABCD is cyclic quadrilateral.

Question 9.

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

Solution:

Given

C (O, r) and C (O_{1}, r_{1}) are two circles. Two lines ABD and PBQ are drawn which intersect at B.

To prove:

∠1 = ∠3

Proof:

∠1 = ∠2 …(1) (Z s on the same segment of a circle are equal.)

∠3 = ∠4 …..(2) (Z s on the same segment of a circle are equal)

∠2 = ∠4 …(3) (OA’s)

From (1), (2) and (3), we get

∠1 = ∠3 …(2)

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

C (O, r) and C (O_{1}, r_{1}) are two Circles in whichAB andAC are diameter. These circles intersect at point A and D.

To prove:

BDC is a line.

Construction: JoinAD.

Proof:

∠ADB = 90° (∠s in a semicircle)

∠ADC) = 90° (∠s in a semicircle)

Adding (1) and (2), we get

∠ADB + ∠ADC = 90° + 90°

∠BDC = 180°

∴ BDC is a line and hence D lies on the third side.

Question 11.

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

Solution:

Given:

ABC and ADC are two right ∆’s on common base AC. ∠B – 90° and ∠D = 90°.

To prove: ∠CAD = ∠CBD

Proof:

∠ABC + ∠ADC = 90° + 90°

= 180°

ABCD is a cyclic quadrilateral.

∠CAD and ∠CBD are angles on the same segment of a circle.

∴ ∠CAD = ∠CBD.

Question 12.

Prove that a cyclic parallelogram is a rectangle.

Solution:

Given: ABCD is a ∥^{gm}

To prove: ABCD is a rectangle.

Proof:

∠A = ∠C

∠A + ∠C =180°

∠A + ∠A = 180°

∠A = \(\frac{180^{\circ}}{2}\) = 90°

∴ ABCD is a rectangle.

∠A = \(\frac{180^{\circ}}{2}\) = 90°

∴ ABCD is a rectangle.