# Circles – Exercise 10.5 – (MATHEMATICS) – 9th Class

**Exercise 10.5**

**1. In the adjoining 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.**

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**Sol.**∠AOC = 60° + 30° = 90°

[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.]

**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.**

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**Sol.**We have, OA = OB = AB Given]

∠OAB is equilateral triangle.

∠AOB = 60°

Also APBQ is a cyclic quadrilateral. Q

∴ ∠P + ∠Q = 180° [Sum of opposite angles of a cyclic quadrilateral is 180°.]

⇒ 30° + ∠Q = 180° ⇒ ∠Q = 150°.

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

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**Sol.**Let

*∠OPR = x, then ∠ORP = x*

*
*and

*∠POR = 180° – 2x.*

∴ Angle formed by arc PXR, at the centre

= 3600 — (180° — 2x)

= 180° + 2x.

Also,

⇒ 100° = 90° + *x* ⇒ *x* = 10°.

**4. In figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.**

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**Sol.**In triangle ABC,

∠A + 69° + 31° = 180° [Sum of angles of a triangle is 180° ]

⇒ ∠A = 180° — 100° = 80°.

Also, ∠D = ∠A = 80° [Angles in the same segment of a circle]

i.e., ∠BDC = 80°.

**5. In 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.**

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**Given ∠BEC = 130°, ∠ECD = 20°.**

Sol.

Sol.

∠DEC + ∠BEC = 180° [Linear pair]

In triangle DEC,

LDEC = 180° — 130° = 50°. A

∠D + 50° + 20° = 180°

[Sum of angles of a triangle is 180°]

⇒ ∠D = 110° …(i)

Also, ∠BAC = ∠D

∴ [Angles in the same segment of a circle are equal]

∠BAC = 110°. [From (i)]

**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.**

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**Sol.**∠CDB = ∠BAC = 30° …(i)

[Angles in the same segment]

In triangle BCD,

∠CBD + ∠BCD + ∠CDB = 180° [Sum of angles of a triangle is 180]

70° + ∠BCD + 30° = 180°

⇒ ∠BCD = 80°

Now, in ΔABC,

if AB = BC, then ∠BCA = ∠BAC = 30° …..(iii)

[Angles opposite to equal sides are equal ]

Now, ∠BCD = ∠BCA + ∠ACD

⇒ 80° = 30° + ∠ECD [ ∠ACD = ∠ECD]

⇒ ∠ECD = 50°. [ From (ii) and (iii)

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

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**Sol.**As AC and BD are the diagonals of a cyclic quadrilateral.

∴ ∠ADC, ∠BAD, ∠ABC and ∠BCD are angles in a semicircle. Hence, each angle is 90°.

As in a quadrilateral each angle is 90°, hence quadrilateral is a rectangle.

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

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**Sol.**

**Construction:**Draw DL and CM perpendi-culars to AB.

**Proof:**In ΔDLA and ΔCMB,

DL = CM

[Distance between parallel lines]

AD = BC [Given]

∠DLA = ∠CMB [90° each] [Construction]

∴ ΔDLA ≅ ΔCMB [RHS]

∴ ∠DAL = ∠CBM ….(i) [CPCT]

Now AB ‖ CD and AD is transversal

∴ ∠CDA + ∠DAL = 180°

⇒ ∠CDA + ∠CBM = 180° [From (i)]

⇒ ∠CBM = ∠CBA = 180° [ ∠CBM = ∠CBA]

As sum of opposite angles of a quadrilateral is 180°, then

Hence, ABCD is a cyclic quadrilateral.

**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 figure). Prove that ∠ACP = ∠QCD.**

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**Sol.**∠ACP = ∠ABP ……………(i)

[Angles in the same segment of a circle are equal]

∠QCD = ∠QBD …(ii) [Reason same as above]

∠ABP = ∠QBD …(iii) [Vertically opposite angles]

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

∠ACP = ∠QCD.

**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.**

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**Sol.**

**Construction:**Join AD.

**Proof:**Let circle with AB as diameter meets BC at D.

Then ∠ADB = 90°. [Angle in a semicircle]

Now ∠ADB + ∠ADC = 180° [Linear pair]

∴ ∠ADC = 90°

As we know angle in a semicircle is 90°, therefore, a

circle with AC as diameter passes through D.

Hence both the circles meet the third side at D.

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

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**Sol.**∠ABC = ∠ADC = 90°

∴ ACDB is a cyclic quadrilateral.

[As if a line segment subtends equal angles at two other points on the same side of the segment, then the four points are concyclic.]

∴ ∠CAD = ∠CBD.

[Angles in the same segment of a circle are equal.]

**12. Prove that a cyclic parallelogram is a rectangle.**

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**Sol.**ABCD is a cyclic parallelogram.

∴ ∠A + ∠C= 180° …..(i)

[Sum of opposite angles of a cyclic quadrilateral is 180°.]

Also, ∠A = ∠C …(ii)

[Opposite angles of a parallelogram]

From (i) and (ii), we have

2∠A = 180° ⇒ ∠A = 90°

As in a parallelogram one angle is 90°, hence it is a rectangle.