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

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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 ADC = {1 \over 2}\angle AOC


[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.]

\angle ADC = {1 \over 2} \times 90^\circ = 45^\circ


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°

\angle APB = {1 \over 2}\angle AOB = {1 \over 2} \times 60^\circ = 30^\circ

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, \angle PQR = {1 \over 2}(180^\circ + 2x)

⇒ 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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Sol.
Given ∠BEC = 130°, ∠ECD = 20°.

∠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.


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