18d1Question 1: In the given circle with diameter AB , find the value of x . [2003]

Answer:

\angle ABD = \angle ACD = 30^o (angles in the same segment)

In \triangle ADB

\angle BAD + \angle ADB + \angle ABD = 180^o

Since AB is the diameter,

\angle ADB = 90^o (angle in the semi circle)

x+ 90+30 = 180 \Rightarrow x = 60^o

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18d2Question 2: In the given figure, O is the center of the circle with radius 5\ cm . OP and OQ are \perp to AB and CD respectively. AB = 8\ cm and CD = 6\ cm . Determine the length of PQ .

Answer:

Given: Radius = 5 \ cm, \ OP \perp AB \ and \ OQ \perp CD

AB = 8 \ cm \ and \ CD = 6 \ cm

P \ and \  Q are mid point of AB \ and \  CD respectively (\perp from the center to a chord will bisect the chord)

d1In \triangle OAP

OA^2= OP^2+AP^2

25 = OP^2 + 16 \Rightarrow OP = 3 \ cm

Similarly, in \triangle OCQ

OC^2= OQ^2+CQ^2

25 = OQ^2+9 \Rightarrow OQ = 4 \ cm

Hence PQ = OP + OQ = 3 + 4 = 7 \ cm

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18d3Question 3: The given figure shows two circles with centers A and B ; and radius 5\ cm and 3\ cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q , find the length of PQ .

Answer:

AB = AC - BC = 5 - 3 = 2 \ cm d2

PQ bisects AB (let the point be R )

AR = 1 \ cm

In \triangle APR

AP^2= RP^2+AR^2

\Rightarrow 25 = RP^2+1 \Rightarrow RP = 24 = 2\sqrt{6} \ cm

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18d4Question 4: In the given figure, \triangle ABC in which \angle BAC = 30^o . Show that BC is equal to the radius of the circumcircle of the \triangle ABC , whose center is O .

Answer:

Given \angle BAC = 30^o

\angle BOC = 2 \angle BAC = 2 . 30 = 60^o (angle at the center of the circle is twice that of the angle subtend at the circumference by the same chord)

In \triangle OBC

OB = OC   (radius of the same circle)

d3\angle OBC + \angle OCB + \angle BOC = 180^o

\angle OBC + \angle OCB = 180-60 = 120^o

2 \angle OBC = 120^o \Rightarrow  \angle OBC = 60^o

Therefore \angle OBC = \angle OCB = \angle BOC = 60^o

Therefore \triangle BOC is equilateral

\Rightarrow BC = OB = OC

Hence BC is equal to the radius of the circle.

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Question 5: Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Answer:

Given: Isosceles \triangle ABC , where AB = AC .

AB is the diameter of the circle.

d4\angle ADB = 90^o (angle in the semi circle)

\angle ADC = 180-90=90^o

Consider \triangle ADB and \triangle ADC

AD is common

\angle ADC = \angle ADB

AB = AC

Therefore  \triangle ADB \cong \triangle ADC (SAS \ axiom)

Therefore BD = DC

Hence D is the midpoint of BC .

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18d5Question 6: In the given figure, chord ED is parallel to diameter AC of the circle. Given \angle CBE = 65^o , calculate \angle DEC .

Answer:

Given AD \parallel AC

\angle CBE = 65^o

\widehat{EC} subtends \angle EOC at center  and \angle EBC at circumference.

\angle ECO = 2 \angle EBC = 2 \times 65 = 130^o

d5In \triangle EOC ,

EO = OC (radius of the same circle)

\angle OEC = \angle OCE

\angle OEC + \angle OCE + \angle EOC = 180^o

2 \angle OEC + 130 = 180 \Rightarrow \angle OEC = 25^o

Given ED \parallel AC

\Rightarrow \angle DEC = \angle OEC \Rightarrow \angle DEC = 25^o

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18d6Question 7: Chords AB and CD of a circle intersect each other at point P such that AP = CP . Show that: AB = CD .

Answer:

Given AB and CD intersect at P

AP = CP (given)

Therefore AP \times PB = CP \times PD

\Rightarrow \frac{AP}{CP} = \frac{PD}{PB}

\Rightarrow PD = PB

Therefore AP + PB = CP + PD \Rightarrow AB = CD 

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Question 8: The quadrilateral formed by the angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.

Answer:

Given: ABCD  is a cyclic quadrilateral and PQRS  is a quadrilateral

d6In \triangle APD  :

PAD + ADP + APD = 180^o  … … … … … (i)

In \triangle BQC 

QBC + BCQ + BQC = 180^o    … … … … … (ii)

Adding (i) and (ii)

PAD + ADP + APD + QBC + BCQ + BQC = 360^o 

\frac{1}{2} (BAD + ADC + DCB + CBA) + APD + BQC = 360^o 

180 + APD + BQC = 360^o \Rightarrow APD + BQC = 180^o

Therefore PQRS  is a cyclic quadrilateral as the opposite angles are supplementary.

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18d7Question 9: In the given diagram, \angle DBC = 58^o , BD is a diameter of the circle. Calculate: (i) \angle BDC (ii) BEC (iii) \angle BAC [2014]

Answer:

Given BD is diameter

\angle DBC = 58^o

(i) \angle DCB = 90^o (angle in semi circle)

In \triangle BDC

58+\angle BDC+90=180^o

\angle BDC = 32^o

(ii) ABEC is a cyclic quadrilateral

\angle BDC = \angle BAC = 32^o (angle in the same segment)

Therefore \angle BEC + 32 = 180^o

\angle BEC = 148^o

(iii) \angle BAC = 32^o (angle in the same segment)

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Question 10: D and E are points on equal sides AB and AC of an isosceles \triangle ABC such that AD = AE . Prove that the points B, C, E and D are concyclic.

d8Answer:

In \triangle ABC AB = AC . Also given AD = AE

Therefore \angle ABC = \angle ACB = x \ (say)

In \triangle ADE, AD = AE

Now in \triangle ABC

\frac{AD}{AB} =\frac{AE}{AC}

\Rightarrow DE \parallel BC

\Rightarrow \angle BCE = \angle DEA = \angle ADE = x

Therefore \angle BCE + \angle BDE = x + (180-x) = 180 

\Rightarrow BDCE is a cyclic quadrilateral.

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18d8Question 11: In the given figure, ABCD is a cyclic quadrilateral. AF \parallel CB and DA is produced to point E . If \angle ADC = 92^o, \angle FAE= 20^o ; Determine \angle BCD . Give reason in support of your answer.

Answer:

Given: ABCD is a cyclic quadrilateral

AF \parallel CB. \angle ADC = 92^o and \angle FAE = 20^o

\angle CDA + \angle CBA = 180^o

\angle CBA = 180-92 = 88^o

Since AF \parallel CB

\angle CBA = \angle BAF (alternate angles)

\angle BAE = 88 + 20 = 108^o

Therefore \angle DBA = 180-108 = 72^o

Therefore \angle BCD = 180-72 = 108^o

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18d9Question 12: If I is the incentre \triangle ABC and AI when produced meets the circumcircle of \triangle ABC at point D . If \angle BAC = 66^o and \angle ABC = 80^o . Calculate (i) \angle DBC , (ii) \angle IBC , (iii) \angle BIC

 

d9Answer:

Given \angle BAC = 66^o and \angle ABC = 80^o

(i) \angle DBC = \angle DAC (angles in the same segment)

Since I is the incenter

\angle DAC = \frac{1}{2} \angle BAC = \frac{1}{2} \times 66 = 33^o 

Therefore \angle DBC = 33^o

(ii) Similarly, \angle IBC = \frac{1}{2} \angle ABC = \frac{1}{2} \times 80 = 40^o

(iii) In \triangle ABC, \angle ACB = 180-\angle ABC - \angle BAC = 180-80-66 = 34^o

Also \angle ICB = \frac{1}{2} \angle BCA = \frac{1}{2} \times 34 = 17

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18d10Question 13: ln the given figure, AB = AD = DC = PB and \angle DBC = x^o . Determine, in terms of x : (i) \angle ABD , (ii) \angle APB . Hence or otherwise, prove that AP \parallel DB .

Answer:

Given AB = AD =DC = PB

\angle DBC = x^o

\angle DAC = \angle DBC (angles in the same segment)

d10Since AC = DC

\angle DAC = \angle DCA = x

Similarly, \angle ABD = \angle DAC ( Since ABD = ACD \Rightarrow ACD = DAC )

In \triangle ABP, \   \angle ABC = \angle BAP + \angle APB

Since AB = BP

\angle BAP = \angle APB

2x = \angle APB + \angle APB = 2 \angle APB

\Rightarrow \angle APB = x^o

Therefore \angle APB = \angle DBC = x^o and hence AP \parallel DB (alternate angles)

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18d11Question 14: In the given figure; ABC, AEQ and CEP are straight lines. Show that \angle APE and \angle CQE are supplementary.

Answer:

Given ABC, AEQ \ and \  CEP are straight lines.

In cyclic quadrilateral ABFP d11

\angle APE + \angle ABE = 180^o … … … … … (i)

Similarly in cyclic quadrilateral BCQE

\angle CQE+\angle CBE = 180^o … … … … … (ii)

Adding (i) and (ii)

\angle APE + \angle ABE + \angle CQE+\angle CBE = 360

\angle APE + \angle CQE + 180 = 360

\Rightarrow \angle APE + \angle CQE = 180

Hence they are supplementary.

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