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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18d12Question 15: In the given figure, AB is the diameter of the circle with center O . If \angle ADC = 32^o , find \angle BOC .

Answer:

\widehat{AC} subtends AOC at the center and ADC at the circumference of the circle.

\angle AOC = 2 \angle ADC

\Rightarrow \angle AOC = 2 \times 32 = 64^o

\angle AOC + \angle BOC = 180 \Rightarrow \angle BOC = 180-64=116^o

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d21Question 16: In a cyclic-quadrilateral PQRS, \angle PQR = 135^o . Sides SP and RQ produced meet at point A whereas sides PQ and SR produced meet at point B . lf \angle A: \angle B = 2 : 1 ; find \angle A and \angle B .

Answer:

Given PQRS is a cyclic quadrilateral.

\angle A = \angle B = 2 :1   or if \angle B = x , then \angle A = 2x

\angle PQR +\angle PSR = 180^o  (sum of the opposite angles in a cyclic quadrilateral is 180^o )

\angle PSR = 180-135 = 45^o

In \triangle PBS; \angle SPB + 45+x = 180^o 

\Rightarrow \angle SPB = 135-x … … … (i)

In \triangle PQA; \angle APQ = 180-2x-45 = 135-2x

Therefore \angle SPB = 180 - (135-2x) = 45+2x   … … … (ii)

Hence 35-x = 45+2x \Rightarrow x = 30^o

Therefore \angle B = 30^o \ and \  \angle A = 60^o

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18d13Question 17: In the following figure, AB is the diameter of a circle with center O and CD is the chord with length equal to radius OA . If AC produced and BD produced meet at point P ; show that \angle APB = 60^o .

Answer:

Given: AB is diameter, CD = OA = Radius

In \triangle CDO, CD = OD = OC

\Rightarrow \triangle CDO is equilateral

Therefore \angle OCD = \angle OCD = \angle COD = 60^o

d22.jpgIn \triangle AOC

OA = OC (radius of the same circle)

Therefore \angle OCA = \angle CAO

Similarly in \triangle BOD

OD = OB (radius of the same circle)

Therefore \angle ODB = \angle BDO

Since ABCD is cyclic quadrilateral

Therefore \angle ACD + \angle OBD = 180^o (opposite angles of a cyclic quadrilateral are supplementary)

60+ \angle ACO + \angle OBD = 180^o

\Rightarrow \angle ACO + \angle OBD = 120^o

In \triangle APB

\angle APB + \angle PBA + \angle BAP = 180^o

\angle APB + 120 = 180^o

\Rightarrow \angle APB = 60^o . Hence proved.

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18d14Question 18: In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC . If the bisector of \angle A meets BC at point E and the given circle at point F , prove that: (i) EF = FC (ii) BF = DF

Answer:

Given: ABCD is a cyclic quadrilateral

AD \parallel BC

\angle BAE = \angle EAD \ (AE is the angle bisector) … … … (i)

(i) \angle AEB = \angle EAD (alternate angles) … … … (ii)

d29In \triangle ABE 

\angle ABE = 180^o - \angle BAE - \angle AEB

Using (i) and (ii) we get  \angle ABE = 180^o - 2 \angle AEB

\angle CEF = \angle AEB (vertically opposite angles)

\angle ADC = 180^o - \angle ABC = 180^o -(180^o-2 \angle AEB)

\Rightarrow \angle ADC = 2 \angle AEB

\angle AFC = 180^o - \angle ADC = 180^o - 2 \angle AEB (since AFCD is a cyclic quadrilateral)

\angle ECF = 180^o - (\angle AFC + \angle CEF)  \\ = 180^o - (180^o-2 \angle AEB + \angle AEB) = \angle AEB

Also \angle AEB = \angle CEF \Rightarrow EF = EC

(ii) \widehat{BF} subtends \angle BAF on circumference

\widehat{OF} subtends \angle FAD on circumference

Given \angle BAF = \angle FAD

Therefore \widehat{BF} = \widehat{OF}  (equal arcs subtends equal angles on circumference)

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Question 19: ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E ; whereas sides BC and AD produced meet at point F . If \angle DCF: \angle F: \angle E = 3: 5 : 4 , find the angles of the cyclic quadrilateral ABCD .

Answer:d23

Given DCF : F : E = 3:4:5

In \triangle CDF: \angle ADC = 3x+5x = 8x

In \triangle BCE: \angle ABC = 3x+4x = 7x

\angle ABC + \angle ADC = 180^o

or 7x+8x= 180 \Rightarrow x = 12^o

Therefore

\angle DAC = 180 -12x = 180 - 144 = 36^o

\angle CBA = 7x = 7 \times 12 = 84^o

\angle DCB = 180-3x = 180-36 = 144^o

\angle ADC = 8x = 8 \times 12 = 96^o

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18d15Question 20: In the following figure shows a circle with PR as its diameter. If PQ =7\ cm and QR=3RS=6cm , find the perimeter of the cyclic quadrilateral PQRS . [1992]

Answer:

Given: PQ = 7 \ cm, QR = 3RS = 6 \ cm

PR = \sqrt{7^2+6^2} = \sqrt{85}

SR = \sqrt{PR^2-PS^2} = \sqrt{85-4} =9

Therefore the perimeter of PQRS = 2+9+6+7 = 24 \ cm

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18d16Question 21: In the following figure, AB is the diameter of a circle with center O . If chord AC = chord AD , prove that: (i) arc BC = arc DB (ii) AB is bisector of \angle CAD . Further, if the length of arc AC = 2 \times arc BC , find : (a) \angle BAC (b) \angle ABC .

Answer:

Given: AC = AD

Consider \triangle ABC and \triangle ABD

AB is commond28

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

AC = AD

Therefore \triangle ABC \cong \triangle ABD

(i) BC = BD (corresponding parts of congruent triangles)

(ii) \angle CAB = \angle BAD (equal chords subtend equal angles on the circumference of the same circle)

Therefore AB bisects \angle CAD

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Question 22: In cyclic quadrilateral ABCD ; AD = BC, \angle BAC = 30^o and \angle CBD = 70^o ; find: (i) \angle BCD (ii) \angle BCA (iii) \angle ABC (iv) \angle ADC

Answer:

d27Given AD = BC, \angle BAC = 30^o and \angle CBD = 70^o

\angle BDC = \angle BAC = 30^o (angles in the same segment)

\angle DAC = \angle DBC = 70^o (angles in the same segment)

\angle BCD + 30^o + 70^o = 180^o \Rightarrow \angle BCD = 80^o (opposite angles in a cyclic quadrilateral are supplementary)

BC = AD \Rightarrow \angle BDC = \angle ADC = 30^o

Therefore \angle BCA = 50^o

70^o + 30^o + 30^o +\angle BDA = 180 \Rightarrow \angle BDA = 50^o

50^o+ 30^o + 70^o+\angle ABD = 180 \Rightarrow \angle ABD = 30^o

\angle ABC = 100^o, \angle BAD = 100^o, \angle BCD = 80^o \ and \  \angle CDA = 80^o

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18d17Question 23: In the given figure \angle ACE = 43^o and \angle CAF = 62^o ; find the values of a, b and c . [2007]

Answer:

Given \angle ACE = 43^o and \angle CAF = 62^o

\angle BAE + \angle BDE = 180^o (ABDE is a cyclic quadrilateral)

\angle BDE = 118^o

\angle c + 118^o = 180^o (straight line) \Rightarrow c = 62^o

In \triangle ACE :

62^o+43^o+\angle CEA = 180^o  \Rightarrow \angle CEA = 75^o

Therefore \angle DEF = 180^o - 75^o = 105^o

Therefore \angle a + 75^o = 180^o \Rightarrow \angle a = 105^o \ (ABDE  is a cyclic quadrilateral)

In \triangle DEF :

62^o+ 105^o + \angle b = 180^o \Rightarrow \angle b = 13^o

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18d18Question 24: In the given figure, AB \parallel DC, \angle BCE = 80^o and \angle BAC = 25^o . Find (i) \angle CAD (ii) \angle CBD (iii) \angle ADC

Answer:

Given \angle = 80^o and \angle BAC = 25^o . Also AB \parallel DC

Therefore \angle CAB = \angle ACD (alternate angles)

d26Hence \angle ACD = 25^o

Therefore \angle ACB = 180^o - 80^o - 25^o = 75^o

and \angle DCB = 25^o + 75^o = 100^o

\angle DAB = 180^o - 100^o = 80^o

Therefore \angle DAC = 80^o - 25^o = 55^o

In \triangle ABC :

\angle ABC = 180^o - 25^o - 75^o = 80^o

Therefore \angle ADC = 180^o = 80^o = 100^o

\angle CBD = \angle DAC = 55^o (angles in the same segment)

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Question 25: ABCD is a cyclic quadrilateral of a circle center O such that AB is a diameter of the circle and the length of the chord CD is equal to the radius of the circle. If AD and BC produced meet at P , show that \angle APB = 60^o .

Answer:

Given CD = OC = Radius d25

In \triangle OCD :

OD = OC = CD  \Rightarrow \angle ODC \\ = \angle OCD = \angle DOC = 60^o

In \triangle AOD, OA = OD (Radius of the same circle)

Let \angle OAD = \angle ADO = x

Therefore \angle AOD = 180^o -2x  … … … (i)

In \triangle COB, OC = OB (Radius of the same circle)

Let \angle OBC = \angle OCB = y

Therefore \angle COB = 180^o - 2y  … … … (ii)

Now, 180^o - 2x + 60^o + 180^o - 2y = 180^o (straight line angle)

\Rightarrow x + y = 120^o

In \triangle APB :

x + y + \angle APB = 180^o \Rightarrow \angle APB = 180^o - 120^o = 60^o

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18d19Question 26: In the figure, given alongside, CP bisects \angle ACB . Show that DP bisects \angle ADB

Answer:

Given CP bisects \angle ACB

Therefore \angle ACP = \angle PCB

\angle ACP = \angle ADP (angles in the same segment)

\angle PCB = \angle PDB (angles in the same segment)

Therefore \angle ADP = \angle PDB

Hence DP \ bisects \  \angle ADB

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18d20Question 27: In the figure shown, AD = BC, \angle BAC = 30^o and \angle CBD=70^o . Find (i) \angle BCD (ii) \angle BCA (iii) \angle ABC (iv) \angle ADB

Answer:

From chord \ CD: \angle DBC = \angle DAC ( angles in the same segment)

\Rightarrow \angle DAC = 70^o

From chord \ CB: \angle CAB = \angle CDB (angles in the same segment)

\Rightarrow \angle CDB = 30^o

In \triangle \angle DCB: 70 + 30 + \angle DCB = 180 \Rightarrow \angle DCB = 80^o

Since AD = BC: \angle DCA = \angle CAB \Rightarrow \angle DCA = 30^o

In \triangle ADC: 70 + \angle ADB + 30 + 30 = 180  \Rightarrow \angle ADB = 50^o

ABCD is a cyclic quadrilateral. \therefore 50 + 30 + \angle DBA + 70 = 180 \Rightarrow DBA = 30^o

Hence

(i) \angle BCD = 80^o

(ii) \angle BCA = 50^o

(iii) \angle ABC = 100^o

(iv) \angle ADB = 50^o

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18d21Question 28: In the figure, given below, AB and CD are two parallel chords and O is the center. If the radius of the circle is 15\ cm , find the distance MN between the two chords of lengths 24\ cm and 18\ cm respectively. [2010]

Answer:

We know that perpendicular drawn from the center of the circle will bisect the chord.

d24Therefore MO^2 = 15^2-12^2 \\ = 225 - 144 = 81 \Rightarrow MO = 81 \ cm

ON^2 = 15^2-9^2 = 225 - 81 = 12 \\ \Rightarrow ON = 12 \ cm

Hence MN = 9 + 12 = 21 \ cm