18d21Question 1: 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

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18d17Question 2: 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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18d15Question 3: 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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18d1Question 4: 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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18d7Question 5: 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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18c1Question 6: In the given diagram, BD is the side of a regular hexagon, DC is the side of a regular pentagon and AD is a diameter. Calculate:

(i) \angle ADC

(ii) \angle BDA

(iii) \angle ABC

(iv) \angle AEC [1984]

Answer:c15.jpg

BD is a side of a regular hexagon

(i) \angle BOD = \frac{360}{6} = 60^o

DC is a side of regular pentagon

\angle DOC = \frac{360}{5} = 72^o

(ii) In \triangle BOD, \angle BOD = 60^o

OB = OD (radius of the same circle)

\therefore \angle OBD = \angle ODB = 60^o

(iii) In \triangle OCD, \angle COD = 72^o \ and \  OC = OD

\angle ODC = \frac{1}{2} (180-72) = 54^o

(iv) In cyclic quadrilateral AECD

\angle AEC + \angle ADC = 180^o (opposite angles of a cyclic quadrilateral are supplementary)

\angle AEC = 180-54 = 126^o

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c12.jpgQuestion 7: In the diagram, O is the center of the circle and the length of \widehat{AB} = 2 \times \widehat{BC} . If \angle AOB=108^o find:

(i) \angle CAB

(ii) \angle ADB [1996]

Answer:

Given \widehat{AB} = 2 \times \widehat{BC}, \angle AOB = 108^o

(i) \angle AOB = 2 \angle BOC 

\Rightarrow \angle BOC = \frac{108}{2} = 54

(ii) \angle BAC = \frac{1}{2} \angle BOC (angle subtended at the center is twice that subtended at the circumference by a chord)

\angle BAC = \frac{1}{2} \times 54 = 27^o

(iii) Similarly, \angle ACB = \frac{1}{2} \angle AOB = \frac{108}{2} = 54

In cyclic quadrilateral ADBC

\angle ADB + \angle ACB = 180^o

\Rightarrow \angle ADB = 180-54=126^o

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18c4Question 8: In the given diagram, AB=BC=CD and \angle ABC = 132^o . Find:

(i) \angle AEB

(ii) \angle AED

(iii) \angle COD [1993]

Answer:

Given AB = BC = CD, ABC = 32^o

\Rightarrow \angle AOB = \angle BOC = \angle COD (equal arcs subtend equal angles at the center of a circle)

(i) In cyclic quadrilateral ABCE c11.jpg

\angle ABC + \angle AEC = 180^o (opposite angles in a cyclic quadrilateral are supplementary)

132+ \angle AEC = 180 \Rightarrow \angle AEC = 48^o

Since AB = BC \Rightarrow \angle AEB = \angle BEC (equal chords subtend equal angles on the circumference)

\therefore \angle AEB = \frac{1}{2} \angle AEC = 24

(ii) Since AB = BC = CD

\angle AEB = \angle BEC = \angle CED

\angle AED = 24+24+24=72

(iii) \widehat{CD} subtends \angle COD at the center and \angle CED on the circumference

\therefore \angle COD = 2 \angle CED = 2 \times 24 = 48^o

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Question 9: In a regular pentagon ABCDE inscribed in a circle, find the ratio of the \angle EDA : \angle ADC [1990]

Answer:c10.jpg

Consider \widehat{AE}

It subtends \angle AOE at the center and \angle ADE at the circumference

Therefore \angle ADE = \frac{1}{2} \angle AOE = \frac{1}{2} \times \frac{360}{5} = 36^o

Similarly, for \widehat{BC} we have \angle BDC = 36^o

\angle ADC = \angle ADB + \angle BDC = 36+36=72^o

\angle ADE : \angle ADC = 36 :72 =  1:2

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c12Question 10: In the given figure, \angle BAD = 65^o, \angle ABD = 70^o and \angle BDC = 45^o

(i) Prove that AC is a diameter of the circle.

(ii) find \angle ACB    [2013]

Answer:

Given \angle BAD = 65^o, \angle ABD = 70^o and \angle BDC = 45^o .

(i) In \triangle ABD

65^o+70^o+ \angle ADB = 180^o

\Rightarrow \angle ADB = 180^o-135^o = 45^o

Therefore \angle ADC = 45^o+45^o = 90^o

Therefore AC is the diameter (Theorem 11)

(ii) \angle ACB = \angle ADB (angles in the same segment)

Therefore \angle ACB = 45^o since \angle ADB = 45^o

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Question 11: In each of the following figures, O is the center of the circle. Find the values of a, b, c \ and \  d .  [2007]

(i)c151 (ii)c152
(iii)c153 (iv)c154

Answer:

(i) BD is the diameter

\therefore \angle DAB = 90^o

\therefore \angle ADB = 180^o-90^o-3^o5=55^o

Since \angle ADB = \angle ACB = 55^o (angle in same segment) Therefore a = 55^o

(ii) \angle ADB = \angle ACB (angle in same segment)

In \triangle ECB ,

\angle BEC = 180^o-120^o = 60^o

\therefore \angle ACB = 180^o - 60^o - 25^o = 95^o

\therefore \angle ADB = 95^o= b

(iii) In \triangle AOB

AO = OB = radius

2 \angle ACB = \angle AOB

\therefore \angle AOB = 100^o

\therefore 2c+ 100^o= 180^o \Rightarrow c = 40^o 

(iv) Since AB is the diameter

\therefore \angle BAP = 180^o-90^o-45^o = 45^o

\angle PAB = \angle PCB = 45^o \Rightarrow d = 45^o (angles in the same segment)

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c65Question 12: In the given figure AB = AC = CD and \angle ADC = 38^o , calculate (i) \angle ABC (ii) \angle BEC [1995]

Answer:

(i) AC = CD

\angle CAD = \angle CDA = 38^o

\angle ACD = 180^o-38^o - 38^o = 104^o

\angle ACB = 180^o-104^o = 76^o

AB = AC

\angle ABC = \angle ACB = 76^o

(ii) \angle BAC = 180^o-76^o-76^o = 38^o

\angle BAC = \angle BEC = 38^o (angles in the same segment)

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c66Question 13: In the given figure AC is the diameter of the circle with center O . Chord BD \perp AC . Write down the angles p, q and r in terms of x . [1996]

Answer:

\angle AOB = 2 \angle ACB = 2 \angle ADB (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

x = 2q \Rightarrow q=\frac{x}{2}

\angle ADB = \frac{x}{2}

\angle ADC = 90^o (angles in a semicircle)

r +\frac{x}{2} = 90^o \Rightarrow r = 90-\frac{x}{2}

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

p = 90^o-q \Rightarrow p = 90^o-\frac{x}{2}

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c67Question 14:  In the given figure AC is the diameter of the circle with center O . $CD \parallel BE &s=0$. \angle AOB = 80^o and \angle ACE = 110^o . Calculate (i) \angle BEC (ii) \angle BCD (iii) \angle CED [1998]

Answer:

(i) \angle BOC = 180^o-80^o=100^o (Since AC is a straight line)

\angle BOC = 2 \angle BEC   (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\Rightarrow \angle BEC = \frac{100}{2} = 50^o

(ii) DC \parallel EB

\angle DCE = \angle BEC = 50^o

\angle AOB = 80^o (given)

\Rightarrow \angle ACB = \frac{1}{2} \angle AOB = 40^o  (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\angle BCD = \angle ACB + \angle ACE + \angle DCE = 40^o+10^o+50^o = 100^o

(iii) \angle BED = 180^o-\angle BCD = 180^o-100^o = 80^o (cyclic quadrilateral)

\angle CED + 50^o = 80^o \Rightarrow \angle CED = 30^o

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c68Question 15: In the given figure, AE is the diameter of the circle. Write down the numerical value of \angle ABC + \angle CDE . Give reasons for your answer. [1998]

Answer:

\angle AOC = \frac{180}{2} = 90^o

\angle AOC = 2 \angle AEC  (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\Rightarrow \angle AEC = \frac{90}{2} = 45^o

ABCE is a cyclic quadrilateral

\therefore \angle ABC + \angle AEC = 180 ^o

\Rightarrow \angle ABC = 180-45 = 135^o

Similarly, \angle CDE = 135^o

\therefore \angle ABC + \angle CDE = 135^o + 135^o = 270^o

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c69.jpgQuestion 16: In the given figure AOC is the diameter and AC \parallel ED . If \angle CBE = 64^o , calculate \angle DEC . [1991]

Answer:

\angle ABC = 90^o (angle in a semi circle)

\angle ABE = 90^o-64^o = 26^o

\angle ABE = \angle ACE = 26^o (angles in the same segment)

AC \parallel ED

\therefore \angle DEC = \angle ACE = 26^o (alternate angles)

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c610Question 17: Use the given figure below to find (i) \angle BAD (ii) \angle DQB  [1987]

Answer:

(i) In \triangle ADP 

\angle BAD = 180^o - 85^o - 40^o = 55^o 

(ii) \angle ABC = 180^o - \angle ADC = 180^o - 85^o = 95^o 

\angle AQB = 180^o - 95^o - 55^o = 30^o 

\Rightarrow \angle DQB = 30^o 

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c79Question 18: In the given figure AOB is the diameter and DC \parallel AB . If \angle CAB = x^o , find in terms of x , the values of: (i) \angle COB (ii) \angle DOC (iii) \angle DAC (iv) \angle ADC [1991]

Answer:

(i) \angle OCB = 2 \angle CAB = 2x (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

(ii) \angle OCD = \angle COB = 2 x (alternate angles)

In \triangle OCD 

OC = OC (radius of the same circle)

\angle ODC = \angle OCD = 2x

\angle DOC = 180^o-2x-2x = 180^o-4x

(iii) \angle DAC = \frac{1}{2} \angle DOC = \frac{1}{2} (180^o-4x) = 90^o-2x  (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

(iv) DC \parallel AO (given)

\therefore \angle ACD = \angle OAC = x (alternate angles)

\therefore \angle ADC = 180^o - \angle DAC-\angle ACD = 180^o-(90^o-2x) - x = 90^o+x

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c78Question 19: In the figure AB is the diameter of the circle with center O . \angle BCD = 130^o .  Find (i) \angle DAB (ii) \angle DBA [2012]

Answer:

(i) \angle DAB = 180^o-\angle DCB = 180^o-130^o = 50^o  \ (ABCD  is a cyclic quadrilateral)

(ii) In \triangle ADB 

\angle DAB + \angle ADB + \angle DBA = 180^o 

\Rightarrow \angle DBA = 180^o-50^o-90^o = 140^o 

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c77Question 20: In the given figure PQ is the diameter of the circle whose center is O . Given \angle ROS=42^o , calculate \angle RTS . [1992]

Answer:

Join P and S as shown in the diagram.

\angle PSQ = 90^o (angle in a semi circle)

c715\angle SPR = \frac{1}{2} \angle ROS

\Rightarrow \angle SPT = \frac{1}{2} \times 42^o = 21^o

In \triangle PST

\angle PST = 90^o-\angle SPT

\Rightarrow \angle RTS = 90^o-21^o = 69^o

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c76Question 21: In the given figure PQ is the diameter. Chord SR \parallel PQ . Given the \angle PQR = 58^o , calculate (i) \angle RPQ (ii) \angle STP [1989]

Answer:

c714Join P and R as shown in the diagram.

(i) \angle PRQ = 90^o (angles in the semi circle)

\therefore \angle RPQ = 90^o- \angle PQR = 90^o-58^o = 32^o

(ii) SP \parallel PQ (given)

\angle PSR = \angle RPQ = 32^o (alternate angles)

\angle SPT = 180^o-\angle PSR = 180^o-32^o = 148^o (PTSR is a cyclic quadrilateral)

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c75Question 22: AB   is the diameter of the circle with center O  . OD \parallel BC   and \angle AOD = 60^o  . Calculate the numerical values of (i) \angle ABD   (ii) \angle DBC   (iii) \angle ADC [1987]

Answer:

Join B and D as shown in the diagram.

(i) \angle ABD = \frac{1}{2} \angle AOD = \frac{1}{2} \times 60^o = 30^o  (angle subtended by an chord on the center is double that subtended by the same chord on the circumference)c713

(ii) \angle BDA = 90^o (angle in semi circle)

Since \angle OAD = 60^o \Rightarrow \triangle OAD  is equilateral

\angle ODB = 90^o-ODA = 90^o-60^o = 30^o

Since OD \parallel BC

\angle DBC = \angle ODB = 30^o (alternate angles)

(ii) \angle ABC = \angle ABD + \angle DBC = 30^o +30^o = 60^o

\angle ADC = 180^o -\angle ABC = 180^o- 60^o = 120^o ( ABCD is a cyclic quadrilateral)

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c74Question 23: In the given figure, the center O of the smaller circle lies on the circumference of the bigger circle. If \angle APB = 75^o and  \angle BCD = 40^o , find: (i) \angle AOB (ii) \angle ACB (iii) \angle ABD (iv) \angle ADB . [1984]

Answer:

Join A and B as shown in the diagram.

(i) \angle AOB = 2 \angle APB = 2 \times 75^o = 150^o  (angle subtended by an chord on the center is double that subtended by the same chord on the circumference)c712

(ii) \angle ACB = 180^o-\angle AOB^o = 180^o-150^o = 30^o  (AOBC  is a cyclic quadrilateral)

(iii) \angle ABD = 180^o - \angle ACD = 180^o-(40^o+30^o) = 110^o  (ABDC  is a cyclic quadrilateral)

(iv) \angle ADB = 180^o - \angle AOB = 180^o-150^o=  30^o  (ADBO is a cyclic quadrilateral)

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Question 24: O is the center of the circle of radius 10 \ cm . P is any point in the circle such that OP=6 \ cm . A is the point travelling along the circumference, x is the distance from A to P . What are the least and the greatest values of x \ in \ cm . What is the position of the points O, P \ and \ A at these values . [1992]

Answer:

When P is on OA , the least distance of x = 4 \ cm

When P is on extended OA , the greatest distance of x = 16 \ cm 

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c11Question 25: In the given figure, O is the center of the circle. AB and CD are two chords of circle. OM \perp AB and ON \perp CD . AB = 24 \ cm, OM = 5 \ cm , ON = 12 \ cm . Find the 

(i) radius of the circle

(ii) length of chord CD [2014]

c20.jpgAnswer:

AB = 24 \ cm , OM = 5 \ cm , ON =12 \ cm

OM \perp AB

\Rightarrow AM = BM

Therefore AM = MB = 12 \ cm

(i) Consider \triangle AOM

AO^2 = AM^2 +OM^2 = 12^2+5^2 = 169

\therefore AO = 13 \ cm = Radius of the circle.

(ii) Now consider \triangle CNO

CO^2 = CN^2+ON^2

\Rightarrow CN^2 = CO^2-ON^2 = 13^2 - 12^2 = 169 - 144 = 25

\therefore CN= 5 \ cm

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2011-1Question 26: In the given figure O is the center of the circle Tangent of A and B meet at C if \angle AOC = 30^o , find (i) \angle BCO (ii) \angle AOB (iii) \angle APB [2011]

Answer:

Consider \triangle AOC and \triangle BOC

\angle OAC = \angle OBC = 90^o

OC is common

AC = BC   (two tangents drawn from a point on a circle are of equal lengths)

Therefore \triangle AOC \cong \triangle BOC

(i) \angle ACO = \angle BCO = 30^o

(ii) \angle AOC = 180^o - 90^o - 30^o = 60^o

\angle BOC = 180^o - 90^o - 30^o = 60^o

\therefore \angle AOB = \angle AOC + \angle BOC = 120^o

(iii) \angle AOB = 2 \angle APB (chord subtends twice the angle at the center than that it subtends on the circumference)

\Rightarrow \angle AOB = 2 \angle APB 

\Rightarrow \angle APB = 60^o

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Question 27: In the following figure O is the center of the circle and AB is a tangent to it at  point B. \angle BDC = 65^o .  Find \angle BAO .     [2010]

2010-2

Answer:

AB is a tangent to the circle.

\Rightarrow \angle ABO = 90^o

\angle BDC = 65^o (given)

\angle BCD = 90^o-65^o = 25^o (angle at the center)

\angle BOE = 2 \times 25^o = 50^o

\angle BAO = 90^o - \angle BOE 

\therefore \angle BAO = 90^o - 50^o = 40^o

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