\displaystyle \textbf{Question 1: }\text{In the figure given below, }AB\text{ and }CD\text{ are two parallel chords}
\displaystyle \text{and }O\text{ is the center. If the radius of the circle is }15\text{ cm, find the distance }MN
\displaystyle \text{between the two chords of lengths }24\text{ cm and }18\text{ cm respectively.}\hfill \text{[\text{ICSE } 2010]}
18d21
\displaystyle \text{Answer:}
\displaystyle \text{The perpendicular from the center of a circle to a chord bisects the chord.}
\displaystyle \therefore AM=MB=12\text{ cm and }CN=ND=9\text{ cm}
\displaystyle \text{In right }\triangle OMB,d24
\displaystyle OB^2=OM^2+MB^2
\displaystyle 15^2=OM^2+12^2
\displaystyle OM^2=225-144=81
\displaystyle OM=9\text{ cm}
\displaystyle \text{In right }\triangle OND,
\displaystyle OD^2=ON^2+ND^2
\displaystyle 15^2=ON^2+9^2
\displaystyle ON^2=225-81=144
\displaystyle ON=12\text{ cm}
\displaystyle MN=OM+ON
\displaystyle =9+12=21\text{ cm}
\displaystyle \therefore MN=21\text{ cm}
\\

\displaystyle \textbf{Question 2: }\text{In the given figure, }\angle ACE=43^\circ\text{ and }\angle CAF=62^\circ;
\displaystyle \text{find the values of }a,\ b\text{ and }c.\hfill \text{[\text{ICSE } 2007]} 18d17\displaystyle \text{Answer:}
\displaystyle \text{Given, }\angle ACE=43^\circ\text{ and }\angle CAF=62^\circ
\displaystyle \angle BAE+\angle BDE=180^\circ\quad (\text{opposite angles of cyclic quadrilateral }ABDE)
\displaystyle \angle BDE=180^\circ-62^\circ=118^\circ
\displaystyle \angle c+\angle BDE=180^\circ\quad (\text{linear pair})
\displaystyle \therefore c=180^\circ-118^\circ=62^\circ
\displaystyle \text{In }\triangle ACE,
\displaystyle \angle CEA=180^\circ-(62^\circ+43^\circ)=75^\circ
\displaystyle \angle DEF=180^\circ-75^\circ=105^\circ\quad (\text{linear pair})
\displaystyle \angle a+\angle CEA=180^\circ\quad (\text{opposite angles of cyclic quadrilateral }ABDE)
\displaystyle \therefore a=180^\circ-75^\circ=105^\circ
\displaystyle \text{In }\triangle DEF,
\displaystyle b+105^\circ+62^\circ=180^\circ
\displaystyle b=13^\circ
\displaystyle \therefore a=105^\circ,\ b=13^\circ\text{ and }c=62^\circ
\\

\displaystyle \textbf{Question 3: }\text{The following figure shows a circle with }PR\text{ as its diameter.}
\displaystyle \text{If }PQ=7\text{ cm and }QR=3RS=6\text{ cm, find the perimeter of the cyclic} 18d15\displaystyle \text{quadrilateral }PQRS.\hfill \text{[\text{ICSE } 1992]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }PQ=7\text{ cm},\ QR=6\text{ cm and }3RS=6\text{ cm}
\displaystyle \therefore RS=2\text{ cm}
\displaystyle \text{Since }PR\text{ is the diameter, }\angle PQR=90^\circ
\displaystyle \text{In right }\triangle PQR,
\displaystyle PR=\sqrt{PQ^2+QR^2}
\displaystyle =\sqrt{7^2+6^2}=\sqrt{49+36}=\sqrt{85}
\displaystyle \text{Also, }\angle PSR=90^\circ
\displaystyle \text{In right }\triangle PSR,
\displaystyle PR^2=PS^2+RS^2
\displaystyle 85=PS^2+2^2
\displaystyle PS^2=81
\displaystyle PS=9\text{ cm}
\displaystyle \therefore \text{Perimeter of }PQRS=PQ+QR+RS+PS
\displaystyle =7+6+2+9=24\text{ cm}
\\

\displaystyle \textbf{Question 4: }\text{In the given circle with diameter }AB,\text{ find the value of }x.\hfill \text{[\text{ICSE } 2003]} 18d1\displaystyle \text{Answer:}
\displaystyle \angle ABD=\angle ACD=30^\circ\quad \text{angles in the same segment}
\displaystyle \text{Since }AB\text{ is the diameter,}
\displaystyle \angle ADB=90^\circ\quad \text{angle in a semicircle}
\displaystyle \text{In }\triangle ADB,
\displaystyle \angle BAD+\angle ADB+\angle ABD=180^\circ
\displaystyle x+90^\circ+30^\circ=180^\circ
\displaystyle x=60^\circ
\displaystyle \therefore x=60^\circ
\\

\displaystyle \textbf{Question 5: }\text{In the given diagram, }\angle DBC=58^\circ,\ BD\text{ is a diameter}
\displaystyle \text{of the circle. Calculate: (i) }\angle BDC\text{ (ii) }\angle BEC\text{ (iii) }\angle BAC.\hfill \text{[\text{ICSE } 2014]} 18d7\displaystyle \text{Answer:}
\displaystyle \text{Given, }BD\text{ is a diameter and }\angle DBC=58^\circ
\displaystyle \text{Since }BD\text{ is a diameter,}
\displaystyle \angle BCD=90^\circ\quad \text{angle in a semicircle}
\displaystyle \text{In }\triangle BDC,
\displaystyle \angle DBC+\angle BDC+\angle BCD=180^\circ
\displaystyle 58^\circ+\angle BDC+90^\circ=180^\circ
\displaystyle \angle BDC=32^\circ
\displaystyle \therefore \text{(i) }\angle BDC=32^\circ
\displaystyle \angle BAC=\angle BDC=32^\circ\quad \text{angles in the same segment}
\displaystyle \therefore \text{(iii) }\angle BAC=32^\circ
\displaystyle \text{Since }ABEC\text{ is a cyclic quadrilateral,}
\displaystyle \angle BAC+\angle BEC=180^\circ
\displaystyle 32^\circ+\angle BEC=180^\circ
\displaystyle \angle BEC=148^\circ
\displaystyle \therefore \text{(ii) }\angle BEC=148^\circ
\\

\displaystyle \textbf{Question 6: }\text{In the given diagram, }BD\text{ is the side of a regular hexagon, }DC\text{ is the side}
\displaystyle \text{of a regular pentagon and }AD\text{ is a diameter. Calculate:} 18c1\displaystyle \text{(i) }\angle ADC\qquad \text{(ii) }\angle BDA\qquad \text{(iii) }\angle ABC\qquad \text{(iv) }\angle AEC.\hfill \text{[\text{ICSE } 1984]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }BD\text{ is the side of a regular hexagon,}
\displaystyle \angle BOD=\frac{360^\circ}{6}=60^\circc15.jpg
\displaystyle \text{Since }DC\text{ is the side of a regular pentagon,}
\displaystyle \angle DOC=\frac{360^\circ}{5}=72^\circ
\displaystyle \text{Since }AD\text{ is a diameter, }\angle AOD=180^\circ
\displaystyle \angle AOC=180^\circ-\angle DOC
\displaystyle =180^\circ-72^\circ=108^\circ
\displaystyle \angle ADC=\frac{1}{2}\angle AOC=\frac{1}{2}\times108^\circ=54^\circ
\displaystyle \therefore \text{(i) }\angle ADC=54^\circ
\displaystyle \text{Also, }\angle AOB=180^\circ-\angle BOD
\displaystyle =180^\circ-60^\circ=120^\circ
\displaystyle \angle BDA=\frac{1}{2}\angle AOB=\frac{1}{2}\times120^\circ=60^\circ
\displaystyle \therefore \text{(ii) }\angle BDA=60^\circ
\displaystyle \angle ABC=\frac{1}{2}\angle AOC
\displaystyle =\frac{1}{2}\times108^\circ=54^\circ
\displaystyle \therefore \text{(iii) }\angle ABC=54^\circ
\displaystyle \text{In cyclic quadrilateral }AECD,
\displaystyle \angle AEC+\angle ADC=180^\circ
\displaystyle \angle AEC+54^\circ=180^\circ
\displaystyle \angle AEC=126^\circ
\displaystyle \therefore \text{(iv) }\angle AEC=126^\circ
\\

\displaystyle \textbf{Question 7: }\text{In the diagram, }O\text{ is the center of the circle and the length of }\widehat{AB}=2\times\widehat{BC}.
\displaystyle \text{If }\angle AOB=108^\circ,\text{ find: (i) }\angle CAB\text{ (ii) }\angle ADB.\hfill \text{[\text{ICSE } 1996]} c12.jpg\displaystyle \text{Answer:}
\displaystyle \text{Given, }\widehat{AB}=2\times\widehat{BC}\text{ and }\angle AOB=108^\circ
\displaystyle \therefore \angle AOB=2\angle BOC
\displaystyle \angle BOC=\frac{108^\circ}{2}=54^\circ
\displaystyle \angle CAB=\frac{1}{2}\angle BOC
\displaystyle =\frac{1}{2}\times54^\circ=27^\circ
\displaystyle \therefore \text{(i) }\angle CAB=27^\circ
\displaystyle \angle ACB=\frac{1}{2}\angle AOB
\displaystyle =\frac{1}{2}\times108^\circ=54^\circ
\displaystyle \text{In cyclic quadrilateral }ADBC,
\displaystyle \angle ADB+\angle ACB=180^\circ
\displaystyle \angle ADB+54^\circ=180^\circ
\displaystyle \angle ADB=126^\circ
\displaystyle \therefore \text{(ii) }\angle ADB=126^\circ
\\

\displaystyle \textbf{Question 8: }\text{In the given diagram, }AB=BC=CD\text{ and }\angle ABC=132^\circ.
\displaystyle \text{Find: (i) }\angle AEB\qquad \text{(ii) }\angle AED\qquad \text{(iii) }\angle COD.\hfill \text{[\text{ICSE } 1993]} 18c4\displaystyle \text{Answer:}
\displaystyle \text{Given, }AB=BC=CD\text{ and }\angle ABC=132^\circ
\displaystyle \text{Equal chords subtend equal arcs.}c11.jpg
\displaystyle \therefore \widehat{AB}=\widehat{BC}=\widehat{CD}
\displaystyle \text{Since }ABCE\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle AEC=180^\circ
\displaystyle 132^\circ+\angle AEC=180^\circ
\displaystyle \angle AEC=48^\circ
\displaystyle \text{Since }AB=BC,\ \angle AEB=\angle BEC
\displaystyle \therefore \angle AEB=\frac{1}{2}\angle AEC=\frac{1}{2}\times48^\circ=24^\circ
\displaystyle \therefore \text{(i) }\angle AEB=24^\circ
\displaystyle \text{Since }AB=BC=CD,
\displaystyle \angle AEB=\angle BEC=\angle CED=24^\circ
\displaystyle \angle AED=\angle AEB+\angle BEC+\angle CED
\displaystyle =24^\circ+24^\circ+24^\circ=72^\circ
\displaystyle \therefore \text{(ii) }\angle AED=72^\circ
\displaystyle \text{Chord }CD\text{ subtends }\angle CED\text{ at the circumference and }\angle COD\text{ at the center.}
\displaystyle \angle COD=2\angle CED
\displaystyle =2\times24^\circ=48^\circ
\displaystyle \therefore \text{(iii) }\angle COD=48^\circ
\\

\displaystyle \textbf{Question 9: }\text{In a regular pentagon }ABCDE\text{ inscribed in a circle,}
\displaystyle \text{find the ratio of }\angle EDA:\angle ADC.\hfill \text{[\text{ICSE } 1990]} c10.jpg\displaystyle \text{Answer:}
\displaystyle \text{In a regular pentagon, each side subtends an angle }\frac{360^\circ}{5}=72^\circ\text{ at the center.}
\displaystyle \text{Arc }AE\text{ subtends }\angle AOE=72^\circ\text{ at the center}
\displaystyle \text{and }\angle EDA\text{ at the circumference.}
\displaystyle \therefore \angle EDA=\frac{1}{2}\times72^\circ=36^\circ
\displaystyle \text{Similarly, }\angle ADB=36^\circ\text{ and }\angle BDC=36^\circ
\displaystyle \angle ADC=\angle ADB+\angle BDC
\displaystyle =36^\circ+36^\circ=72^\circ
\displaystyle \therefore \angle EDA:\angle ADC=36^\circ:72^\circ=1:2
\\

\displaystyle \textbf{Question 10: }\text{In the given figure, }\angle BAD=65^\circ,\ \angle ABD=70^\circ
\displaystyle \text{and }\angle BDC=45^\circ.\text{ (i) Prove that }AC\text{ is a diameter of the circle.}
\displaystyle \text{(ii) Find }\angle ACB.\hfill \text{[\text{ICSE } 2013]} c12\displaystyle \text{Answer:}
\displaystyle \text{Given, }\angle BAD=65^\circ,\ \angle ABD=70^\circ\text{ and }\angle BDC=45^\circ
\displaystyle \text{(i) In }\triangle ABD,
\displaystyle \angle BAD+\angle ABD+\angle ADB=180^\circ
\displaystyle 65^\circ+70^\circ+\angle ADB=180^\circ
\displaystyle \angle ADB=45^\circ
\displaystyle \angle ADC=\angle ADB+\angle BDC
\displaystyle =45^\circ+45^\circ=90^\circ
\displaystyle \text{Since }\angle ADC=90^\circ,\ AC\text{ is a diameter of the circle.}
\displaystyle \text{(ii) }\angle ACB=\angle ADB\quad \text{angles in the same segment}
\displaystyle \therefore \angle ACB=45^\circ
\\

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

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

\displaystyle \text{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)
\\

\displaystyle \textbf{Question 12: }\text{In the given figure }AB=AC=CD\text{ and }\angle ADC=38^\circ,
\displaystyle \text{calculate (i) }\angle ABC\text{ (ii) }\angle BEC.\hfill \text{[\text{ICSE } 1995]} c65\displaystyle \text{Answer:}
\displaystyle \text{Given, }AC=CD\text{ and }\angle ADC=38^\circ
\displaystyle \therefore \angle CAD=\angle ADC=38^\circ
\displaystyle \text{In }\triangle ACD,
\displaystyle \angle ACD=180^\circ-38^\circ-38^\circ=104^\circ
\displaystyle \angle ACB=180^\circ-104^\circ=76^\circ
\displaystyle \text{Since }AB=AC,
\displaystyle \angle ABC=\angle ACB=76^\circ
\displaystyle \therefore \text{(i) }\angle ABC=76^\circ
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle BAC=180^\circ-76^\circ-76^\circ=28^\circ
\displaystyle \angle BEC=\angle BAC\quad \text{angles in the same segment}
\displaystyle \therefore \text{(ii) }\angle BEC=28^\circ
\\

\displaystyle \textbf{Question 13: }\text{In the given figure }AC\text{ is the diameter of the circle with center }O.
\displaystyle \text{Chord }BD\perp AC.\text{ Write down the angles }p,\ q\text{ and }r\text{ in terms of }x.\hfill \text{[1996]} c66\displaystyle \text{Answer:}
\displaystyle \angle AOB=x
\displaystyle \text{The angle subtended by chord }AB\text{ at the center is double the angle subtended at the circumference.}
\displaystyle \therefore \angle ACB=\angle ADB=\frac{x}{2}
\displaystyle \therefore q=\frac{x}{2}
\displaystyle \text{Since }AC\text{ is a diameter,}
\displaystyle \angle ADC=90^\circ
\displaystyle \angle ADB+r=90^\circ
\displaystyle \frac{x}{2}+r=90^\circ
\displaystyle \therefore r=90^\circ-\frac{x}{2}
\displaystyle \text{Also, }\angle ABC=90^\circ\quad \text{angle in a semicircle}
\displaystyle \text{In }\triangle ABC,
\displaystyle p+q+90^\circ=180^\circ
\displaystyle p+\frac{x}{2}=90^\circ
\displaystyle \therefore p=90^\circ-\frac{x}{2}
\displaystyle \therefore p=90^\circ-\frac{x}{2},\ q=\frac{x}{2}\text{ and }r=90^\circ-\frac{x}{2}
\\

\displaystyle \textbf{Question 14: }\text{In the given figure }AC\text{ is the diameter of the circle with center }O.
\displaystyle CD\parallel BE,\ \angle AOB=80^\circ\text{ and }\angle ACE=10^\circ.\text{ Calculate (i) }\angle BEC c67\displaystyle \text{(ii) }\angle BCD\text{ (iii) }\angle CED.\hfill \text{[1998]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }AC\text{ is a diameter, }\angle AOC=180^\circ
\displaystyle \angle BOC=180^\circ-80^\circ=100^\circ
\displaystyle \angle BEC=\frac{1}{2}\angle BOC
\displaystyle =\frac{1}{2}\times100^\circ=50^\circ
\displaystyle \therefore \text{(i) }\angle BEC=50^\circ
\displaystyle \text{Since }CD\parallel BE,
\displaystyle \angle DCE=\angle BEC=50^\circ
\displaystyle \angle ACB=\frac{1}{2}\angle AOB
\displaystyle =\frac{1}{2}\times80^\circ=40^\circ
\displaystyle \angle BCD=\angle BCA+\angle ACE+\angle DCE
\displaystyle =40^\circ+10^\circ+50^\circ=100^\circ
\displaystyle \therefore \text{(ii) }\angle BCD=100^\circ
\displaystyle \text{Since }BCDE\text{ is a cyclic quadrilateral,}
\displaystyle \angle BED+\angle BCD=180^\circ
\displaystyle \angle BED=180^\circ-100^\circ=80^\circ
\displaystyle \angle CED+\angle BEC=\angle BED
\displaystyle \angle CED+50^\circ=80^\circ
\displaystyle \angle CED=30^\circ
\displaystyle \therefore \text{(iii) }\angle CED=30^\circ
\\

\displaystyle \textbf{Question 15: }\text{In the given figure, }AE\text{ is the diameter of the circle. Write down}
\displaystyle \text{the numerical value of }\angle ABC+\angle CDE.\text{ Give reasons for your answer.}\hfill \text{[1998]} c68\displaystyle \text{Answer:}
\displaystyle \text{Since }A,\ B,\ C,\ D,\ E\text{ lie on the same circle, }ABCDE\text{ is a cyclic pentagon.}
\displaystyle \text{Join }AC.
\displaystyle \text{In cyclic quadrilateral }ABCE,
\displaystyle \angle ABC+\angle AEC=180^\circ
\displaystyle \text{In cyclic quadrilateral }ACDE,
\displaystyle \angle CDE+\angle CAE=180^\circ
\displaystyle \text{Adding,}
\displaystyle \angle ABC+\angle CDE+\angle AEC+\angle CAE=360^\circ
\displaystyle \text{Since }AE\text{ is a diameter,}
\displaystyle \angle ACE=90^\circ\quad \text{angle in a semicircle}
\displaystyle \text{In }\triangle ACE,
\displaystyle \angle CAE+\angle AEC=90^\circ
\displaystyle \therefore \angle ABC+\angle CDE+90^\circ=360^\circ
\displaystyle \angle ABC+\angle CDE=270^\circ
\displaystyle \therefore \text{The numerical value of }\angle ABC+\angle CDE\text{ is }270^\circ
\\

\displaystyle \textbf{Question 16: }\text{In the given figure, }AOC\text{ is the diameter and }AC\parallel ED.
\displaystyle \text{If }\angle CBE=64^\circ,\text{ calculate }\angle DEC.\hfill \text{[1991]} c69.jpg\displaystyle \text{Answer:}
\displaystyle \text{Since }AOC\text{ is the diameter,}
\displaystyle \angle ABC=90^\circ\quad \text{angle in a semicircle}
\displaystyle \angle ABE=90^\circ-64^\circ=26^\circ
\displaystyle \angle ACE=\angle ABE=26^\circ\quad \text{angles in the same segment}
\displaystyle \text{Since }AC\parallel ED,
\displaystyle \angle DEC=\angle ACE=26^\circ\quad \text{alternate angles}
\displaystyle \therefore \angle DEC=26^\circ
\\

\displaystyle \textbf{Question 17: }\text{Use the given figure below to find}
\displaystyle \text{(i) }\angle BAD\qquad \text{(ii) }\angle DQB.\hfill \text{[1987]} c610\displaystyle \text{Answer:}
\displaystyle \text{Since }A,\ B,\ P\text{ are collinear and }D,\ C,\ P\text{ are collinear,}
\displaystyle \angle APD=40^\circ
\displaystyle \text{In }\triangle ADP,
\displaystyle \angle BAD=180^\circ-85^\circ-40^\circ
\displaystyle =55^\circ
\displaystyle \therefore \text{(i) }\angle BAD=55^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle ADC=180^\circ
\displaystyle \angle ABC=180^\circ-85^\circ=95^\circ
\displaystyle \text{Since }A,\ D,\ Q\text{ are collinear and }B,\ C,\ Q\text{ are collinear,}
\displaystyle \angle QAB=\angle BAD=55^\circ
\displaystyle \angle QBA=\angle ABC=95^\circ
\displaystyle \text{In }\triangle AQB,
\displaystyle \angle AQB=180^\circ-55^\circ-95^\circ=30^\circ
\displaystyle \therefore \angle DQB=30^\circ
\displaystyle \therefore \text{(ii) }\angle DQB=30^\circ
\\

\displaystyle \textbf{Question 18: }\text{In the given figure, }AOB\text{ is the diameter and }DC\parallel AB.
\displaystyle \text{If }\angle CAB=x^\circ,\text{ find in terms of }x,\text{ the values of:} c79\displaystyle \text{(i) }\angle COB\qquad \text{(ii) }\angle DOC\qquad \text{(iii) }\angle DAC\qquad \text{(iv) }\angle ADC.\hfill \text{[1991]}
\displaystyle \text{Answer:}
\displaystyle \angle COB=2\angle CAB
\displaystyle \therefore \angle COB=2x^\circ
\displaystyle \text{Since }DC\parallel AB,
\displaystyle \angle OCD=\angle COB=2x^\circ\quad \text{alternate angles}
\displaystyle \text{In }\triangle OCD,\ OC=OD
\displaystyle \therefore \angle ODC=\angle OCD=2x^\circ
\displaystyle \angle DOC=180^\circ-2x^\circ-2x^\circ
\displaystyle =180^\circ-4x^\circ
\displaystyle \therefore \angle DOC=180^\circ-4x^\circ
\displaystyle \angle DAC=\frac{1}{2}\angle DOC
\displaystyle =\frac{1}{2}(180^\circ-4x^\circ)=90^\circ-2x^\circ
\displaystyle \text{Since }DC\parallel AB,
\displaystyle \angle ACD=\angle CAB=x^\circ\quad \text{alternate angles}
\displaystyle \text{In }\triangle ACD,
\displaystyle \angle ADC=180^\circ-\angle DAC-\angle ACD
\displaystyle =180^\circ-(90^\circ-2x^\circ)-x^\circ
\displaystyle =90^\circ+x^\circ
\displaystyle \therefore \angle COB=2x^\circ,\ \angle DOC=180^\circ-4x^\circ,
\displaystyle \angle DAC=90^\circ-2x^\circ\text{ and }\angle ADC=90^\circ+x^\circ
\\

\displaystyle \textbf{Question 19: }\text{In the figure, }AB\text{ is the diameter of the circle with center }O.
\displaystyle \angle BCD=130^\circ.\text{ Find (i) }\angle DAB\text{ (ii) }\angle DBA.\hfill \text{[2012]} c78\displaystyle \text{Answer:}
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle DAB+\angle DCB=180^\circ
\displaystyle \angle DAB+130^\circ=180^\circ
\displaystyle \angle DAB=50^\circ
\displaystyle \therefore \text{(i) }\angle DAB=50^\circ
\displaystyle \text{Since }AB\text{ is the diameter,}
\displaystyle \angle ADB=90^\circ\quad \text{angle in a semicircle}
\displaystyle \text{In }\triangle ADB,
\displaystyle \angle DAB+\angle ADB+\angle DBA=180^\circ
\displaystyle 50^\circ+90^\circ+\angle DBA=180^\circ
\displaystyle \angle DBA=40^\circ
\displaystyle \therefore \text{(ii) }\angle DBA=40^\circ
\\

\displaystyle \textbf{Question 20: }\text{In the given figure, }PQ\text{ is the diameter of the circle whose center is }O.
\displaystyle \text{Given }\angle ROS=42^\circ,\text{ calculate }\angle RTS.\hfill \text{[1992]} c77\displaystyle \text{Answer:}
\displaystyle \text{Join }PS.
\displaystyle \text{Since }PQ\text{ is the diameter,}c715
\displaystyle \angle PSQ=90^\circ\quad \text{angle in a semicircle}
\displaystyle \text{Since }Q,\ S,\ T\text{ are collinear,}
\displaystyle \angle PST=90^\circ
\displaystyle \angle SPR=\frac{1}{2}\angle ROS
\displaystyle =\frac{1}{2}\times42^\circ=21^\circ
\displaystyle \text{Since }P,\ R,\ T\text{ are collinear,}
\displaystyle \angle SPT=\angle SPR=21^\circ
\displaystyle \text{In }\triangle PST,
\displaystyle \angle PTS=180^\circ-90^\circ-21^\circ=69^\circ
\displaystyle \therefore \angle RTS=69^\circ
\\

\displaystyle \textbf{Question 21: }\text{In the given figure, }PQ\text{ is the diameter. Chord }SR\parallel PQ.
\displaystyle \text{Given }\angle PQR=58^\circ,\text{ calculate (i) }\angle RPQ\text{ (ii) }\angle STP.\hfill \text{[1989]} c76\displaystyle \text{Answer:}
\displaystyle \text{Join }PR.
\displaystyle \text{Since }PQ\text{ is the diameter,}
\displaystyle \angle PRQ=90^\circ\quad \text{angle in a semicircle}
\displaystyle \text{In }\triangle PQR,c714
\displaystyle \angle RPQ=180^\circ-90^\circ-58^\circ=32^\circ
\displaystyle \therefore \text{(i) }\angle RPQ=32^\circ
\displaystyle \text{Since }SR\parallel PQ,
\displaystyle \angle SRP=\angle RPQ=32^\circ\quad \text{alternate angles}
\displaystyle \text{Since }PTSR\text{ is a cyclic quadrilateral,}
\displaystyle \angle STP+\angle SRP=180^\circ
\displaystyle \angle STP+32^\circ=180^\circ
\displaystyle \angle STP=148^\circ
\displaystyle \therefore \text{(ii) }\angle STP=148^\circ
\\

\displaystyle \textbf{Question 22: }AB\text{ is the diameter of the circle with center }O.\ OD\parallel BC
\displaystyle \text{and }\angle AOD=60^\circ.\text{ Calculate the numerical values of (i) }\angle ABD
\displaystyle \text{(ii) }\angle DBC\text{ (iii) }\angle ADC.\hfill \text{[1987]} c75\displaystyle \text{Answer:}
\displaystyle \text{Join }BD.
\displaystyle \angle ABD=\frac{1}{2}\angle AOD
\displaystyle =\frac{1}{2}\times60^\circ=30^\circc713
\displaystyle \therefore \text{(i) }\angle ABD=30^\circ
\displaystyle \text{Since }AB\text{ is a diameter,}
\displaystyle \angle BDA=90^\circ\quad \text{angle in a semicircle}
\displaystyle \text{Since }OA=OD\text{ and }\angle AOD=60^\circ,
\displaystyle \triangle AOD\text{ is equilateral.}
\displaystyle \therefore \angle ODA=60^\circ
\displaystyle \angle ODB=90^\circ-60^\circ=30^\circ
\displaystyle \text{Since }OD\parallel BC,
\displaystyle \angle DBC=\angle ODB=30^\circ\quad \text{alternate angles}
\displaystyle \therefore \text{(ii) }\angle DBC=30^\circ
\displaystyle \angle ABC=\angle ABD+\angle DBC
\displaystyle =30^\circ+30^\circ=60^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ADC+\angle ABC=180^\circ
\displaystyle \angle ADC+60^\circ=180^\circ
\displaystyle \angle ADC=120^\circ
\displaystyle \therefore \text{(iii) }\angle ADC=120^\circ
\\

\displaystyle \textbf{Question 23: }\text{In the given figure, the center }O\text{ of the smaller circle lies}
\displaystyle \text{on the circumference of the bigger circle. If }\angle APB=75^\circ\text{ and }\angle BCD=40^\circ,
\displaystyle \text{find: (i) }\angle AOB\text{ (ii) }\angle ACB\text{ (iii) }\angle ABD\text{ (iv) }\angle ADB.\hfill \text{[1984]} c74\displaystyle \text{Answer:}
\displaystyle \angle AOB=2\angle APB
\displaystyle =2\times75^\circ=150^\circ
\displaystyle \therefore \text{(i) }\angle AOB=150^\circ
\displaystyle \text{Since }AOBC\text{ is a cyclic quadrilateral,}
\displaystyle \angle ACB+\angle AOB=180^\circc712
\displaystyle \angle ACB=180^\circ-150^\circ=30^\circ
\displaystyle \therefore \text{(ii) }\angle ACB=30^\circ
\displaystyle \angle ACD=\angle ACB+\angle BCD
\displaystyle =30^\circ+40^\circ=70^\circ
\displaystyle \text{Since }ABDC\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABD+\angle ACD=180^\circ
\displaystyle \angle ABD=180^\circ-70^\circ=110^\circ
\displaystyle \therefore \text{(iii) }\angle ABD=110^\circ
\displaystyle \text{Since }ADBO\text{ is a cyclic quadrilateral,}
\displaystyle \angle ADB+\angle AOB=180^\circ
\displaystyle \angle ADB=180^\circ-150^\circ=30^\circ
\displaystyle \therefore \text{(iv) }\angle ADB=30^\circ
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\displaystyle \textbf{Question 24: }O\text{ is the center of the circle of radius }10\text{ cm. }P\text{ is any point}
\displaystyle \text{in the circle such that }OP=6\text{ cm. }A\text{ is a point travelling along the circumference,}
\displaystyle x\text{ is the distance from }A\text{ to }P.\text{ What are the least and greatest values of }x\text{ in cm?}
\displaystyle \text{What is the position of the points }O,\ P\text{ and }A\text{ at these values?}\hfill \text{[1992]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the circle }OA=10\text{ cm and }OP=6\text{ cm.}
\displaystyle \text{The least value of }AP\text{ occurs when }O,\ P\text{ and }A\text{ are collinear,}
\displaystyle \text{with }P\text{ between }O\text{ and }A.
\displaystyle \therefore AP=OA-OP=10-6=4\text{ cm}
\displaystyle \text{The greatest value of }AP\text{ occurs when }O,\ P\text{ and }A\text{ are collinear,}
\displaystyle \text{with }O\text{ between }A\text{ and }P.
\displaystyle \therefore AP=OA+OP=10+6=16\text{ cm}
\displaystyle \therefore \text{Least value of }x=4\text{ cm and greatest value of }x=16\text{ cm}
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\displaystyle \textbf{Question 25: }\text{In the given figure, }O\text{ is the center of the circle. }AB\text{ and }CD
\displaystyle \text{are two chords of the circle. }OM\perp AB\text{ and }ON\perp CD.
\displaystyle AB=24\text{ cm},\ OM=5\text{ cm},\ ON=12\text{ cm. Find the}
\displaystyle \text{(i) radius of the circle}\qquad \text{(ii) length of chord }CD.\hfill \text{[2014]} c11\displaystyle \text{Answer:}
\displaystyle \text{Since the perpendicular from the center of a circle to a chord bisects the chord,}
\displaystyle AM=MB=\frac{AB}{2}=\frac{24}{2}=12\text{ cm}
\displaystyle \text{In right }\triangle AOM,c20.jpg
\displaystyle AO^2=AM^2+OM^2
\displaystyle =12^2+5^2=144+25=169
\displaystyle AO=13\text{ cm}
\displaystyle \therefore \text{(i) Radius of the circle }=13\text{ cm}
\displaystyle \text{Also, }ON\perp CD
\displaystyle \therefore CN=ND
\displaystyle \text{In right }\triangle CNO,
\displaystyle CO^2=CN^2+ON^2
\displaystyle 13^2=CN^2+12^2
\displaystyle CN^2=169-144=25
\displaystyle CN=5\text{ cm}
\displaystyle CD=2CN=2\times5=10\text{ cm}
\displaystyle \therefore \text{(ii) Length of chord }CD=10\text{ cm}
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\displaystyle \textbf{Question 26: }\text{In the given figure }O\text{ is the center of the circle. Tangents }CA\text{ and }CB
\displaystyle \text{meet at }C.\text{ If }\angle ACO=30^\circ,\text{ find (i) }\angle BCO\text{ (ii) }\angle AOB\text{ (iii) }\angle APB.\hfill \text{[2011]} 2011-1\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle AOC\text{ and }\triangle BOC.
\displaystyle \angle OAC=\angle OBC=90^\circ\quad \text{radius is perpendicular to tangent}
\displaystyle OC=OC\quad \text{common side}
\displaystyle AC=BC\quad \text{tangents from an external point are equal}
\displaystyle \therefore \triangle AOC\cong\triangle BOC
\displaystyle \therefore \angle BCO=\angle ACO=30^\circ
\displaystyle \therefore \text{(i) }\angle BCO=30^\circ
\displaystyle \text{In }\triangle AOC,
\displaystyle \angle AOC=180^\circ-90^\circ-30^\circ=60^\circ
\displaystyle \text{Similarly, }\angle BOC=60^\circ
\displaystyle \angle AOB=\angle AOC+\angle BOC
\displaystyle =60^\circ+60^\circ=120^\circ
\displaystyle \therefore \text{(ii) }\angle AOB=120^\circ
\displaystyle \angle AOB=2\angle APB\quad \text{angle at the center is double the angle at the circumference}
\displaystyle 120^\circ=2\angle APB
\displaystyle \angle APB=60^\circ
\displaystyle \therefore \text{(iii) }\angle APB=60^\circ
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\displaystyle \textbf{Question 27: }\text{In the following figure, }O\text{ is the center of the circle and }AB\text{ is a tangent}
\displaystyle \text{to it at point }B.\ \angle BDC=65^\circ.\text{ Find }\angle BAO.\hfill \text{[2010]} 2010-2\displaystyle \text{Answer:}
\displaystyle AB\text{ is a tangent to the circle at }B.
\displaystyle \therefore OB\perp AB
\displaystyle \angle ABO=90^\circ
\displaystyle \text{In }\triangle BDC,
\displaystyle \angle DBC=90^\circ
\displaystyle \therefore \angle BCD=180^\circ-90^\circ-65^\circ=25^\circ
\displaystyle \angle BCD=\angle BCE=25^\circ
\displaystyle \angle BOE=2\angle BCE
\displaystyle =2\times25^\circ=50^\circ
\displaystyle \text{Since }A,\ O\text{ and }E\text{ are collinear,}
\displaystyle \angle AOB=\angle BOE=50^\circ
\displaystyle \text{In }\triangle AOB,
\displaystyle \angle BAO=180^\circ-90^\circ-50^\circ
\displaystyle =40^\circ
\displaystyle \therefore \angle BAO=40^\circ
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