Class 9: Circles – Exercise 17

$\displaystyle \text{Exercise - 17(A)}$

$\displaystyle \textbf{Question 1: } \text{A chord of length }6\text{ cm is drawn in a circle of radius }5\text{ cm.}$
$\displaystyle \text{Calculate its distance from the centre of the circle.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }AB\text{ be the chord and }O\text{ be the centre.}$
$\displaystyle \text{Draw }OP\perp AB.$
$\displaystyle \text{Since the perpendicular from the centre bisects the chord,}$
$\displaystyle AP=PB=\frac{1}{2}\times6=3\text{ cm.}$
$\displaystyle OA=5\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle 5^2=OP^2+3^2$
$\displaystyle OP^2=25-9=16$
$\displaystyle OP=4\text{ cm.}$
$\displaystyle \therefore \text{The distance of the chord from the centre is }4\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: } \text{A chord of length }8\text{ cm is drawn at a distance of }3\text{ cm}$
$\displaystyle \text{from the centre of a circle. Calculate the radius of the circle.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }AB\text{ be the chord and }O\text{ be the centre.}$
$\displaystyle \text{Draw }OP\perp AB.$
$\displaystyle OP=3\text{ cm.}$
$\displaystyle \text{Since the perpendicular from the centre bisects the chord,}$
$\displaystyle AP=PB=\frac{1}{2}\times8=4\text{ cm.}$
$\displaystyle \text{Let the radius }OA=r\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle r^2=3^2+4^2$
$\displaystyle r^2=9+16=25$
$\displaystyle r=5\text{ cm.}$
$\displaystyle \therefore \text{The radius of the circle is }5\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: } \text{The radius of a circle is }17.0\text{ cm and the length of the perpendicular}$
$\displaystyle \text{drawn from its centre to a chord is }8.0\text{ cm. Calculate the length of the chord.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }AB\text{ be the chord, }O\text{ be the centre and }OP\perp AB.$
$\displaystyle OA=17\text{ cm and }OP=8\text{ cm.}$
$\displaystyle \text{Since the perpendicular from the centre bisects the chord, }AP=PB.$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle 17^2=8^2+AP^2$
$\displaystyle AP^2=289-64=225$
$\displaystyle AP=15\text{ cm.}$
$\displaystyle AB=2AP=2\times15=30\text{ cm.}$
$\displaystyle \therefore \text{The length of the chord is }30\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: } \text{A chord of length }24\text{ cm is at a distance of }5\text{ cm from the centre}$
$\displaystyle \text{of a circle. Find the length of the chord of the same circle which is at a}$
$\displaystyle \text{distance of }12\text{ cm from the centre.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }AB=24\text{ cm and }OP\perp AB,\text{ where }O\text{ is the centre.}$
$\displaystyle OP=5\text{ cm.}$
$\displaystyle \text{Since the perpendicular from the centre bisects the chord,}$
$\displaystyle AP=PB=\frac{1}{2}\times24=12\text{ cm.}$
$\displaystyle \text{Let the radius }OA=r\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle r^2=5^2+12^2=25+144=169$
$\displaystyle r=13\text{ cm.}$
$\displaystyle \text{Let }CD\text{ be the required chord and }OQ\perp CD.$
$\displaystyle OQ=12\text{ cm and }OC=13\text{ cm.}$
$\displaystyle \text{Since the perpendicular from the centre bisects the chord, }CQ=QD.$
$\displaystyle \text{In right-angled }\triangle OQC,$
$\displaystyle OC^2=OQ^2+CQ^2$
$\displaystyle 13^2=12^2+CQ^2$
$\displaystyle CQ^2=169-144=25$
$\displaystyle CQ=5\text{ cm.}$
$\displaystyle CD=2CQ=2\times5=10\text{ cm.}$
$\displaystyle \therefore \text{The length of the required chord is }10\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{In the following figure, }AD\text{ is a straight line. }OP\perp AD\text{ and }O$
$\displaystyle \text{is the centre of both the circles. If }OA=34\text{ cm, }OB=20\text{ cm and }OP=16\text{ cm,}$
$\displaystyle \text{find the length of }AB.$

$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }OP\perp AD,\text{ the perpendicular from the centre bisects the chords.}$
$\displaystyle \therefore AP=PD\text{ and }BP=PC.$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle 34^2=16^2+AP^2$
$\displaystyle AP^2=1156-256=900$
$\displaystyle AP=30\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OPB,$
$\displaystyle OB^2=OP^2+BP^2$
$\displaystyle 20^2=16^2+BP^2$
$\displaystyle BP^2=400-256=144$
$\displaystyle BP=12\text{ cm.}$
$\displaystyle AB=AP-BP=30-12=18\text{ cm.}$
$\displaystyle \therefore \text{The length of }AB\text{ is }18\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{In a circle of radius }17\text{ cm, two parallel chords of lengths }30\text{ cm}$
$\displaystyle \text{and }16\text{ cm are drawn. Find the distance between the chords, if both the chords are:}$
$\displaystyle \text{(i) on the opposite sides of the centre,}$
$\displaystyle \text{(ii) on the same side of the centre.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }AB=30\text{ cm and }CD=16\text{ cm be the two parallel chords.}$
$\displaystyle \text{Draw }OP\perp AB\text{ and }OQ\perp CD,\text{ where }O\text{ is the centre.}$
$\displaystyle OA=OC=17\text{ cm.}$
$\displaystyle \text{Since a perpendicular from the centre bisects a chord,}$
$\displaystyle AP=PB=\frac{1}{2}\times30=15\text{ cm.}$
$\displaystyle CQ=QD=\frac{1}{2}\times16=8\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle 17^2=OP^2+15^2$
$\displaystyle OP^2=289-225=64$
$\displaystyle OP=8\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OQC,$
$\displaystyle OC^2=OQ^2+CQ^2$
$\displaystyle 17^2=OQ^2+8^2$
$\displaystyle OQ^2=289-64=225$
$\displaystyle OQ=15\text{ cm.}$
$\displaystyle \text{(i) When the chords are on the opposite sides of the centre,}$
$\displaystyle \text{Distance between the chords}=OP+OQ=8+15=23\text{ cm.}$
$\displaystyle \therefore \text{The distance between the chords is }23\text{ cm.}$
$\displaystyle \text{(ii) When the chords are on the same side of the centre,}$
$\displaystyle \text{Distance between the chords}=OQ-OP=15-8=7\text{ cm.}$
$\displaystyle \therefore \text{The distance between the chords is }7\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Two parallel chords are drawn in a circle of diameter }30.0\text{ cm. The length}$
$\displaystyle \text{of one chord is }24.0\text{ cm and the distance between the two chords is }21.0\text{ cm.}$
$\displaystyle \text{Find the length of the other chord.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Radius of the circle}=\frac{30}{2}=15\text{ cm.}$
$\displaystyle \text{Let }AB=24\text{ cm and }OP\perp AB,\text{ where }O\text{ is the centre.}$
$\displaystyle \text{Since a perpendicular from the centre bisects a chord,}$
$\displaystyle AP=PB=\frac{1}{2}\times24=12\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle 15^2=OP^2+12^2$
$\displaystyle OP^2=225-144=81$
$\displaystyle OP=9\text{ cm.}$
$\displaystyle \text{Since the distance between the chords is }21\text{ cm, they lie on opposite sides}$
$\displaystyle \text{of the centre. Let }CD\text{ be the other chord and }OQ\perp CD.$
$\displaystyle OP+OQ=21$
$\displaystyle 9+OQ=21$
$\displaystyle OQ=12\text{ cm.}$
$\displaystyle \text{Since a perpendicular from the centre bisects a chord, }CQ=QD.$
$\displaystyle \text{In right-angled }\triangle OQC,$
$\displaystyle OC^2=OQ^2+CQ^2$
$\displaystyle 15^2=12^2+CQ^2$
$\displaystyle CQ^2=225-144=81$
$\displaystyle CQ=9\text{ cm.}$
$\displaystyle CD=2CQ=2\times9=18\text{ cm.}$
$\displaystyle \therefore \text{The length of the other chord is }18\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{A chord }CD\text{ of a circle, whose centre is }O,\text{ is bisected at }P$
$\displaystyle \text{by a diameter }AB.\text{ Given }OA=OB=15\text{ cm and }OP=9\text{ cm.}$
$\displaystyle \text{Calculate the lengths of: (i) }CD\qquad\text{(ii) }AD\qquad\text{(iii) }CB.$

$\displaystyle \text{Answer:}$
$\displaystyle \text{Since diameter }AB\text{ bisects chord }CD\text{ at }P,\text{ therefore }AB\perp CD.$
$\displaystyle \therefore CP=PD.$
$\displaystyle \text{In right-angled }\triangle OPD,$
$\displaystyle OD^2=OP^2+PD^2$
$\displaystyle 15^2=9^2+PD^2$
$\displaystyle PD^2=225-81=144$
$\displaystyle PD=12\text{ cm.}$
$\displaystyle \text{(i) }CD=2PD=2\times12=24\text{ cm.}$
$\displaystyle \therefore CD=24\text{ cm.}$
$\displaystyle \text{(ii) }AP=AO+OP=15+9=24\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle APD,$
$\displaystyle AD^2=AP^2+PD^2$
$\displaystyle AD^2=24^2+12^2=576+144=720$
$\displaystyle AD=\sqrt{720}=12\sqrt{5}\text{ cm.}$
$\displaystyle \therefore AD=12\sqrt{5}\text{ cm.}$
$\displaystyle \text{(iii) }PB=OB-OP=15-9=6\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle BPC,$
$\displaystyle CB^2=PB^2+PC^2$
$\displaystyle CB^2=6^2+12^2=36+144=180$
$\displaystyle CB=\sqrt{180}=6\sqrt{5}\text{ cm.}$
$\displaystyle \therefore CB=6\sqrt{5}\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The figure given below shows a circle with centre }O\text{ in which diameter }AB$
$\displaystyle \text{bisects chord }CD\text{ at point }E.\text{ If }CE=ED=8\text{ cm and }EB=4\text{ cm, find the}$
$\displaystyle \text{radius of the circle.}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the circle }OB=OC=r\text{ cm.}$
$\displaystyle \text{Since diameter }AB\text{ bisects chord }CD\text{ at }E,\text{ therefore }AB\perp CD.$
$\displaystyle OE=OB-EB=(r-4)\text{ cm.}$
$\displaystyle CE=8\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OEC,$
$\displaystyle OC^2=OE^2+CE^2$
$\displaystyle r^2=(r-4)^2+8^2$
$\displaystyle r^2=r^2-8r+16+64$
$\displaystyle 8r=80$
$\displaystyle r=10\text{ cm.}$
$\displaystyle \therefore \text{The radius of the circle is }10\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{In the given figure, }O\text{ is the centre of the circle. }AB\text{ and }CD$
$\displaystyle \text{are two chords of the circle. }OM\perp AB\text{ and }ON\perp CD.$
$\displaystyle \text{If }AB=24\text{ cm, }OM=5\text{ cm and }ON=12\text{ cm, find:}$
$\displaystyle \text{(i) the radius of the circle}\qquad\text{(ii) the length of chord }CD.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since the perpendicular from the centre bisects a chord,}$
$\displaystyle AM=MB=\frac{1}{2}AB=\frac{1}{2}\times24=12\text{ cm.}$
$\displaystyle \text{Let the radius }OA=OC=r\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OMA,$
$\displaystyle OA^2=OM^2+AM^2$
$\displaystyle r^2=5^2+12^2=25+144=169$
$\displaystyle r=13\text{ cm.}$
$\displaystyle \therefore \text{The radius of the circle is }13\text{ cm.}$
$\displaystyle \text{Since }ON\perp CD,\text{ therefore }CN=ND.$
$\displaystyle \text{In right-angled }\triangle ONC,$
$\displaystyle OC^2=ON^2+CN^2$
$\displaystyle 13^2=12^2+CN^2$
$\displaystyle CN^2=169-144=25$
$\displaystyle CN=5\text{ cm.}$
$\displaystyle CD=2CN=2\times5=10\text{ cm.}$
$\displaystyle \therefore \text{The length of chord }CD\text{ is }10\text{ cm.}$
$\displaystyle \\$

$\displaystyle \text{Exercise - 17(B)}$

$\displaystyle \textbf{Question 1: }\text{The figure shows two concentric circles and }AD\text{ is a chord of the}$
$\displaystyle \text{larger circle. Prove that }AB=CD.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Let }O\text{ be the common centre of the two circles.}$
$\displaystyle \text{Draw }OP\perp AD.$
$\displaystyle \text{Since the perpendicular from the} \\ \text{centre bisects a chord,}$
$\displaystyle AP=PD\qquad\text{and}\qquad BP=PC.$
$\displaystyle AB=AP-BP$
$\displaystyle CD=PD-PC$
$\displaystyle \text{But }AP=PD\text{ and }BP=PC.$
$\displaystyle \therefore AP-BP=PD-PC$
$\displaystyle \therefore AB=CD.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A straight line is drawn cutting two equal circles and passing through the}$
$\displaystyle \text{mid-point }M\text{ of the line joining their centres }O\text{ and }O'.$
$\displaystyle \text{Prove that the chords }AB\text{ and }CD,\text{ intercepted by the two circles, are equal.}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Draw }OP\perp AD\text{ and }O'Q\perp AD.$
$\displaystyle \text{Since }M\text{ is the mid-point of }OO',$
$\displaystyle OM=O'M.$
$\displaystyle \text{In }\triangle OMP\text{ and }\triangle O'MQ,$
$\displaystyle \angle OPM=\angle O'QM=90^\circ$
$\displaystyle \angle OMP=\angle O'MQ\qquad\text{[Vertically opposite angles]}$
$\displaystyle OM=O'M\qquad\text{[Given]}$
$\displaystyle \therefore \triangle OMP\cong\triangle O'MQ\qquad\text{[A.A.S.]}$
$\displaystyle \therefore OP=O'Q\qquad\text{[C.P.C.T.C.]}$
$\displaystyle \text{Since the circles are equal, their radii are equal.}$
$\displaystyle OA=O'C.$
$\displaystyle \text{Since the perpendicular from the centre bisects a chord,}$
$\displaystyle AP=PB\qquad\text{and}\qquad CQ=QD.$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle AP^2=OA^2-OP^2$
$\displaystyle \text{In right-angled }\triangle O'QC,$
$\displaystyle CQ^2=O'C^2-O'Q^2$
$\displaystyle \text{Since }OA=O'C\text{ and }OP=O'Q,$
$\displaystyle AP^2=CQ^2$
$\displaystyle \therefore AP=CQ.$
$\displaystyle AB=2AP\qquad\text{and}\qquad CD=2CQ$
$\displaystyle \therefore AB=CD.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{M and N are the mid-points of two equal chords }AB\text{ and }CD$
$\displaystyle \text{respectively, of a circle with centre }O.\text{ Prove that:}$
$\displaystyle \text{(i) }\angle BMN=\angle DNM\qquad\text{(ii) }\angle AMN=\angle CNM.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Join }OM\text{ and }ON.$
$\displaystyle \text{Since }M\text{ and }N \\ \text{ are the mid-points of chords }AB\text{ and }CD,$
$\displaystyle OM\perp AB\qquad\text{and}\qquad ON\perp CD.$
$\displaystyle \text{Since equal chords are} \\ \text{equidistant from the centre, }OM=ON.$
$\displaystyle \therefore \triangle OMN\text{ is an isosceles triangle.}$
$\displaystyle \therefore \angle OMN=\angle ONM.$

$\displaystyle \text{(i) Since }OM\perp AB,$
$\displaystyle \angle BMN=90^\circ-\angle OMN.$
$\displaystyle \text{Since }ON\perp CD,$
$\displaystyle \angle DNM=90^\circ-\angle ONM.$
$\displaystyle \text{But }\angle OMN=\angle ONM.$
$\displaystyle \therefore \angle BMN=\angle DNM.$

$\displaystyle \text{(ii) Since }A,M,B\text{ are collinear,}$
$\displaystyle \angle AMN=180^\circ-\angle BMN.$
$\displaystyle \text{Since }C,N,D\text{ are collinear,}$
$\displaystyle \angle CNM=180^\circ-\angle DNM.$
$\displaystyle \text{But }\angle BMN=\angle DNM.$
$\displaystyle \therefore \angle AMN=\angle CNM.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{In the following figure, }P\text{ and }Q\text{ are the points of intersection}$
$\displaystyle \text{of two circles with centres }O\text{ and }O'.\text{ If the straight lines }APB\text{ and }CQD$
$\displaystyle \text{are parallel to }OO',\text{ prove that:}$
$\displaystyle \text{(i) }OO'=\frac{1}{2}AB\qquad\text{(ii) }AB=CD.$

$\displaystyle \text{Answer:}$
$\displaystyle \text{Draw }OM\perp AB\text{ and }O'N\perp AB.$
$\displaystyle \text{Since }AB\parallel OO'\text{ and }OM\perp AB,$
$\displaystyle OM\perp OO'.$
$\displaystyle \text{Similarly, }O'N\perp OO'.$
$\displaystyle \therefore OMNO'\text{ is a rectangle.}$
$\displaystyle \therefore MN=OO'.$
$\displaystyle \text{Since a perpendicular from the centre bisects a chord,}$
$\displaystyle AM=MP\qquad\text{and}\qquad PN=NB.$

$\displaystyle \text{(i) }AB=AM+MP+PN+NB$
$\displaystyle AB=MP+MP+PN+PN$
$\displaystyle AB=2(MP+PN)$
$\displaystyle AB=2MN=2OO'$
$\displaystyle \therefore OO'=\frac{1}{2}AB.$

$\displaystyle \text{(ii) Draw }OR\perp CD\text{ and }O'S\perp CD.$
$\displaystyle \text{Since }CD\parallel OO',\text{ quadrilateral }ORSO'\text{ is a rectangle.}$
$\displaystyle \therefore RS=OO'.$
$\displaystyle \text{Since a perpendicular from the centre bisects a chord,}$
$\displaystyle CR=RQ\qquad\text{and}\qquad QS=SD.$
$\displaystyle CD=CR+RQ+QS+SD$
$\displaystyle CD=2(RQ+QS)=2RS=2OO'.$
$\displaystyle \text{But }AB=2OO'.$
$\displaystyle \therefore AB=CD.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Two equal chords }AB\text{ and }CD\text{ of a circle with centre }O\text{ intersect}$
$\displaystyle \text{each other at point }P\text{ inside the circle. Prove that:}$
$\displaystyle \text{(i) }AP=CP\qquad\text{(ii) }BP=DP.$ $\displaystyle \text{Answer:}$ $\displaystyle \text{Draw }OM\perp AB\text{ and }ON\perp CD.\text{ Also, join }OP.$
$\displaystyle \text{Since a perpendicular from the centre bisects a chord,}$
$\displaystyle AM=MB=\frac{1}{2}AB\qquad\text{and}\qquad CN=ND=\frac{1}{2}CD.$
$\displaystyle \text{But }AB=CD.$
$\displaystyle \therefore AM=MB=CN=ND.$
$\displaystyle \text{Equal chords of a circle are equidistant from the centre.}$
$\displaystyle \therefore OM=ON.$
$\displaystyle \text{In }\triangle OMP\text{ and }\triangle ONP,$
$\displaystyle OM=ON\qquad\text{[Equal chords are equidistant from the centre]}$
$\displaystyle OP=OP\qquad\text{[Common]}$
$\displaystyle \angle OMP=\angle ONP=90^\circ$
$\displaystyle \therefore \triangle OMP\cong\triangle ONP\qquad\text{[R.H.S.]}$
$\displaystyle \therefore MP=NP\qquad\text{[C.P.C.T.C.]}$

$\displaystyle \text{(i) }AP=AM-MP$
$\displaystyle CP=CN-NP$
$\displaystyle \text{But }AM=CN\text{ and }MP=NP.$
$\displaystyle \therefore AP=CP.$

$\displaystyle \text{(ii) }BP=BM+MP$
$\displaystyle DP=DN+NP$
$\displaystyle \text{But }BM=DN\text{ and }MP=NP.$
$\displaystyle \therefore BP=DP.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{In the following figure, }OABC\text{ is a square. A circle is drawn with }O$
$\displaystyle \text{as centre which meets }OC\text{ at }P\text{ and }OA\text{ at }Q.\text{ Prove that:}$
$\displaystyle \text{(i) }\triangle OPA\cong\triangle OQC\qquad\text{(ii) }\triangle BPC\cong\triangle BQA.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }OABC\text{ is a square,}$
$\displaystyle OA=OC,\qquad AB=BC\qquad\text{and}\qquad OA\perp OC.$
$\displaystyle \text{Also, }OP=OQ\qquad\text{[Radii of the same circle]}$

$\displaystyle \text{(i) In }\triangle OPA\text{ and }\triangle OQC,$
$\displaystyle OP=OQ\qquad\text{[Radii of the same circle]}$
$\displaystyle OA=OC\qquad\text{[Sides of the square]}$
$\displaystyle \angle POA=\angle QOC=90^\circ$
$\displaystyle \therefore \triangle OPA\cong\triangle OQC\qquad\text{[S.A.S.]}$

$\displaystyle \text{(ii) }PC=OC-OP$
$\displaystyle QA=OA-OQ$
$\displaystyle \text{But }OC=OA\text{ and }OP=OQ.$
$\displaystyle \therefore PC=QA.$
$\displaystyle \text{In }\triangle BPC\text{ and }\triangle BQA,$
$\displaystyle BC=BA\qquad\text{[Sides of the square]}$
$\displaystyle PC=QA\qquad\text{[Proved above]}$
$\displaystyle \angle BCP=\angle BAQ=90^\circ$
$\displaystyle \therefore \triangle BPC\cong\triangle BQA\qquad\text{[S.A.S.]}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{The length of the common chord of two intersecting circles is }30\text{ cm.}$
$\displaystyle \text{If the diameters of the circles are }50\text{ cm and }34\text{ cm, calculate the distance}$
$\displaystyle \text{between their centres.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }AB\text{ be the common chord and }O\text{ and }O'\text{ be the centres.}$
$\displaystyle \text{Let }OO'\text{ intersect }AB\text{ at }P.$
$\displaystyle \text{The line joining the centres bisects the common chord perpendicularly.}$
$\displaystyle \therefore AP=PB=\frac{1}{2}AB=\frac{1}{2}\times30=15\text{ cm.}$
$\displaystyle OA=\frac{50}{2}=25\text{ cm and }O'A=\frac{34}{2}=17\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle OP^2=25^2-15^2=625-225=400$
$\displaystyle OP=20\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle O'PA,$
$\displaystyle O'A^2=O'P^2+AP^2$
$\displaystyle O'P^2=17^2-15^2=289-225=64$
$\displaystyle O'P=8\text{ cm.}$
$\displaystyle \text{Since the centres lie on opposite sides of the common chord,}$
$\displaystyle OO'=OP+PO'=20+8=28\text{ cm.}$
$\displaystyle \therefore \text{The distance between the centres is }28\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{The line joining the mid-points of two chords of a circle passes through its}$
$\displaystyle \text{centre. Prove that the chords are parallel.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }M\text{ and }N\text{ be the mid-points of chords }AB\text{ and }CD\text{ respectively.}$
$\displaystyle \text{Let }O\text{ be the centre of the circle, where }M,O,N\text{ are collinear.}$
$\displaystyle \text{Since }M\text{ is the mid-point of chord }AB,$
$\displaystyle OM\perp AB.$
$\displaystyle \text{Since }N\text{ is the mid-point of chord }CD,$
$\displaystyle ON\perp CD.$
$\displaystyle \text{But }OM\text{ and }ON\text{ lie on the same straight line }MN.$
$\displaystyle \therefore AB\perp MN\qquad\text{and}\qquad CD\perp MN.$
$\displaystyle \text{Since two lines perpendicular to the same line are parallel,}$
$\displaystyle \therefore AB\parallel CD.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{In the following figure, the line }ABCD\text{ is perpendicular to }PQ,$
$\displaystyle \text{where }P\text{ and }Q\text{ are the centres of the circles. Show that:}$
$\displaystyle \text{(i) }AB=CD\qquad\text{(ii) }AC=BD.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Let }PQ\text{ intersect }ABCD\text{ at }M.$
$\displaystyle \text{Since }PQ\perp ABCD,\text{ therefore }PM\perp BD\text{ and }QM\perp AC.$
$\displaystyle \text{The perpendicular from the centre of a circle bisects the chord.}$
$\displaystyle \therefore BM=CM\qquad\text{and}\qquad AM=DM.$

$\displaystyle \text{(i) }AB=AM-BM$
$\displaystyle CD=DM-CM$
$\displaystyle \text{But }AM=DM\text{ and }BM=CM.$
$\displaystyle \therefore AB=CD.$

$\displaystyle \text{(ii) }AC=AM+CM$
$\displaystyle BD=BM+DM$
$\displaystyle \text{But }AM=DM\text{ and }CM=BM.$
$\displaystyle \therefore AC=BD.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{AB and }CD\text{ are two equal chords of a circle with centre }O,$
$\displaystyle \text{which intersect each other at right angle at point }P.\text{ If }OM\perp AB$
$\displaystyle \text{and }ON\perp CD,\text{ show that }OMPN\text{ is a square.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Since }OM\perp AB\text{ and }ON\perp CD,$
$\displaystyle M\text{ and }N\text{ are the mid-points of chords }AB\text{ and }CD\text{ respectively.}$
$\displaystyle \text{Since }AB=CD,\text{ equal chords are equidistant from the centre.}$
$\displaystyle \therefore OM=ON.$
$\displaystyle \text{Since }AB\perp CD\text{ and }OM\perp AB,\text{ therefore }OM\parallel CD.$
$\displaystyle \therefore OM\parallel PN.$
$\displaystyle \text{Since }AB\perp CD\text{ and }ON\perp CD,\text{ therefore }ON\parallel AB.$
$\displaystyle \therefore ON\parallel MP.$
$\displaystyle \therefore OMPN\text{ is a parallelogram.}$
$\displaystyle \text{Also, }\angle OMP=90^\circ.$
$\displaystyle \therefore OMPN\text{ is a rectangle.}$
$\displaystyle \text{In parallelogram }OMPN,$
$\displaystyle MP=ON\qquad\text{and}\qquad PN=OM.$
$\displaystyle \text{But }OM=ON.$
$\displaystyle \therefore OM=MP=PN=NO.$
$\displaystyle \text{Thus, }OMPN\text{ is a rectangle with all its sides equal.}$
$\displaystyle \therefore OMPN\text{ is a square.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \text{Exercise - 17(C)}$

$\displaystyle \textbf{Question 1: }\text{In the given figure, an equilateral triangle }ABC\text{ is inscribed in a circle}$
$\displaystyle \text{with centre }O.\text{ Find: (i) }\angle BOC\qquad\text{(ii) }\angle OBC.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }ABC\text{ is an equilateral triangle, its three sides are equal.}$
$\displaystyle \therefore \text{The equal chords }AB,\ BC\text{ and }CA\text{ subtend equal angles at the centre.}$
$\displaystyle \angle AOB=\angle BOC=\angle COA$
$\displaystyle \angle AOB+\angle BOC+\angle COA=360^\circ$
$\displaystyle 3\angle BOC=360^\circ$

$\displaystyle \text{(i) }\angle BOC=120^\circ.$

$\displaystyle \text{(ii) In }\triangle BOC,$
$\displaystyle OB=OC\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OBC=\angle OCB.$
$\displaystyle \angle OBC+\angle OCB+\angle BOC=180^\circ$
$\displaystyle 2\angle OBC+120^\circ=180^\circ$
$\displaystyle 2\angle OBC=60^\circ$
$\displaystyle \therefore \angle OBC=30^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{In the given figure, a square is inscribed in a circle with centre }O.$
$\displaystyle \text{Find: (i) }\angle BOC\quad\text{(ii) }\angle OCB\quad\text{(iii) }\angle COD\quad\text{(iv) }\angle BOD.$
$\displaystyle \text{Is }BD\text{ a diameter of the circle?}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }ABCD\text{ is a square, its four sides are equal chords of the circle.}$
$\displaystyle \therefore \angle AOB=\angle BOC=\angle COD=\angle DOA.$
$\displaystyle \angle AOB+\angle BOC+\angle COD+\angle DOA=360^\circ$
$\displaystyle 4\angle BOC=360^\circ$

$\displaystyle \text{(i) }\angle BOC=90^\circ.$

$\displaystyle \text{(ii) In }\triangle BOC,$
$\displaystyle OB=OC\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OBC=\angle OCB.$
$\displaystyle \angle OBC+\angle OCB+\angle BOC=180^\circ$
$\displaystyle 2\angle OCB+90^\circ=180^\circ$
$\displaystyle \therefore \angle OCB=45^\circ.$

$\displaystyle \text{(iii) }\angle COD=90^\circ.$

$\displaystyle \text{(iv) }\angle BOD=\angle BOC+\angle COD$
$\displaystyle \angle BOD=90^\circ+90^\circ=180^\circ.$

$\displaystyle \text{Since }\angle BOD=180^\circ,\text{ the points }B,\ O\text{ and }D\text{ are collinear.}$
$\displaystyle \text{Also, }B\text{ and }D\text{ lie on the circumference of the circle.}$
$\displaystyle \therefore BD\text{ is a diameter of the circle.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{In the given figure, }AB\text{ is a side of a regular pentagon and }BC\text{ is a}$
$\displaystyle \text{side of a regular hexagon. Find:}$
$\displaystyle \text{(i) }\angle AOB\quad\text{(ii) }\angle BOC\quad\text{(iii) }\angle AOC\quad\text{(iv) }\angle OBA$
$\displaystyle \text{(v) }\angle OBC\quad\text{(vi) }\angle ABC.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{The angle subtended at the centre by a side of a regular }n\text{-sided polygon is }\frac{360^\circ}{n}.$

$\displaystyle \text{(i) Since }AB\text{ is a side of a regular pentagon,}$
$\displaystyle \angle AOB=\frac{360^\circ}{5}=72^\circ.$

$\displaystyle \text{(ii) Since }BC\text{ is a side of a regular hexagon,}$
$\displaystyle \angle BOC=\frac{360^\circ}{6}=60^\circ.$

$\displaystyle \text{(iii) }\angle AOC=\angle AOB+\angle BOC$
$\displaystyle \angle AOC=72^\circ+60^\circ=132^\circ.$

$\displaystyle \text{(iv) In }\triangle AOB,$
$\displaystyle OA=OB\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OAB=\angle OBA.$
$\displaystyle 2\angle OBA+\angle AOB=180^\circ$
$\displaystyle 2\angle OBA+72^\circ=180^\circ$
$\displaystyle 2\angle OBA=108^\circ$
$\displaystyle \therefore \angle OBA=54^\circ.$

$\displaystyle \text{(v) In }\triangle BOC,$
$\displaystyle OB=OC\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OBC=\angle OCB.$
$\displaystyle 2\angle OBC+\angle BOC=180^\circ$
$\displaystyle 2\angle OBC+60^\circ=180^\circ$
$\displaystyle 2\angle OBC=120^\circ$
$\displaystyle \therefore \angle OBC=60^\circ.$

$\displaystyle \text{(vi) }\angle ABC=\angle ABO+\angle OBC$
$\displaystyle \angle ABC=54^\circ+60^\circ=114^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{In the given figure, arc }AB\text{ and arc }BC\text{ are equal in length.}$
$\displaystyle \text{If }\angle AOB=48^\circ,\text{ find:}$
$\displaystyle \text{(i) }\angle BOC\quad\text{(ii) }\angle OBC\quad\text{(iii) }\angle AOC\quad\text{(iv) }\angle OAC.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Equal arcs of a circle subtend equal angles at the centre.}$

$\displaystyle \text{(i) Since arc }AB=\text{arc }BC,$
$\displaystyle \angle AOB=\angle BOC.$
$\displaystyle \therefore \angle BOC=48^\circ.$

$\displaystyle \text{(ii) In }\triangle BOC,$
$\displaystyle OB=OC\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OBC=\angle OCB.$
$\displaystyle 2\angle OBC+\angle BOC=180^\circ$
$\displaystyle 2\angle OBC+48^\circ=180^\circ$
$\displaystyle 2\angle OBC=132^\circ$
$\displaystyle \therefore \angle OBC=66^\circ.$

$\displaystyle \text{(iii) }\angle AOC=\angle AOB+\angle BOC$
$\displaystyle \angle AOC=48^\circ+48^\circ=96^\circ.$

$\displaystyle \text{(iv) In }\triangle AOC,$
$\displaystyle OA=OC\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OAC=\angle OCA.$
$\displaystyle 2\angle OAC+\angle AOC=180^\circ$
$\displaystyle 2\angle OAC+96^\circ=180^\circ$
$\displaystyle 2\angle OAC=84^\circ$
$\displaystyle \therefore \angle OAC=42^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{In the given figure, the lengths of arcs }AB\text{ and }BC\text{ are in the ratio }$
$\displaystyle 3:2. \ \text{If }\angle AOB=96^\circ,\text{ find: (i) }\angle BOC\qquad\text{(ii) }\angle ABC.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Angles subtended at the centre are proportional to the lengths of their arcs.}$
$\displaystyle \therefore \angle AOB:\angle BOC=3:2.$

$\displaystyle \text{(i) }\frac{\angle AOB}{\angle BOC}=\frac{3}{2}$
$\displaystyle \frac{96^\circ}{\angle BOC}=\frac{3}{2}$
$\displaystyle \angle BOC=\frac{96^\circ\times2}{3}=64^\circ.$

$\displaystyle \text{(ii) In }\triangle AOB,$
$\displaystyle OA=OB\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle ABO=\angle OAB.$
$\displaystyle 2\angle ABO+\angle AOB=180^\circ$
$\displaystyle 2\angle ABO+96^\circ=180^\circ$
$\displaystyle \therefore \angle ABO=42^\circ.$
$\displaystyle \text{In }\triangle BOC,$
$\displaystyle OB=OC\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OBC=\angle OCB.$
$\displaystyle 2\angle OBC+\angle BOC=180^\circ$
$\displaystyle 2\angle OBC+64^\circ=180^\circ$
$\displaystyle \therefore \angle OBC=58^\circ.$
$\displaystyle \angle ABC=\angle ABO+\angle OBC$
$\displaystyle \angle ABC=42^\circ+58^\circ=100^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{In the given figure, arcs }AB=BC=DC\text{ and }\angle AOB=50^\circ.$
$\displaystyle \text{Find: (i) }\angle AOC\quad\text{(ii) }\angle AOD\quad\text{(iii) }\angle BOD$
$\displaystyle \text{(iv) }\angle OAC\quad\text{(v) }\angle ODA.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Equal arcs of a circle subtend equal angles at the centre.}$
$\displaystyle \text{Since arcs }AB=BC=DC,$
$\displaystyle \angle AOB=\angle BOC=\angle COD=50^\circ.$

$\displaystyle \text{(i) }\angle AOC=\angle AOB+\angle BOC$
$\displaystyle \angle AOC=50^\circ+50^\circ=100^\circ.$

$\displaystyle \text{(ii) }\angle AOD=\angle AOB+\angle BOC+\angle COD$
$\displaystyle \angle AOD=50^\circ+50^\circ+50^\circ=150^\circ.$

$\displaystyle \text{(iii) }\angle BOD=\angle BOC+\angle COD$
$\displaystyle \angle BOD=50^\circ+50^\circ=100^\circ.$

$\displaystyle \text{(iv) In }\triangle AOC,$
$\displaystyle OA=OC\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OAC=\angle OCA.$
$\displaystyle 2\angle OAC+\angle AOC=180^\circ$
$\displaystyle 2\angle OAC+100^\circ=180^\circ$
$\displaystyle 2\angle OAC=80^\circ$
$\displaystyle \therefore \angle OAC=40^\circ.$

$\displaystyle \text{(v) In }\triangle AOD,$
$\displaystyle OA=OD\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OAD=\angle ODA.$
$\displaystyle 2\angle ODA+\angle AOD=180^\circ$
$\displaystyle 2\angle ODA+150^\circ=180^\circ$
$\displaystyle 2\angle ODA=30^\circ$
$\displaystyle \therefore \angle ODA=15^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{In the given figure, }AB\text{ is a side of a regular hexagon and }AC\text{ is a}$
$\displaystyle \text{side of a regular eight-sided polygon. Find:}$
$\displaystyle \text{(i) }\angle AOB\quad\text{(ii) }\angle AOC\quad\text{(iii) }\angle BOC\quad\text{(iv) }\angle OBC.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{The angle subtended at the centre by a side of a regular }n\text{-sided polygon is }\frac{360^\circ}{n}.$

$\displaystyle \text{(i) Since }AB\text{ is a side of a regular hexagon,}$
$\displaystyle \angle AOB=\frac{360^\circ}{6}=60^\circ.$

$\displaystyle \text{(ii) Since }AC\text{ is a side of a regular eight-sided polygon,}$
$\displaystyle \angle AOC=\frac{360^\circ}{8}=45^\circ.$

$\displaystyle \text{(iii) }\angle BOC=\angle BOA+\angle AOC$
$\displaystyle \angle BOC=60^\circ+45^\circ=105^\circ.$

$\displaystyle \text{(iv) In }\triangle BOC,$
$\displaystyle OB=OC\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OBC=\angle OCB.$
$\displaystyle 2\angle OBC+\angle BOC=180^\circ$
$\displaystyle 2\angle OBC+105^\circ=180^\circ$
$\displaystyle 2\angle OBC=75^\circ$
$\displaystyle \therefore \angle OBC=37.5^\circ=37^\circ30'.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{In the given figure, }O\text{ is the centre of the circle and the length of arc }AB$
$\displaystyle \text{is twice the length of arc }BC.\text{ If }\angle AOB=100^\circ,\text{ find:}$
$\displaystyle \text{(i) }\angle BOC\qquad\text{(ii) }\angle OAC.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Angles subtended at the centre are proportional to the lengths of their arcs.}$
$\displaystyle \text{Since arc }AB=2\times\text{arc }BC,$
$\displaystyle \angle AOB=2\angle BOC.$

$\displaystyle \text{(i) }100^\circ=2\angle BOC$
$\displaystyle \therefore \angle BOC=50^\circ.$

$\displaystyle \text{(ii) }\angle AOC=\angle AOB+\angle BOC$
$\displaystyle \angle AOC=100^\circ+50^\circ=150^\circ.$
$\displaystyle \text{In }\triangle AOC,$
$\displaystyle OA=OC\qquad\text{[Radii of the same circle]}$
$\displaystyle \therefore \angle OAC=\angle OCA.$
$\displaystyle 2\angle OAC+\angle AOC=180^\circ$
$\displaystyle 2\angle OAC+150^\circ=180^\circ$
$\displaystyle 2\angle OAC=30^\circ$
$\displaystyle \therefore \angle OAC=15^\circ.$
$\displaystyle \\$

$\displaystyle \text{Exercise - 17(D)}$

$\displaystyle \textbf{Question 1: }\text{The radius of a circle is }13\text{ cm and the length of one of its chords is}$
$\displaystyle \text{24 cm. Find the distance of the chord from the centre.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }AB\text{ be the chord and }O\text{ be the centre of the circle.}$
$\displaystyle \text{Draw }OP\perp AB.$
$\displaystyle \text{Since the perpendicular from the centre bisects the chord,}$
$\displaystyle AP=PB=\frac{1}{2}AB=\frac{1}{2}\times24=12\text{ cm.}$
$\displaystyle OA=13\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OPA,$
$\displaystyle OA^2=OP^2+AP^2$
$\displaystyle 13^2=OP^2+12^2$
$\displaystyle OP^2=169-144=25$
$\displaystyle OP=5\text{ cm.}$
$\displaystyle \therefore \text{The distance of the chord from the centre is }5\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Prove that equal chords of congruent circles subtend equal angles at their}$
$\displaystyle \text{centres.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }AB\text{ and }CD\text{ be equal chords of two congruent circles with centres }O\text{ and }O'.$
$\displaystyle \text{Join }OA,\ OB,\ O'C\text{ and }O'D.$
$\displaystyle \text{In }\triangle AOB\text{ and }\triangle CO'D,$
$\displaystyle OA=O'C\qquad\text{[Radii of congruent circles]}$
$\displaystyle OB=O'D\qquad\text{[Radii of congruent circles]}$
$\displaystyle AB=CD\qquad\text{[Given]}$
$\displaystyle \therefore \triangle AOB\cong\triangle CO'D\qquad\text{[S.S.S.]}$
$\displaystyle \therefore \angle AOB=\angle CO'D\qquad\text{[C.P.C.T.C.]}$
$\displaystyle \therefore \text{Equal chords of congruent circles subtend equal angles at their centres.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Draw two circles of different radii. How many points can these circles have}$
$\displaystyle \text{in common? What is the maximum number of common points?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Two circles of different radii may have }0,\ 1\text{ or }2\text{ common points.}$
$\displaystyle \text{They have no common point when they do not intersect.}$
$\displaystyle \text{They have one common point when they touch each other.}$
$\displaystyle \text{They have two common points when they intersect each other.}$
$\displaystyle \therefore \text{The maximum number of common points is }2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Suppose you are given a circle. Describe a method by which you can find}$
$\displaystyle \text{the centre of this circle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Draw any two non-parallel chords }AB\text{ and }CD\text{ of the given circle.}$
$\displaystyle \text{Construct the perpendicular bisectors of chords }AB\text{ and }CD.$
$\displaystyle \text{Let the two perpendicular bisectors intersect at }O.$
$\displaystyle \text{The perpendicular bisector of every chord of a circle passes through its centre.}$
$\displaystyle \therefore O\text{ is the centre of the given circle.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Given two equal chords }AB\text{ and }CD\text{ of a circle with centre }O,$
$\displaystyle \text{intersecting each other at point }P.\text{ Prove that:}$
$\displaystyle \text{(i) }AP=CP\qquad\text{(ii) }BP=DP.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Draw }OM\perp AB\text{ and }ON\perp CD.\text{ Join }OP.$ $\displaystyle \text{Since the perpendicular from the centre bisects a chord,}$
$\displaystyle AM=MB=\frac{1}{2}AB\qquad\text{and}\qquad CN=ND=\frac{1}{2}CD.$
$\displaystyle \text{But }AB=CD.$
$\displaystyle \therefore AM=MB=CN=ND.$
$\displaystyle \text{Since equal chords are equidistant from the centre,}$
$\displaystyle OM=ON.$
$\displaystyle \text{In right-angled }\triangle OMP\text{ and }\triangle ONP,$
$\displaystyle OM=ON\qquad\text{[Equal chords are equidistant from the centre]}$
$\displaystyle OP=OP\qquad\text{[Common]}$
$\displaystyle \angle OMP=\angle ONP=90^\circ$
$\displaystyle \therefore \triangle OMP\cong\triangle ONP\qquad\text{[R.H.S.]}$
$\displaystyle \therefore MP=NP\qquad\text{[C.P.C.T.C.]}$

$\displaystyle \text{(i) }AP=AM+MP$
$\displaystyle CP=CN+NP$
$\displaystyle \text{But }AM=CN\text{ and }MP=NP.$
$\displaystyle \therefore AP=CP.$

$\displaystyle \text{(ii) }BP=BM-MP$
$\displaystyle DP=DN-NP$
$\displaystyle \text{But }BM=DN\text{ and }MP=NP.$
$\displaystyle \therefore BP=DP.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{In a circle of radius }10\text{ cm, }AB\text{ and }CD\text{ are two parallel chords}$
$\displaystyle \text{of lengths }16\text{ cm and }12\text{ cm respectively. Calculate the distance between}$
$\displaystyle \text{the chords, if they are on:}$
$\displaystyle \text{(i) the same side of the centre}\qquad\text{(ii) the opposite sides of the centre.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }O\text{ be the centre of the circle. Draw }OM\perp AB\text{ and }ON\perp CD.$
$\displaystyle \text{Since the perpendicular from the centre bisects a chord,}$
$\displaystyle AM=MB=\frac{1}{2}\times16=8\text{ cm.}$
$\displaystyle CN=ND=\frac{1}{2}\times12=6\text{ cm.}$
$\displaystyle OA=OC=10\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OMA,$
$\displaystyle OA^2=OM^2+AM^2$
$\displaystyle 10^2=OM^2+8^2$
$\displaystyle OM^2=100-64=36$
$\displaystyle OM=6\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle ONC,$
$\displaystyle OC^2=ON^2+CN^2$
$\displaystyle 10^2=ON^2+6^2$
$\displaystyle ON^2=100-36=64$
$\displaystyle ON=8\text{ cm.}$

$\displaystyle \textbf{Question 7: }\text{In the given figure, }O\text{ is the centre of the circle with radius }20\text{ cm}$
$\displaystyle \text{and }OD\text{ is perpendicular to }AB.\text{ If }AB=32\text{ cm, find the length of }CD.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }OD\perp AB,\text{ the perpendicular from the centre bisects the chord.}$
$\displaystyle \therefore AC=CB=\frac{1}{2}AB=\frac{1}{2}\times32=16\text{ cm.}$
$\displaystyle OA=OD=20\text{ cm.}$
$\displaystyle \text{In right-angled }\triangle OCA,$
$\displaystyle OA^2=OC^2+AC^2$
$\displaystyle 20^2=OC^2+16^2$
$\displaystyle OC^2=400-256=144$
$\displaystyle OC=12\text{ cm.}$
$\displaystyle CD=OD-OC=20-12=8\text{ cm.}$
$\displaystyle \therefore \text{The length of }CD\text{ is }8\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{In the given figure, }AB\text{ and }CD\text{ are two equal chords of a circle}$
$\displaystyle \text{with centre }O.\text{ If }P\text{ is the mid-point of chord }AB,\ Q\text{ is the mid-point of}$
$\displaystyle \text{chord }CD\text{ and }\angle POQ=150^\circ,\text{ find }\angle APQ.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }P\text{ and }Q\text{ are the mid-points of chords }AB\text{ and }CD,$
$\displaystyle OP\perp AB\qquad\text{and}\qquad OQ\perp CD.$
$\displaystyle \text{Since }AB=CD,\text{ equal chords are equidistant from the centre.}$
$\displaystyle \therefore OP=OQ.$
$\displaystyle \therefore \triangle POQ\text{ is an isosceles triangle.}$
$\displaystyle \angle OPQ=\angle OQP$
$\displaystyle \angle OPQ+\angle OQP+\angle POQ=180^\circ$
$\displaystyle 2\angle OPQ+150^\circ=180^\circ$
$\displaystyle 2\angle OPQ=30^\circ$
$\displaystyle \angle OPQ=15^\circ.$
$\displaystyle \text{Since }OP\perp AB,\ \angle APO=90^\circ.$
$\displaystyle \angle APQ=\angle APO-\angle OPQ$
$\displaystyle \angle APQ=90^\circ-15^\circ=75^\circ.$
$\displaystyle \therefore \angle APQ=75^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{In the given figure, }AOC\text{ is the diameter of the circle with centre }O.$
$\displaystyle \text{If arc }AXB\text{ is half of arc }BYC,\text{ find }\angle BOC.$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since }AOC\text{ is a diameter, arc }AXC\text{ is a semicircle.}$
$\displaystyle \therefore \text{arc }AXB+\text{arc }BYC=180^\circ.$
$\displaystyle \text{Let the measure of arc }AXB=x^\circ.$
$\displaystyle \therefore \text{The measure of arc }BYC=2x^\circ.$
$\displaystyle x+2x=180^\circ$
$\displaystyle 3x=180^\circ$
$\displaystyle x=60^\circ$
$\displaystyle \therefore \text{The measure of arc }BYC=2\times60^\circ=120^\circ.$
$\displaystyle \text{The angle at the centre is equal to the measure of its corresponding arc.}$
$\displaystyle \therefore \angle BOC=120^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{The circumference of a circle with centre }O\text{ is divided into three arcs}$
$\displaystyle APB,\ BQC\text{ and }CRA\text{ such that}$
$\displaystyle \frac{\text{arc }APB}{2}=\frac{\text{arc }BQC}{3}=\frac{\text{arc }CRA}{4}.\text{ Find }\angle BOC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\frac{\text{arc }APB}{2}=\frac{\text{arc }BQC}{3}=\frac{\text{arc }CRA}{4}=x.$
$\displaystyle \therefore \text{arc }APB=2x,\qquad\text{arc }BQC=3x,\qquad\text{arc }CRA=4x.$
$\displaystyle \text{The sum of the measures of all the arcs of a circle is }360^\circ.$
$\displaystyle 2x+3x+4x=360^\circ$
$\displaystyle 9x=360^\circ$
$\displaystyle x=40^\circ$
$\displaystyle \therefore \text{The measure of arc }BQC=3x=3\times40^\circ=120^\circ.$
$\displaystyle \text{The angle at the centre is equal to the measure of its corresponding arc.}$
$\displaystyle \therefore \angle BOC=120^\circ.$
$\displaystyle \\$