NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2

Q.1 From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm (B) 12 cm
(C) 15 cm (D) 24.5 cm

Ans.

$\begin{array}{l}\text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent.}\\ \therefore \text{OP}\perp \text{PQ}\\ \text{By applying Pythagoras theorem in Δ\hspace{0.17em}OPQ, we get}\\ {\text{OP}}^{\text{2}}{\text{+PQ}}^{\text{2}}{\text{= OQ}}^{\text{2}}\\ {\text{or OP}}^{\text{2}}{\text{+24}}^{\text{2}}{\text{= 25}}^{\text{2}}\\ {\text{or OP}}^{\text{2}}\text{= 625}-\text{576 = 49}\\ \text{or OP = 7 cm.}\\ \text{Therefore, radius of the circle is 7 cm.}\\ \text{Hence, the correct option is (A).}\end{array}$

Q.2

$\begin{array}{l}\text{In the following figure, if TP and TQ are the two tangents to a circle with centre O so that}\\ \angle \text{POQ}=\text{110°, then}\angle \text{PTQ is equal to}\\ \text{(A) 60° (B) 70°}\\ \text{(C) 80° (D) 90°}\end{array}$

Ans.

\begin{array}{l}\text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent}\text{.}\\ \therefore \text{OP}\perp \text{PT and OQ}\perp \text{TQ}\\ \text{In quadrilateral OPTQ,}\angle \text{P =}\angle \text{Q = 90°}\text{.}\\ \text{Now, we have}\\ \angle \text{P+}\angle \text{Q+}\angle \text{T+}\angle \text{O = 360°}\\ \text{or 90°+90°+110°+}\angle \text{T = 360°}\\ \text{or 290°+}\angle \text{T = 360°}\\ \text{or}\angle \text{T = 360°-290° = 70°}\\ \text{Therefore,}\angle \text{PTQ = 70°}\\ \text{Hence, the correct option is (B)}\text{.}\end{array}

Q.3

$\begin{array}{l}\text{If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°,}\\ \text{then}\angle \text{POA is equal to}\\ \text{(A) 50° (B) 60°}\\ \text{(C) 70° (D) 80°}\end{array}$

Ans.

$\begin{array}{l}\text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent.}\\ \therefore \text{OA}\perp \text{AP and OB}\perp \text{PB}\\ \text{In quadrilateral OAPB,}\angle \text{A =}\angle \text{B = 90°.}\\ \text{Now, we have}\\ \angle \text{P+}\angle \text{A+}\angle \text{B+}\angle \text{O = 360°}\\ \text{or 80°+90°+90°+}\angle \text{O = 360°}\\ \text{or 260°+}\angle \text{O = 360°}\\ \text{or}\angle \text{O = 360°}-\text{260° = 100°}\\ \text{Therefore,}\angle \text{AOB = 100°}\\ \text{Now, in Δ\hspace{0.17em}AOP and Δ\hspace{0.17em}BOP, we have}\\ \text{AO = BO [Radius]}\\ \text{AP = BP [Tangents drawn from P]}\\ \text{OP = OP}\\ \therefore \text{Δ\hspace{0.17em}AOP}\cong \text{Δ\hspace{0.17em}BOP [By SSS congruence criterion]}\\ \text{So,}\angle \text{AOP =}\angle \text{BOP \hspace{0.33em}\hspace{0.33em} \hspace{0.17em}[By CPCT]}\\ \angle \text{AOB =}\angle \text{AOP+}\angle \text{BOP = 100°}\\ \text{or}\angle \text{AOP+}\angle \text{AOP = 100° [}\angle \text{AOP =}\angle \text{BOP]}\\ \text{or \hspace{0.17em}\hspace{0.17em}2}\angle \text{AOP = 100°}\\ \text{or \hspace{0.17em}\hspace{0.17em}}\angle \text{AOP = 50°}\\ \text{or \hspace{0.17em}\hspace{0.17em}}\angle \text{POA = 50°}\\ \text{Hence, the correct option is (A).}\end{array}$

Q.4 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Ans.

$\begin{array}{l}\text{Let AB is a diameter and PQ and RS be two tangents at}\\ \text{the ends of diameter AB.}\\ \text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent.}\\ \therefore \text{OA}\perp \text{RS and OB}\perp \text{PQ}\\ \text{Thus,we have}\\ \angle \text{OAR =}\angle \text{OAS = 90° =}\angle \text{OBP =}\angle \text{OBQ}\\ \text{or}\angle \text{OAR =}\angle \text{OBQ and}\angle \text{OAS =}\angle \text{OBP}\\ \text{i.e., alternate interior angles are equal.}\\ \text{Therefore, by converse of parallel lines axiom, RS||PQ.}\end{array}$

Q.5 Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Ans.
Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

Let the perpendicular to AB at P does not pass through the centre O. Let it pass through another point R. We join OP and PR.

\begin{array}{l}\text{Perpendicular to AB at P passes through R}\text{.}\\ \therefore \angle \text{RPB = 90°}\mathrm{\dots }\text{(1)}\\ \text{We know that the line joining the centre and point of}\\ \text{contact to the tangent to a circle are perpendicular to}\\ \text{each other}\text{.}\\ \therefore \angle \text{OPB = 90°}\mathrm{\dots }\text{(2)}\\ \text{From (1) and (2), we get}\\ \angle \text{OPB =}\angle \text{RPB}\\ \text{From figure, we find that}\\ \angle \text{RPB <}\angle \text{OPB}\\ \therefore \angle \text{OPB =}\angle \text{RPB is not possible}\text{. It is only possible when the}\\ \text{line RP coincides with OP}\text{.}\\ \text{Therefore, the perpendicular at the point of contact to}\\ \text{the tangent to a circle passes through the centre}\text{.}\end{array}

Q.6 The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Ans.
Let centre of the circle be at O and point of contact of the tangent from A be P.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
null.

Q.7  Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Ans.
Let centre of the circles be at O and point of contact of the tangent PQ be A.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

\begin{array}{l}\text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent}\text{.}\\ \therefore \text{OA}\perp \text{PQ}\\ \text{It is given that OP = OQ = 5}\text{cm and OA = 3 cm}\\ \text{By applying Pythagoras theorem in Δ}\text{OAP, we get}\\ {\text{OA}}^{\text{2}}{\text{+PA}}^{\text{2}}{\text{= OP}}^{\text{2}}\\ {\text{or 3}}^{\text{2}}{\text{+PA}}^{\text{2}}{\text{= 5}}^{\text{2}}\\ {\text{or PA}}^{\text{2}}\text{= 25}-\text{16 = 9}\\ \text{or PA = 3 cm}\\ \text{Similarly, we have AQ = 3 cm}\\ \text{PQ = PA+AQ = 3+3 = 6 cm}\\ \text{Therefore, length of the chord PQ is 6 cm}\text{.}\end{array}

Q.8

$\begin{array}{l}\text{A quadrilateral ABCD is drawn to circumscribe a circle}\\ \text{(see the following figure). Prove that}\\ \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC}.\end{array}$

Ans.

\begin{array}{l}\text{We have to prove that}\\ \text{AB+CD = AD+BC}\\ \text{We know}\text{that lengths of tangents drawn from a point to a}\\ \text{circle are equal}\text{.}\\ \text{Therefore, from figure, we have}\\ \text{DR = DS, CR = CQ,}\text{AS = AP,}\text{BP = BQ}\\ \text{Now,}\\ \text{LHS = AB+CD = (AP+BP)+(CR+DR)}\\ \text{= (AS+BQ)+(CQ+DS)}\\ \text{= AS+DS+BQ+CQ}\\ \text{= AD+BC}\\ \text{= RHS}\end{array}

Q.9

$\begin{array}{l}\text{In\hspace{0.33em}the\hspace{0.33em}following\hspace{0.33em}figure.\hspace{0.33em}XY\hspace{0.33em}and\hspace{0.33em}X’Y’\hspace{0.33em}are\hspace{0.33em}two\hspace{0.33em}parallel}\\ \text{Tangents\hspace{0.33em}to\hspace{0.33em}a\hspace{0.33em}circle\hspace{0.33em}with\hspace{0.33em}center\hspace{0.33em}O\hspace{0.33em}and\hspace{0.33em}another\hspace{0.33em}}\\ \text{tangent\hspace{0.33em}AB\hspace{0.33em}with\hspace{0.33em}point\hspace{0.33em}of\hspace{0.33em}contact\hspace{0.33em}C\hspace{0.33em}intersecting\hspace{0.33em}XY\hspace{0.33em}at}\\ \text{A\hspace{0.33em}and\hspace{0.33em}X’Y’\hspace{0.33em}at\hspace{0.33em}B.\hspace{0.33em}Prove\hspace{0.33em}that\hspace{0.33em}}\angle {\text{AOB\hspace{0.33em}=\hspace{0.33em}90}}^{\circ }\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{Let us join points O and C.}\\ \text{In Δ\hspace{0.17em}OPA and Δ\hspace{0.17em}OCA,}\\ \text{OP = OC (Radii of the given circle)}\\ \text{AP = AC (Tangents from point A)}\\ \text{AO = AO (Common side)}\\ \therefore \text{Δ\hspace{0.17em}OPA}\cong \text{Δ\hspace{0.17em}OCA (SSS congruence criterion)}\\ \text{So,}\angle \text{POA =}\angle \text{COA}...\text{(1)}\\ \text{Similarly, we have Δ\hspace{0.17em}OQB}\cong \text{Δ\hspace{0.17em}OCB and}\\ \angle \text{QOB =}\angle \text{COB}...\text{(2)}\\ \text{Now, POQ is a straight line. So, we have}\\ \angle \text{POA+}\angle \text{COA+}\angle \text{QOB+}\angle \text{COB = 180°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{COA+}\angle \text{COA+}\angle \text{COB+}\angle \text{COB = 180° [From (1) and (2)]}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\left(\angle \text{COA+}\angle \text{COB}\right)\text{= 180°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{COA+}\angle \text{COB =}\frac{\text{180°}}{\text{2}}\text{= 90°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{AOB = 90°}\end{array}$

Q.10 Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Ans.

$\begin{array}{l}\text{We have the above figure as per given information.}\\ \text{We know that a line from the centre of a circle is perpendicular}\\ \text{to the point of contact of a tangent.}\\ \text{Therefore, in quadrilateral PAOB, we have}\\ \angle \text{OAP+}\angle \text{OBP+}\angle \text{AOB+}\angle \text{APB = 360°}\\ \text{or 90°+90°+}\angle \text{O+}\angle \text{P = 360°}\\ \text{or}\angle \text{AOB+}\angle \text{APB = 360°-180° = 180°}\\ \text{or}\angle \text{AOB+}\angle \text{APB = 180°}\end{array}$

Q.11 Prove that the parallelogram circumscribing a circle is a rhombus.

Ans.

$\begin{array}{l}\text{It is given that ABCD is a rhombus. Therefore, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB = CD}...\text{(1)}\\ \text{and \hspace{0.17em}AD = BC}...\text{(2)}\\ \text{We know\hspace{0.33em}that lengths of tangents drawn from a point to a}\\ \text{circle are equal.}\\ \text{Therefore, from figure, we have}\\ \text{DR = DS, CR = CQ,\hspace{0.17em}\hspace{0.17em}AS = AP,\hspace{0.17em}\hspace{0.17em}BP = BQ}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+CD = (AP+BP)+(CR+DR)}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+CD\hspace{0.17em}\hspace{0.17em}= (AS+BQ)+(CQ+DS)}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+CD\hspace{0.17em}\hspace{0.17em}= AS+DS+BQ+CQ}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+CD\hspace{0.17em}\hspace{0.17em}= AD+BC}...\text{(3)}\\ \text{From equations (1), (2) and (3), we get}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+AB = AD+AD}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2AB = 2AD}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB = AD}...\text{(4)}\\ \text{On comparing equations (1), (2) and (4), we get}\\ \text{AB = BC = CD = AD}\\ \text{Hence,ABCD is a rhombus.}\end{array}$

Q.12 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see the following figure). Find the sides AB and AC.


Ans.

$\begin{array}{l}\text{Let the given circle touch the sides AB and AC of the triangle}\\ \text{at points E and F respectively and length of the line segment}\\ \text{AF be x.}\\ \text{In Δ\hspace{0.17em}ABC, we have}\\ \text{CF = CD = 6\hspace{0.17em}cm [Tangents on the circle from point C]}\\ \text{BE = BD = 8\hspace{0.17em}cm \hspace{0.17em}[Tangents on the circle from point B]}\\ \text{AE = AF = x \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Tangents on the circle from point A]}\\ \text{AB = AE+EB = x+8}\\ \text{BC = BD+DC = 8+6 = 14 cm}\\ \text{CA = CF+FA = 6+x}\\ \text{Now, perimeter of Δ\hspace{0.17em}ABC = 2s = AB+BC+CA}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}= (x+8)+14+(x+6)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}= 2x+28 = 2(x+14)}\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}s = x+14}\\ \text{Area of Δ\hspace{0.17em}ABC =}\sqrt{\text{s(s}-\text{a)(s}-\text{b)(s}-\text{c)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=}\sqrt{\text{(x+14)(x+14}-\text{x}-\text{8)(x+14}-\text{14)(x+14}-\text{6}-\text{x)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=}\sqrt{\text{(x+14)48x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}= 4}\sqrt{{\text{3(x}}^{\text{2}}\text{+14x)}}\\ \text{Area of Δ\hspace{0.17em}OBC =}\frac{\text{1}}{\text{2}}\text{OD×BC =}\frac{\text{1}}{\text{2}}{\text{×4×14 = 28 cm}}^{\text{2}}\\ \text{Area of Δ\hspace{0.17em}OCA =}\frac{\text{1}}{\text{2}}\text{OF×AC =}\frac{\text{1}}{\text{2}}\text{×4×(6+x) = 12+2x}\\ \text{Area of Δ\hspace{0.17em}OAB =}\frac{\text{1}}{\text{2}}\text{OE×AB =}\frac{\text{1}}{\text{2}}\text{×4×(8+x) = 16+2x}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area of Δ\hspace{0.17em}ABC = Area of Δ\hspace{0.17em}OBC+Area of Δ\hspace{0.17em}OCA+Area of Δ\hspace{0.17em}OAB}\\ \text{or 4}\sqrt{{\text{3(x}}^{\text{2}}\text{+14x)}}\text{= 28+12+2x+16+2x = 56+4x = 4(x+14)}\\ {\text{or 3x}}^{\text{2}}\text{+42x =}{\left(\text{x+14}\right)}^{\text{2}}{\text{= x}}^{\text{2}}\text{+28x+196}\\ {\text{or 3x}}^{\text{2}}\text{+42x}-{\text{x}}^{\text{2}}-\text{28x = 196}\\ {\text{or 2x}}^{\text{2}}\text{+14x = 196}\\ {\text{or 2(x}}^{\text{2}}\text{+7x}-\text{98) = 0}\\ {\text{or x}}^{\text{2}}\text{+7x}-\text{98=0}\\ {\text{or x}}^{\text{2}}\text{+14x}-\text{7x-98=0}\\ \text{or x(x+14)}-\text{7(x+14)=0}\\ \text{or (x+14)(x}-\text{7)=0}\\ \text{or x=7 [Length can not be negative, so x¹}-\text{14.]}\\ \text{Hence, \hspace{0.17em}AB=x+8=7+8=15 cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}CA=6+x=6+7=13 cm}\end{array}$

Q.13 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Ans.

$\begin{array}{l}\text{Let ABCD be a quadrilateral circumscribing a circle centred at O}\\ \text{such that it touches the circle at points P, Q, R and S. Let us join}\\ \text{the vertices of the quadrilateral ABCD to the centre of the circle.}\\ \text{In ΔOAP and Δ\hspace{0.17em}OAS, we have}\\ \text{AP = AS [Tangents drawn from point A]}\\ \text{OP = OS \hspace{0.17em}[Radii of the same circle]}\\ \text{OA = OA [Common side]}\\ \text{Δ\hspace{0.17em}OAP}\cong \text{Δ\hspace{0.17em}OAS [SSS congruence criterion]}\\ \text{Thus,}\angle \text{POA =}\angle \text{AOS or \hspace{0.17em}\hspace{0.17em}}\angle \text{1 =}\angle \text{2}\\ \text{Similarly,}\angle \text{3 =}\angle \text{4,}\angle \text{5 =}\angle \text{6,\hspace{0.17em} \hspace{0.17em}}\angle \text{7 =}\angle \text{8}\\ \text{Now, from the above figure, we have}\\ \angle \text{1+}\angle \text{2+}\angle \text{3+}\angle \text{4+}\angle \text{5+}\angle \text{6+\hspace{0.17em}}\angle \text{7+}\angle \text{8 = 360°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(}\angle \text{1+}\angle \text{2)+(}\angle \text{3+}\angle \text{4)+(}\angle \text{5+}\angle \text{6)+(}\angle \text{7+}\angle \text{8) = 360°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(}\angle \text{2+}\angle \text{2)+(}\angle \text{3+}\angle \text{3)+(}\angle \text{6+}\angle \text{6)+(\hspace{0.17em}}\angle \text{7+}\angle \text{7) = 360°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\left(\angle \text{2+}\angle \text{3+}\angle \text{6+}\angle \text{7}\right)\text{= 360°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\angle \text{2+}\angle \text{3}\right)\text{+}\left(\angle \text{6+}\angle \text{7}\right)\text{=}\frac{\text{360°}}{\text{2}}\text{= 180°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\angle \text{2+}\angle \text{3}\right)\text{+}\left(\angle \text{6+}\angle \text{7}\right)\text{= 180°}\\ \text{or}\angle \text{AOB+}\angle \text{COD = 180°}\\ \text{Similarly,\hspace{0.17em}we\hspace{0.17em}\hspace{0.17em}have}\\ \left(\angle \text{1+}\angle \text{8}\right)\text{+}\left(\angle \text{4+}\angle \text{5}\right)\text{= 180°}\\ \text{or}\angle \text{AOD+}\angle \text{BOC = 180°}\\ \therefore \angle \text{AOB+}\angle \text{COD = 180° =}\angle \text{AOD+}\angle \text{BOC}\\ \text{i.e., opposite sides of a quadrilateral circumscribing a circle}\\ \text{subtend supplementary angles at the centre of the circle.}\end{array}$