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**NCERT TEXTBOOK QUESTIONS SOLVED****Page No. 213****EXERCISE 10.2****In Q.1 to 3, choose the correct option and give justification.**

**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****Sol. **From figure

OQ^{2} = OP^{2} + PQ^{2}

(25)^{2} = OP^{2 }+ (24)^{2}

⇒ 625 - 576 = OP^{2}

⇒ 49 = OP^{2}

⇒

OP = 7cm

Radius of the circle = 7cm.

Hence, correct option is (a).**Q.2. ****In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to ****(A) 60° ****(B) 70° ****(C) 80° ****(D) 90°**

**Sol.** ∠OPT = 90°; ∠OQT = 90°; ∠POQ = 110^{0}

TPOQ is a quadrilateral

⇒ ∠PTQ + ∠POQ = 180°

⇒ ∠PTQ + 110^{0} = 180^{0}

⇒ ∠PTQ = 180^{0} - 100^{0} = 70^{0}**Q.3. Choose the correct option: If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to****(A) 50° ****(B) 60° ****(C) 70° ****(D) 80°****Sol.** In ΔOAP and ΔOBP

OA = OB [Radii]

PA = PB

[Length of tangents from an external point are equal]

OP = OP [common]

Hence, Option (a) is correct**Page No. 214****Q.4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.****Sol.** AB is a diameter of the circle, p and q are two tangents

OA ⊥ p and OB ⊥ q and ∠1 = ∠2 = 90^{0}

⇒ p ║q [∠1 and ∠2 are alternate angle]

**Q.5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.****Sol.** XY tangent to the circle C(O, r) at B and AB ⊥ XY

Join OB

∠ABY = 90º^{ }

∠OBY = 90º

[Radius through point of contact is ⊥ to the tangent]

∴ ∠ABY + ∠OBY = 180^{0}

⇒ ABO is collinear.

∴ AB passes through centre of the circle.**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.****Sol. **OP = Radius of the circle

OA = 5 cm; AP = 4 cm

∠OPA = 90° [Radius and tangent are ⊥ar]

OA^{2} = AP^{2} + OP^{2} [By Pythagoras theorem]

5^{2} = 4^{2} + OP^{2}

⇒ 25 = 16 + OP^{2}

⇒ 25 - 16 = OP^{2}

⇒ 9 = OP^{2}

⇒ OP = √ 9 = 3

Radius = 3 cm**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.****Sol.** The radius of larger circle = 5 cm and radius of smaller circle = 3 cm

OP ⊥ AB [Radius of circle is perpendicular to the tangent]

AB is a chord of the larger circle

∴ OP bisect AB AP = BP

In ΔOAP OA^{2} = AP^{2} + OP^{2}

⇒ (5)2 = AP^{2} + (3)^{2}

⇒ AP^{2} = 25 - 9 = 16

⇒

AB = 2AP = 2 x 4 = 8 cm**Q.8. A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that: ****AB + CD = AD + BC**

**Sol.**

AP = AS ...(i) [Lengths of tangents from an external point are equal]

BP = BQ ...(ii)

CR = CQ ...(iii)

DR = DS ...(iv)

Adding equations (i), (ii), (iii) and (iv), we get

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ (AP + BP) + (CR + DR) = (AS + DS) +(BQ + CQ

⇒ AB + CD = AD + BC**Q.9. In the figure, XY and X′ Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠AOB = 90°.**

**Sol**. Given: Two parallel tangents to a circle with centre O. Tangent AB with point of contact C intersects XY at A and X'Y' at B**To Prove**: ∠AOB = 90°**Construction:** Join OA, OB and OC**Proof:** In ΔAOP and ΔAOC

AP = AC [Lengths of tangents]

OP = OC [Radii]

OA = OA

⇒ ΔAOP ≅ ΔAOC [SSS congruence rule]

⇒ ∠PAO = ∠CAO [C.P.C.T]

∠PAC = 2 ∠OAC ...(i)

Similarly ∠QBC = 2 ∠OBC ...(ii)

Adding (i) and (ii), we get

∠PAC + ∠QBC = 2 [∠OAC + ∠OBC]

∠PAC + ∠QBC = 180°

[interior consecutive angle on same side of transversal]

2 = 180° [∠OAC + ∠OBC]

⇒ ∠OAC + ∠OBC = 90°

In ΔAOB, ∠AOB + [∠OAC + ∠OBC] = 180°

⇒ ∠AOB + 90° = 180°

⇒ ∠AOB = 90°**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.****Sol. **PA and PB are two tangents, A and B are the points of contact of the tangents.

OA ⊥ AP and OB ⊥ BP

∠OAP = ∠OBP = 90°

[Radius and tangent are perpendicular to each other]

In quadrilateral OAPB

(∠OAP + ∠OBP) + ∠APB + ∠AOB = 360^{0}

⇒ 180° + ∠APB + ∠AOB = 360°

∠APB + ∠AOB = 360° - 180° = 180°**Q 11. Prove that the parallelogram circumscribing a circle is a rhombus. ****Sol.** Parallelogram ABCD circumscribing a circle with centre O.

OP⊥ AB and OS ⊥ AD

In ΔOPB and ΔOSD, ∠OPB = ∠OSD [Each 90°]

OB = OD

[Diagonals of ║gm bisect each other]

OP = OS [Radii]

⇒ ΔOPB ≅ ΔOSD [RHS congruence rule]

PB = SD [C.P.C.T] ...(i)

AP = AS [Lengths of tangents]... (ii)

Adding (i) and (ii)

AP + PB = AS + DS

⇒ AB AD

Similarly AB = BC = CD = DA

∴ ║gm ABCD is a rhombus**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 figure). Find the sides AB and AC.****Sol. **BD = 8 cm and DC = 6 cm

BE = BD = 8 cm; CD= CF = 6 cm;

Let AE=AF = x cm

In ΔABC a = 6 + 8 = 14 cm;

b = (x + 6) cm;

c = (x +8)

ar ΔABC = ar ΔOBC + ar ΔOCA + ar ΔOAB

From (i) and (ii)

⇒ 3x(x+14) = (x + 14)^{2}

⇒ 3x^{2} + 42x = x^{2} + 196 + 28x

⇒ 2x^{2} + 14x - 196 = 0

⇒ x^{2} + 7x - 98 = 0

⇒ x^{2} + 14x - 7x - 98 = 0

⇒ x(x + 14) - 7(x + 14) = 0

⇒ (x - 7)(x + 14) = 0 ⇒ x = 7

AB = x + 8 = 7 + 8 = 15 cm

AC = x + 6 = 7 + 6 = 13 cm**Q 13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.****Sol. **

AB touches at P and BC, CD and DA touch the circle at Q, R and S.**Construction**: Join OA, OB, OC, OD and OP, OO, OR, OS

∴ ∠1 = ∠2 [OA bisects ∠POS]

Similarly

Similarly ∠AOB + ∠COD = 180°

Hence, opposite sides of quadrilateral circumscribing a circle subtend supplementary angles at the centre of a circle.

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