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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}

â‡’ 625 - 576 = OP^{2}

â‡’ 49 = OP^{2}

â‡’

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 âˆ POQ = 110Â°, then âˆ PTQ is equal to ****(A) 60Â° ****(B) 70Â° ****(C) 80Â° ****(D) 90Â°**

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

TPOQ is a quadrilateral

â‡’ âˆ PTQ + âˆ POQ = 180Â°

â‡’ âˆ PTQ + 110^{0} = 180^{0}

â‡’ âˆ 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 80Â°, then âˆ POA is equal to****(A) 50Â° ****(B) 60Â° ****(C) 70Â° ****(D) 80Â°****Sol.** In Î”OAP and Î”OBP

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 âŠ¥ p and OB âŠ¥ q and âˆ 1 = âˆ 2 = 90^{0}

â‡’ p â•‘q [âˆ 1 and âˆ 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 âŠ¥ XY

Join OB

âˆ ABY = 90Âº^{ }

âˆ OBY = 90Âº

[Radius through point of contact is âŠ¥ to the tangent]

âˆ´ âˆ ABY + âˆ OBY = 180^{0}

â‡’ ABO is collinear.

âˆ´ 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

âˆ OPA = 90Â° [Radius and tangent are âŠ¥ar]

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

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

â‡’ 25 = 16 + OP^{2}

â‡’ 25 - 16 = OP^{2}

â‡’ 9 = OP^{2}

â‡’ OP = âˆš 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 âŠ¥ AB [Radius of circle is perpendicular to the tangent]

AB is a chord of the larger circle

âˆ´ OP bisect AB AP = BP

In Î”OAP OA^{2} = AP^{2} + OP^{2}

â‡’ (5)2 = AP^{2} + (3)^{2}

â‡’ AP^{2} = 25 - 9 = 16

â‡’

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

â‡’ (AP + BP) + (CR + DR) = (AS + DS) +(BQ + CQ

â‡’ AB + CD = AD + BC**Q.9. In the figure, XY and Xâ€² Yâ€² are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and Xâ€²Yâ€² at B. Prove that âˆ AOB = 90Â°.**

**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**: âˆ AOB = 90Â°**Construction:** Join OA, OB and OC**Proof:** In Î”AOP and Î”AOC

AP = AC [Lengths of tangents]

OP = OC [Radii]

OA = OA

â‡’ Î”AOP â‰… Î”AOC [SSS congruence rule]

â‡’ âˆ PAO = âˆ CAO [C.P.C.T]

âˆ PAC = 2 âˆ OAC ...(i)

Similarly âˆ QBC = 2 âˆ OBC ...(ii)

Adding (i) and (ii), we get

âˆ PAC + âˆ QBC = 2 [âˆ OAC + âˆ OBC]

âˆ PAC + âˆ QBC = 180Â°

[interior consecutive angle on same side of transversal]

2 = 180Â° [âˆ OAC + âˆ OBC]

â‡’ âˆ OAC + âˆ OBC = 90Â°

In Î”AOB, âˆ AOB + [âˆ OAC + âˆ OBC] = 180Â°

â‡’ âˆ AOB + 90Â° = 180Â°

â‡’ âˆ AOB = 90Â°**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 âŠ¥ AP and OB âŠ¥ BP

âˆ OAP = âˆ OBP = 90Â°

[Radius and tangent are perpendicular to each other]

In quadrilateral OAPB

(âˆ OAP + âˆ OBP) + âˆ APB + âˆ AOB = 360^{0}

â‡’ 180Â° + âˆ APB + âˆ AOB = 360Â°

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

OPâŠ¥ AB and OS âŠ¥ AD

In Î”OPB and Î”OSD, âˆ OPB = âˆ OSD [Each 90Â°]

OB = OD

[Diagonals of â•‘gm bisect each other]

OP = OS [Radii]

â‡’ Î”OPB â‰… Î”OSD [RHS congruence rule]

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

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

Adding (i) and (ii)

AP + PB = AS + DS

â‡’ AB AD

Similarly AB = BC = CD = DA

âˆ´ â•‘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 Î”ABC a = 6 + 8 = 14 cm;

b = (x + 6) cm;

c = (x +8)

ar Î”ABC = ar Î”OBC + ar Î”OCA + ar Î”OAB

From (i) and (ii)

â‡’ 3x(x+14) = (x + 14)^{2}

â‡’ 3x^{2} + 42x = x^{2} + 196 + 28x

â‡’ 2x^{2} + 14x - 196 = 0

â‡’ x^{2} + 7x - 98 = 0

â‡’ x^{2} + 14x - 7x - 98 = 0

â‡’ x(x + 14) - 7(x + 14) = 0

â‡’ (x - 7)(x + 14) = 0 â‡’ 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

âˆ´ âˆ 1 = âˆ 2 [OA bisects âˆ POS]

Similarly

Similarly âˆ AOB + âˆ COD = 180Â°

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

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