Q. 28

# In Fig. 10.90, PR =

A. 20 cm

B. 26 cm

C. 24 cm

D. 28 cm

Answer :

Given:

QP = 4 cm

OQ = 3 cm

SR = 12 cm

SO’ = 5 cm

__Property:__*The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.*

By above property, ∆OPQ is right-angled at ∠OQP (i.e., ∠OQP = 90°) and ∆O’SR is right-angled at ∠O’SR (i.e., ∠O’SR = 90°).

By Pythagoras theorem in ∆OPQ,

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

⇒ OP^{2} = 4^{2} + 3^{2}

⇒ OP^{2} = 16 + 9

⇒ OP^{2} = 25

⇒ OP = √25

⇒ OP = 5 cm

By Pythagoras theorem in ∆O’SR,

O’R^{2} = SR^{2} + O’S^{2}

⇒ O’R^{2} = 12^{2} + 5^{2}

⇒ O’R^{2} = 144 + 25

⇒ O’R^{2} = 169

⇒ O’R = √169

⇒ O’R^{2} = 13 cm

Now,

PR = PO + ON + NO’ + O’R

⇒ PR = 5 cm + 3 cm + 5 cm + 13 cm

⇒ PR = 26 cm

Hence, PR = 26 cm

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PREVIOUSIn Fig. 10.89, if tangents PA and PB are drawn to a circle such that ∠APS = 30° and chord AC is drawn parallel to the tangent PB, then ∠ABC =NEXTTwo circles of same radii r and centres O and O' touch each other at P as shown in Fig. 10.91. If 00' is produced to meet the circle C (O', r) at A and AT is a tangent to the circle C(O, r) such that O'Q ⊥ AT. Then AO: AO' =

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