Question

In Fig. PR =

A. 20 cm 

B. 26 cm 

C. 24 cm 

D. 28 cm

Answer 1

Answer is B. 26 cm

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, 

OP² = QP² + OQ² 

'⇒ OP² = 4² + 3² 

⇒ OP² = 16 + 9 

⇒ OP² = 25 

⇒ OP = √25 

⇒ OP = 5 cm 

By Pythagoras theorem in ∆O’SR, 

O’R² = SR² + O’S² 

⇒ O’R ² = 12² + 5² 

⇒ O’R² = 144 + 25 

⇒ O’R² = 169 

⇒ O’R = √169 

⇒ O’R² = 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