Q. 28

# In Fig. 10.90, PR

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 OSR is right-angled at OSR (i.e., OSR = 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

OR2 = 122 + 52

OR2 = 144 + 25

OR2 = 169

OR = 169

OR2 = 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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