RRB NTPC Mains or Stage 2 Model Paper 9 with solutions

Question : 16 of 120

Marks: +1, -0

OA is perpendicular to the chord PQ of a circle with centre O. If QR is a diametre, AQ = 4 cm, OQ = 5 cm, then PR is equal to

Solution:

(a) AO

= √{OQ^2 - AQ^2} = √{5^2 - 4^2} =√ 9 = 3

Now, from similar Δs QAO and QOR
OR = 2OA = 2× 3 = 6 cm.

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