Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C that passes through points A and B where A is located 4 meters north of O and B is located 3 meters east of O. Then, the length of PQ, in meters, is nearest to

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Question :

Started 4 days ago by Shashank in

0 2

A.
6.6
B.
7.2
C.
8.8
D.
7.8

Option C is the correct answer.

Explanatory Answer

Since AOB is a right-angled triangle and AO = 4m, BO = 3m
AB = 5m

Dropping a perpendicular from O on AB at D, we know that:
OA * OB = AB * OD
4*3 = 5*OD
OD = 12/5

We know that OD bisects the chord PQ. Hence, PD = QD
Also, in triangle OPD:
OD 2 + PD 2 = OP 2
(12/5) 2 + PD 2 = 5 2 = 25
PD 2 = 25 – 144/25 = (625 – 144)/25 = 481/25
PD = Ö 481/5 ̴ 21.9/5 = 4.4
PQ = 2*PD = 8.8