Question

6. In Fig. 10.36, O is the centre of the circle with radius 5 cm. OP 1 AB, OQ 1 CD, AB || CD, AB = 6 cm
and CD = 8 cm. Determine PQ​

Answer 1

Step-by-step explanation:

Join OA and OC.

Since, the perpendicular from the centre of the circle to a chord bisects the chord. Therefore, P and Q are mid- points of AB and CD respectively.

Consequently, AP = PB = ½ AB = 3 cm

and CQ = QD = ½ CD = 4 cm

In right triangles OAP and OCQ, we have

OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2

⇒ 52 = OP2 + 32 and 52 = OQ2 + 42

⇒ OP2 = 52 – 32 and OQ2 = 52 - 42

⇒ OP2 = 16 and OQ2 = 9

⇒ OP = 4 and OQ = 3

∴ PQ = OP + OQ = (4 + 3) cm = 7 cm.