Question

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

Answer 1

Join OA and OC.
Since the perpendicular from the centre of the circle to a chord bisects the chord. Therefore, P and Q are midpoints of AB and CD respectively.
Consequently, AP = PB = `1/2"AB"` = 3 cm.
and CQ = QD = `1/2`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.