Question

In Fig. O is the centre of the circle of radius 5 cm. OP ⊥ AB, OQ ⊥ CD, AB || CD, AB = 6 cm and CD = 8 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 the right-angled triangle OAP, we have
OA² = OP² + AP²
⇒ 5² = OP² + 3²
⇒ OP² = 5² - 3² = 16 cm 
⇒ OP = 4 cm²

In the right angled triangle OCQ we have
OC² = OQ² + CQ²
⇒ 5² = OQ² + 4²
⇒ OQ² = 5² - 4² = 9 cm² 
⇒ OQ = 3 cm
∴ PQ = PO - QO
∴ PQ = OP - OQ = (4 - 3) cm = 1 cm