Question

O is the centre of the circle of radius 5 cm, OP is perpendicular to AB, OQ is perpendicular to CD, AB || CD, AB = 6 cm and CD = 8 cm. Determine PQ.

• A. 1cm
• B. 2cm
• C. 3cm
• D. 4cm

Answer 1

The correct option is A

1cm

Given: A circle with centre O, such that chord AB = 6 cm, chord CD = 8 cm

AB $\parallel$ CD and radius of the circle = 5 cm

OP $\perp$ AB and OQ $\perp$ CD

To find: PQ

Construction: Join OA and OC

AP = PB = 3 cm,

CQ = QD = 4 cm

[Perpendicular from the centre of the circle bisects the chord]

In right $△$OPA,

${OA}^2$ = ${OP}^2$ + ${AP}^2$ [By pythagoras Theorem]

$5^2$ = ${OP}^2$ + $3^2$

$5^2$ - $3^2$ = ${OP}^2$

25 – 9 = ${OP}^2$

16 = ${OP}^2$

4 = OP

In right $△$OQC,

${OC}^2$ = ${OQ}^2$ + ${CQ}^2$ [By Pythagoras Theorem]

$5^2$ = ${OQ}^2$ + $4^2$

$5^2$ - $4^2$ = ${OQ}^2$

25 – 16 = ${OQ}^2$

9 = ${OQ}^2$

3 = OQ

PQ = OP – OQ = 4 – 3 = 1 cm.