Question

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

• A. 1 cm
• B. 2 cm
• C. 3 cm
• D. 4 cm

Answer 1

The correct option is A

1 cm

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

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

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

To find: $PQ$

Construction: Join $OA$ and $OC$

$AP=PB=\frac{AB}{2}=3cm,$

$CQ=QD=\frac{CD}{2}=4cm$

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

In right triangle $△OPA,$

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

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

$\Rightarrow O{P}^2=5^2-3^2$

$\Rightarrow {OP}^2=16$

$\Rightarrow OP=4cm$

In right $△OQC,$

$O{C}^2=O{Q}^2+C{Q}^2$ [Pythagoras Theorem]

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

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

$\Rightarrow {OQ}^2=25–16$

$\Rightarrow OQ=3cm$

$PQ=OP–OQ=4–3=1cm$