Question

$O$ is the centre of the circle with radius $5$ cm. Chords $AB$ and $CD$ are parallel. $AB=6$ cm and $CD=8$ cm. If $PQ$ is distance between $AB$ and $CD$, then the length of $PQ$ is

• A. $10$ cm
• B. $8$ cm
• C. $7$ cm
• D. $7\sqrt{2}$ cm

Answer 1

Answer: (C)

$7$ cm

$OC=OA=5cm,AB=6cm$ and $CD=8cm$

$CP=4cm$ and $AQ=3cm$ [perpendicular from the centre bisects the chord]

In right angled triangle OMPC and ONA ,by Pythagoras theorem

${OC}^2={OP}^2+{PC}^2$

$OP=\sqrt{{OC}^2-{PC}^2}$

$OP=\sqrt{5^2-4^2}$

$OP=3cm$

${OA}^2={OQ}^2+{AQ}^2$

$OQ=\sqrt{{OA}^2-{AQ}^2}$

$OQ=\sqrt{5^2-3^2}$

$OQ=4cm$

$PQ=OP+OQ$

$PQ=4+3$ (Since chords AB and CD are parallel, P-O-Q is a straight line.)

$PQ=7cm$

Therefore length of PQ is 7cm.

So, option C is the answer.