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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

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B

2cm

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C

3cm

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D

4cm

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Solution

The correct option is A

1cm


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

ABCD and radius of the circle =5 cm

OPAB and OQCD

To find: PQ

Construction: Join OA and OC

AP=PB=AB2=3 cm,

CQ=QD=CD2=4 cm

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

In right triangle OPA,

OA2=OP2+AP2 [Pythagoras Theorem]

52=OP2+32

OP2=5232

OP2=16

OP=4 cm

In right OQC,

OC2=OQ2+CQ2 [Pythagoras Theorem]

52=OQ2+42

OQ2=5242

OQ2=2516

OQ=3 cm

PQ=OPOQ=43=1 cm


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