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Question
A chord of length 14 cm is at a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is
Options
12 cm
12 cm
12 cm
18 cm
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Solution
18 cm
We are given the chord of length 14 cm and perpendicular distance from the centre to the chord is 6 cm. We are asked to find the length of another chord at a distance of 2 cm from the centre.
We have the following figure
We are given AB = 14 cm, OD = 6 cm, MO = 2 cm, PQ = ?
Since, perpendicular from centre to the chord divide the chord into two equal parts
Therefore
`AQ^2 = AD^2 +OD^2`
= ` 7^2 + 6^2`
= 49 + 36
`= sqrt( 85)`
Now consider the ΔOPQ in which OM = 2 cm
So using Pythagoras Theorem in ΔOPM
`PM^2 = OP^2 -OM^2`
`=(sqrt(85))^2 - 2^2` (∵ OP = AO = radius)
= 81
= 9 cm
Hence PQ = 2 PM
= 2 × 9
= 18 cm
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