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The radius of a circle is $8\ cm$ and the length of one of its chords is $12\ cm$. Find the distance of the chord from the centre.
Given:
The radius of a circle is $8\ cm$ and the length of one of its chords is $12\ cm$.
To do:
We have to find the distance of the chord from the centre.
Solution:
Let the radius of the circle with centre $O$ be $OA = 8\ cm$
Length of the chord $AB = 12\ cm$
$OC \perp AB$ which bisects $AB$ at $C$
Therefore,
$AC = CB = 12 \times \frac{1}{2} = 6\ cm$
In $\triangle OAC$,
$OA^2 = OC^2 + AC^2$ (Pythagoras Theorem)
$(8)^2 = OC^2 + (6)^2$
$64 = OC^2 + 36$
$OC^2 = 64 - 36 = 28$
$\Rightarrow OC = \sqrt{28}$
$ = 2\sqrt{7}\ cm$
$= 2 \times 2.6457$
$= 5.291\ cm$
The distance of the chord from the centre is $5.291\ cm$.
Updated on: 10-Oct-2022
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