A chord $ 6 \mathrm{~cm} $ long is drawn in a circle with a diameter equal to $ 10 \mathrm{~cm} $. Find its perpendicular distance from the centre.

Given:

A chord \( 6 \mathrm{~cm} \) long is drawn in a circle with a diameter equal to \( 10 \mathrm{~cm} \).

To do:

We have to find the distance of the perpendicular from the centre.
 Solution:

Let $AB$ be the chord, $O$ be the centre of the circle and $OC$ be the perpendicular drawn from $O$ to $AB$.

We know that,

The perpendicular to a chord from the centre of a circle bisects the chord. 

Therefore,

$AC=CB$

$=\frac{6}{2}$

$=3\ cm$

$\triangle OCA$ is a right-angled triangle.

Therefore, by Pythagoras theorem,

$OA^2 =OC^2 + AC^2$

$OC^2 = 5^2-3^2$

$=25-9$

$=16$

$\Rightarrow OC=\sqrt{16}=4\ cm$

Hence, the distance of the chord from the centre of the circle is $4\ cm$.

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Updated on: 10-Oct-2022

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