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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 $4\ cm$.