Question

A chord of length $16cm$ is drawn in a circle of radius $10cm$. Find the distance of the chord from the centre of the circle.

Answer 1

Consider $AB$ as the chord with $O$ as the centre and radius $10cm$

So we get

$OA=10cm$ and $AB=16cm$

Construct $OL\perp AB$

Perpendicular from the centre of a circle to a chord bisect the chord

So we get

$AL=\frac{1}{2}\times AB$

By substituting the values

$AL=\frac{1}{2}\times 16$

So we get

$Al=8cm$

Consider $△OLA$

Using the Pythagoras theorem it can be written as

$O{A}^2=O{L}^2+A{L}^2$

By substituting the values we get

${10}^2=O{L}^2+8^2$

On further calculation

$O{L}^2={10}^2-8^2$

So we get

$O{L}^2=100-64$

By subtraction

$O{L}^2=36$

By taking the square root

$OL=\sqrt{36}$

So we get

$OL=6cm$

Therefore, the distance of the chord from the centre of the circle is $6cm$