Question

The radius of a circle is $13$cm. and the length of one of its chords is $24$cm. Find the distance of the chord from the centre.(Pythagoras theorem: One of the most important result in elementary geometry is Pythagoras theorem. In a right-angled triangle ABC with $\angle$B$={90}^o$, we have $A{B}^2+B{C}^2=C{A}^2$).

Answer 1

We solve the problem by using Pythagoras theorem.

In the figure let $AB$ be a chord of length $2l=24$.

Let $C$ be the mid point of chord $AB$.

Draw $OC$ perpendicular to $AB$.

Let $r=13$ be the radius of the circle and $OC=t$.

Thus $AC=CB=l=12$ and $OA=r=13$.

By Pythagorastheorem we have,

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

$⟹O{C}^2=O{A}^2-A{C}^2$

$⟹O{C}^2={13}^2-{12}^2=169-144=25$.

Hence, $OC=\sqrt{25}=5$.

Thus, the distance of the chord from the centre is $5$ cm.