Two concentric circles are of diameters $ 30 \mathrm{~cm} $ and $ 18 \mathrm{~cm} $. Find the length of the chord of the larger circle which touches the smaller circle.

Given:

Two concentric circles are of diameters \( 30 \mathrm{~cm} \) and \( 18 \mathrm{~cm} \).

To do:

We have to find the length of the chord of the larger circle which touches the smaller circle.

Solution:

Let $R$ be the radius of outer circle and $r$ be the radius of small circle of two concentric circles.
$AB$ is the chord of the outer circle and touches the smaller circle at $P$
Join $OP$ and $OA$.

This implies,

$OP\ \perp\ AB$ and bisects it at $P$.

$OA=R$ and $OP=r$

$OA=\frac{30}{2}=15\ cm$, $OP=\frac{18}{2}=9\ cm$

In right angled triangle $OAP$,

$AP=\sqrt{OA^{2}-OP^{2}}$

$=\sqrt{15^{2}-9^{2}}$

$=\sqrt{225-81}$

$=\sqrt{144}$

$=12\ cm$

$AB=2AP$

$=2 \times 12\ cm$

$=24\ cm$

The length of the chord of the larger circle is 24 cm.

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Updated on: 10-Oct-2022

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