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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.