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Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Given:
Two concentric circles are of radii \( 5 \mathrm{~cm} \) and \( 3 \mathrm{~cm} \).
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 $OP=r$
$OA=5\ cm$
$OP=3\ cm$
In right angled triangle $OAP$,
$AP=\sqrt{OA^{2}-OP^{2}}$
$=\sqrt{5^{2}-3^{2}}$
$=\sqrt{25-9}$
$=\sqrt{16}$
$=4\ cm$
$AB=2AP$
$=2 \times 4\ cm$
$=8\ cm$
The length of the chord of the larger circle is 8 cm.
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