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In the figure, $ A B $ is a chord of length $ 16 \mathrm{~cm} $ of a circle of radius $ 10 \mathrm{~cm} $. The tangents at $ A $ and $ B $ intersect at a point $ P $. Find the length of $ P A $."
Given:
In the figure, \( A B \) is a chord of length \( 16 \mathrm{~cm} \) of a circle of radius \( 10 \mathrm{~cm} \). The tangents at \( A \) and \( B \) intersect at a point \( P \).
To do:
We have to find the length of \( P A \).
Solution:
Join $OA$.
$OA = 10\ cm$
$OL$ is perpendicular to $AB$.
$AL=LB= \frac{16}{2}=8\ cm$ (AL and LB are the radii of the circle)
In right angled triangle $OLA$,
By Pythagoras theorem,
$OA^2=OL^2 +LA^2$
$OL^2=OA^2-LA^2$
$OL^2=(10)^2-8^2$
$=100-64$
$=36$
$\Rightarrow OL=6\ cm$
$\tan\ AOL=\frac{AL}{OL}$
$=\frac{8}{6}$
$=\frac{4}{3}$
From $\triangle OAP$,
$\tan\ AOL=\frac{PA}{OA}$
$\frac{4}{3}=\frac{PA}{10}$
$PA=\frac{10\times4}{3}$
$PA=\frac{40}{3}\ cm$
The length of PA is $\frac{40}{3}\ cm$.