In the figure, there are two concentric circles with centre $ O $ of radii $ 5 \mathrm{~cm} $ and $ 3 \mathrm{~cm} $. From an external point $ P $, tangents $ P A $ and $ P B $ are drawn to these circles. If $ A P=12 \mathrm{~cm} $, find the length of $ B P $. "
Given:
In the figure, there are two concentric circles with centre \( O \) of radii \( 5 \mathrm{~cm} \) and \( 3 \mathrm{~cm} \). From an external point \( P \), tangents \( P A \) and \( P B \) are drawn to these circles.
\( A P=12 \mathrm{~cm} \).
To do:
We have to find the length of \( B P \).
Solution:
From the figure,
$AP = 12\ cm$
In right angled triangle $OAP$,
By Pythagoras theorem,
$OP^2 = OA^2 + AP^2$
$= 5^2 + (12)^2$
$= 25 + 144$
$= 169$
$= (13)^2$
$\Rightarrow OP = 13\ cm$
In right angled triangle $OBP$,
$OP^2 = OB^2 + BP^2$
$(13)² = 3^2 + BP^2$
$169 = 9 + BP^2$
$BP^2 = 169 - 9$
$= 160$
$= 16 \times 10$
$\Rightarrow BP = \sqrt{16 \times 10}$
$= 4\sqrt{10}\ cm$
The length of \( B P \) is $ 4\sqrt{10}\ cm$.
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