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From an external point $ P $, tangents $ P A $ and $ P B $ are drawn to a circle with centre $ O $. At one point $ E $ on the circle tangent is drawn, which intersects $ P A $ and $ P B $ at $ C $ and $ D $ respectively. If $ P A=14 \mathrm{~cm} $, find the perimeter of $ \triangle P C D $.
Given:
From an external point \( P \), tangents \( P A \) and \( P B \) are drawn to a circle with centre \( O \). At one point \( E \) on the circle tangent is drawn, which intersects \( P A \) and \( P B \) at \( C \) and \( D \) respectively.
\( P A=14 \mathrm{~cm} \).
To do:
We have to find the perimeter of \( \triangle P C D \).
Solution:
$PA$ and $PB$ are the tangents drawn from a point $P$ outside the circle with centre $O$.
$CD$ is another tangent to the circle at point $E$ which intersects $PA$ and $PB$ at $C$ and $D$ respectively.
$PA = 14\ cm$
$PA$ and $PB$ are the tangents to the circle from $P$
$PA = PB = 14\ cm$
$CA$ and $CE$ are the tangents from $C$
This implies,
$CA = CE$...….(i)
Similarly,
$DB$ and $DE$ are the tangents from $D$
$DB = DE$....….(ii)
Perimeter of $\triangle PCD= PC + PD + CD$
$= PC + PD + CE + DE$
$= PC + CE + PD + DE$
$= PC + CA + PD + DB$ [From (i) and (ii)]
$= PA + PB$
$= 14 + 14$
$= 28\ cm$
The perimeter of \( \triangle P C D \) is $28\ cm$.
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