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In the figure, $ P A $ and $ P B $ are tangents to the circle from an external point $ P $. $ C D $ is another tangent touching the circle at $ Q $. If $ PA=12\ cm, QC=QD=3\ cm, $ then find $ P C+P D $."
Given:
In the figure, \( P A \) and \( P B \) are tangents to the circle from an external point \( P \). \( C D \) is another tangent touching the circle at \( Q \).
\( PA=12\ cm, QC=QD=3\ cm. \)
To do:
We have to find \( P C+P D \).
Solution:
We know that,
Tangents drawn from an external point are equal.
This implies,
$PA = PB = 12\ cm$.....…(i)
$QC = AC = 3\ cm$......(ii)
$QD = BD = 3\ cm$.......(iii)
Therefore,
$PC + PD = (PA - AC) + (PB - BD)$
$= (12 - 3) + (12 - 3)$ [From (i), (ii) and (iii)]
$= 9 + 9$
$= 18\ cm$
Therefore, \( P C+P D = 18\ cm \).
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