A circle is inscribed in $△ A B C$ and $P . Q, R$ are points of contact. If $P A = 4 c m, P B = 6 c m$ and $A C = 12 c m$ , find the measure of $B C$ .

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Solution

$G i v e n :$ $P A = 4 c m$ , $P B = 6 c m$ and $A C = 12 c m$

AP and AR are tangents from point A to the circle. Hence, $A P = A R$

BP and BQ are tangents from B to the circle, Hence, $B P = B Q$

CQ and CR are tangents from C to the circle. Hence, $C Q = C R$

$A R = A P = 4 c m$

$R C = A C - A R$
$= 12 - 4$
$= 8 c m$

$C R = C Q = 8 c m$

$B P = B Q = 6 c m$

From the figure, $B C = B Q + C Q$
$= 8 + 6 = 14 c m$

Therefore, $B C = 14 c m$

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