A triangle $ P Q R $ is drawn to circumscribe a circle of radius $ 8 \mathrm{~cm} $ such that the segments $ Q T $ and $ T R $, into which $ Q R $ is divided by the point of contact $ T $, are of lengths $ 14 \mathrm{~cm} $ and $ 16 \mathrm{~cm} $ respectively. If area of $ \Delta P Q R $ is $ 336 \mathrm{~cm}^{2} $, find the sides $ P Q $ and $ P R $.
Given:
A triangle \( P Q R \) is drawn to circumscribe a circle of radius \( 8 \mathrm{~cm} \) such that the segments \( Q T \) and \( T R \), into which \( Q R \) is divided by the point of contact \( T \), are of lengths \( 14 \mathrm{~cm} \) and \( 16 \mathrm{~cm} \) respectively.
Area of \( \Delta P Q R \) is \( 336 \mathrm{~cm}^{2} \).
To do:
We have to find the sides \( P Q \) and \( P R \).
Solution:
$\triangle PQR$ is circumscribed by a circle with centre $O$ and radius $8\ cm$.
$T$ is the point of contact that divides the line segment $OT$ into two parts such that
$QT = 14\ cm$ and $TR = 16\ cm$.
Area of $\triangle PQR = 336\ cm^2$
Let $PS = x\ cm$
$QT$ and $QS$ are tangents to the circle from $Q$.
$QS = QT = 14\ cm$
Similarly,
$RU$ and $RT$ are tangents to the circle
$RT = RU = 16\ cm$
$PS$ and $PU$ are tangents from $P$
$PS = PU = x\ cm$
$PQ = x + 14$ and $PR = x + 16$ and $QR = 14 + 16 = 30\ cm$
Area of $\triangle PQR =$ Area of $\triangle POQ +$ Area of $\triangle QOR +$ Area of $\triangle POR$
$\Rightarrow 336=\frac{1}{2}(\mathrm{QR}) \times 8+\frac{1}{2}(14+x) \times 8+\frac{1}{2}(16+x) \times 8$
$\Rightarrow 336=\frac{1}{2} \times 30 \times 8+4(14+x)+4(16+x)$
$\Rightarrow 336=120+56+4 x+64+4 x$
$\Rightarrow 336=8 x+240$
$\Rightarrow 8 x=336-240$
$\Rightarrow 8x=96$
$\Rightarrow x=\frac{96}{8}=12$
Therefore, $\mathrm{PQ}=x+14=12+14=26 \mathrm{~cm}$
$\mathrm{PR}=x+16=12+16=28 \mathrm{~cm}$
The sides \( P Q \) and \( P R \) are $26\ cm$ and $28\ cm$ respectively.
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