A triangle PQR is drawn to circumscribe a circle of radius $8$ cm such that the segments QT and TR, into which QR is divided by the point of contact T, are or length $14$ cm and $16$ cm respectively. If area of $△ P Q R$ is $336$ ${c m}^{2}$ , find the sides PQ and PR.

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Solution

Triangle $P Q R$ circumscribes of radius $8 c m$ take center of circle as $O$
join $P O, Q O, R O$ and $O D, O E, O T$
$D$ and $E$ are points on $P Q$ and $P R$
ar of triangle $P O Q + a r$ tri. $Q O R + a r$ tri. $P O R = 336$
take height as $8 c m, P Q = 14 + a, P R = 16 + a, Q R = 14 + 16 = 30$
find value of $a$
Solve the eqn formed answer $- P Q = 12 + 14 = 26$
$P R = 12 + 16 = 28$

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Q. A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆PQR is 336 cm², find the sides PQ and PR. [CBSE 2014]

A triangle PQR is drawn to circumsribe a circle of radius 8 cm such that the segements QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of $Δ P Q R$ is $336 c m^{2}$ , find the sides PQ and PR.