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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 ΔPQR is 336 cm2, find the sides PQ and PR.

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Let PQ and PR touch the circle at points S and U respectively. Join O with P, Q, R, S and U

We have OS = OT = OU = 6cm

QT = 12 cm and TR = 9cm 

QR = QT +TR = 12cm + 9cm = 21 cm 

Now QT = QS = 12 cm (tangent from the same point) 

TR = RU = 9cm 

Let PS = PU = x cm 

Then PQ=PS+SQ- (12 + x)cm and PR = PU+RU= (9+x)cm 

It is clear that

Thus PQ = (12 + 10.5) cm and PR = (9 + 10.5) cm = 19.5 cm

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