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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 ∆PQR is 336 cm^{2}, find the sides PQ and PR.

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#### Solution

Here, T, S and U are the points of contact of the circle with the sides QR, PQ and PR, respectively.

OT = OS = OU = 8 cm (Radii of the circle)

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴ QS = QT = 14 cm

RU = RT = 16 cm

PS = PU = *x* cm (say)

So, QR = QT + TR = 14 cm + 16 cm = 30 cm

PQ = PS + SQ = *x* cm + 14 cm = (*x* + 14) cm

PR = PU + UR = *x* cm + 16 cm = (*x* + 16) cm

Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OT ⊥ QR, OS ⊥ PQ and OU ⊥ PR

Now,

ar(∆OQR) + ar(∆OPQ) + ar(∆OPR) = ar(∆PQR)

\[\therefore \frac{1}{2} \times QR \times OT + \frac{1}{2} \times PQ \times OS + \frac{1}{2} \times PR \times OU = 336 {cm}^2 \]

\[ \Rightarrow \frac{1}{2} \times 30 \times 8 + \frac{1}{2} \times \left( x + 14 \right) \times 8 + \frac{1}{2} \times \left( x + 16 \right) \times 8 = 336\]

\[ \Rightarrow 120 + 4x + 56 + 4x + 64 = 336\]

\[ \Rightarrow 8x + 240 = 336\]

\[\Rightarrow 8x = 336 - 240 = 96\]

\[ \Rightarrow x = 12\]

∴ PQ = (*x* + 14) cm = (12 + 14) cm = 26 cm

PR = (*x* + 16) cm = (12 + 16) cm = 28 cm

Hence, the lengths of sides PQ and PR are 26 cm and 28 cm, respectively.

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