Question

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]

Answer 1

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)

$$\begin{gathered} \therefore \frac{1}{2}\times \mathrm{QR}\times \mathrm{OT}+\frac{1}{2}\times \mathrm{PQ}\times \mathrm{OS}+\frac{1}{2}\times \mathrm{PR}\times \mathrm{OU}=336{\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times 30\times 8+\frac{1}{2}\times \left(x+14\right)\times 8+\frac{1}{2}\times \left(x+16\right)\times 8=336 \\ \Rightarrow 120+4x+56+4x+64=336 \\ \Rightarrow 8x+240=336 \end{gathered}$$
$$\begin{gathered} \Rightarrow 8x=336-240=96 \\ \Rightarrow x=12 \end{gathered}$$

∴ 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.