Question

In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contant T, are of lengths 12 cm and 9 cm respectively. If the area fo $\Delta$ PQR = 189 cm${}^2$ then the length of side PQ is _____

• A. 10.5 cm
• B. 22.5 cm
• C. 19.5 cm
• D. Can't be found

Answer 1

The correct option is B 22.5 cm

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 = 6 cm (Radii of the circle)
QT = 12 cm and TR = 9 cm
QR = QT + TR = 12 cm + 9 cm = 21 cm

$\because$ Length of tangents from a point is equal
$\therefore$ QT = QS = 12 cm
and TR = RU = 9 cm
Let PS= PU = x cm
Then, PQ = PS + SQ = (12 + x) cm and PR = PU + RU = (9 + x) cm
$\because$ ar (∆OQR) + ar (∆OPR) + ar (∆OPQ) = ar (∆PQR)
$\Rightarrow [\frac{1}{2}\times QR\times OT]+[\frac{1}{2}\times PR\times OU]+[\frac{1}{2}\times PQ\times OS]=189$
$\Rightarrow \frac{1}{2}\left(\right[(12+x)\times 6]+[(9+x)\times 6]+[21\times 6\left]\right)=189$
$\Rightarrow 3(42+2x)=189\Rightarrow 2x=21\Rightarrow x=10.5$

$\therefore$ PQ = x + 12
= 10.5 + 12
= 22.5 cm