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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 Δ PQR = 189 cm2 then the length of side PQ is _____

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

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

Length of tangents from a point is equal
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
ar (∆OQR) + ar (∆OPR) + ar (∆OPQ) = ar (∆PQR)
[12×QR×OT]+[12×PR×OU]+[12×PQ×OS]=189
12([(12+x)×6]+[(9+x)×6]+[21×6])=189
3(42+2x)=1892x=21x=10.5

PQ = x + 12
= 10.5 + 12
= 22.5 cm

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