    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 cmLet 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)=189⇒2x=21⇒x=10.5 ∴ PQ = x + 12 = 10.5 + 12 = 22.5 cm  Suggest Corrections  0      Related Videos     