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]

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) $⇒8x=336-240=96\phantom{\rule{0ex}{0ex}}⇒x=12$ ∴ 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.MathematicsRD Sharma (2016)Standard X

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