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. - Mathematics

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Short Note

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.

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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)

\[\therefore \frac{1}{2} \times QR \times OT + \frac{1}{2} \times PQ \times OS + \frac{1}{2} \times PR \times OU = 336 {cm}^2 \]
\[ \Rightarrow \frac{1}{2} \times 30 \times 8 + \frac{1}{2} \times \left( x + 14 \right) \times 8 + \frac{1}{2} \times \left( x + 16 \right) \times 8 = 336\]
\[ \Rightarrow 120 + 4x + 56 + 4x + 64 = 336\]
\[ \Rightarrow 8x + 240 = 336\]

\[\Rightarrow 8x = 336 - 240 = 96\]
\[ \Rightarrow x = 12\]

∴ 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.

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Chapter 8: Circles - Exercise 8.2 [Page 38]

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RD Sharma Class 10 Maths
Chapter 8 Circles
Exercise 8.2 | Q 39 | Page 38

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