QJEEAdv18Mab6d3: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Mab6d3 Solution: In a PQR = 30 and the sides PQ and QR have lengths 10 and 10,…

Question

Verified Solution • 2025–26 Session moderate

In a

\Delta

PQR = 30

^\circ

and the sides PQ and QR have lengths 10

\sqrt 3

and 10, respectively. Then, which of the following statement(s) is(are) TRUE?

Accepted Answer

We have,

In

\Delta

PQR

\angle PQR = 30^\circ

PQ = 10\sqrt 3

QR = 10

By cosine rule

\cos 30^\circ = {{P{Q^2} + Q{R^2} - P{R^2}} \over {2PQ.QR}}

\Rightarrow {{\sqrt 3 } \over 2} = {{300 + 100 - P{R^2}} \over {200\sqrt 3 }}

\Rightarrow 300 = 300 + 100 - P{R^2}

\Rightarrow PR = 10

Since,

PR = QR = 10

\therefore

\angle QPR = 30^\circ

and

\angle QRP = 120^\circ

Area of

\Delta PQR = {1 \over 2}PQ.QR.\sin 30^\circ

= {1 \over 2} \times 10\sqrt 3 \times 10 \times {1 \over 2} = 25\sqrt 3

Radius of incircle of

\Delta PQR = {{Area\,of\,\Delta PQR} \over {Semi - perimetre\,of\,\Delta PQR}}

i.e.

r = {\Delta \over s} = {{25\sqrt 3 } \over {{{10\sqrt 3 + 10 + 10} \over 2}}} = {{25\sqrt 3 } \over {5(\sqrt 3 + 2)}}

\Rightarrow r = 5\sqrt 3 (2 - \sqrt 3 ) = 10\sqrt 3 - 15

and radius of circumcircle

(R) = {{abc} \over {4\Delta }} = {{10\sqrt 3 \times 10 \times 10} \over {4 \times 25\sqrt 3 }} = 10

\therefore

Area of circumcircle of

\Delta PQR = \pi {R^2} = 100\pi

Hence, option (b), (c) and (d) are correct answer.

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