JEE Advanced 2018 · Paper 1 · Mathematics Q2

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MathematicsMultiple correct+4 / -2Question 38 of 54

In a triangle $PQR$ , let $\angle PQR = 30^\circ$ and the sides $PQ$ and $QR$ have lengths $10\sqrt{3}$ and $10$ , respectively. Then, which of the following statement(s) is (are) TRUE?

\angle QPR = 45^\circ

The area of the triangle

PQR

25\sqrt{3}

and

\angle QRP = 120^\circ

The radius of the incircle of the triangle

PQR

10\sqrt{3} - 15

The area of the circumcircle of the triangle

PQR

100\pi

Answer and worked solution

Official answer: B, C, D

Third side. By the cosine rule,
$PR^2=300+100-2(10\sqrt3)(10)\cos30^\circ=400-300=100\implies PR=10.$
So $PR=QR=10$ , and the triangle is isosceles with $\angle QPR=\angle PQR=30^\circ$ and $\angle QRP=120^\circ$ .

(A) $\angle QPR=30^\circ$ , not $45^\circ$ . False.
(B) Area $=\tfrac12(10\sqrt3)(10)\sin30^\circ=25\sqrt3$ , and $\angle QRP=120^\circ$ . True.
(C) $s=\dfrac{10+10+10\sqrt3}2=10+5\sqrt3$ , so
$r=\frac\Delta s=\frac{25\sqrt3}{10+5\sqrt3}=\frac{5\sqrt3}{2+\sqrt3}=5\sqrt3(2-\sqrt3)=10\sqrt3-15.$
True.
(D) $R=\dfrac{PR}{2\sin Q}=\dfrac{10}{2\cdot\frac12}=10$ , so the circumcircle's area is $100\pi$ . True.

(B), (C), (D)

Topics:Properties of triangles

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