JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Properties of Triangle

1. (A)$\angle QPR=45^{\circ }$
2. (B)The area of the triangle $PQR$ is $25\sqrt{3}$ and $\angle QRP=120^{\circ }$✓ Correct
3. (C)The radius of the incircle of the triangle $PQR$ is $10\sqrt{3}-15$✓ Correct
4. (D)The area of the circumcircle of the triangle $PQR$ is $100\pi$✓ Correct

Answer: (B), (C), (D)

Solution

Using the cosine rule with $PQ=10\sqrt{3}$, $QR=10$, $\angle PQR=30^{\circ }$:

$P{R}^2=(10\sqrt{3}{)}^2+10^2-2(10\sqrt{3})(10)\cos 30^{\circ }=300+100-300=100$

So $PR=10$.

Area $\Delta =\frac{1}{2}(10\sqrt{3})(10)\sin 30^{\circ }=25\sqrt{3}$.

Using cosine rule at $R$: $\cos \angle QRP=\frac{100+100-300}{200}=-\frac{1}{2}$, so $\angle QRP=120^{\circ }$.

Then $\angle QPR=30^{\circ }$ (since angles sum to $180^{\circ }$). (A) is FALSE.

Inradius: $r=\frac{\Delta }{s}=\frac{25\sqrt{3}}{10+5\sqrt{3}}=\frac{5\sqrt{3}}{2+\sqrt{3}}=5\sqrt{3}(2-\sqrt{3})=10\sqrt{3}-15$. (C) is TRUE.

Circumradius: $2R=\frac{PR}{\sin 30^{\circ }}=\frac{10}{1/2}=20$, so $R=10$. Area $=\pi R^2=100\pi$. (D) is TRUE.

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · One or more correct options