JEE Advanced 2016 · Paper 1 · Mathematics Q45

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MathematicsMultiple correct+4 / -2Question 45 of 54

In a triangle $XYZ$ , let $x,y,z$ be the lengths of sides opposite to the angles $X,Y,Z$ , respectively, and $2s=x+y+z$ . If $\dfrac{s-x}{4}=\dfrac{s-y}{3}=\dfrac{s-z}{2}$ and area of incircle of the triangle $XYZ$ is $\dfrac{8\pi}{3}$ , then

area of the triangle

XYZ

6\sqrt6

the radius of circumcircle of the triangle

XYZ

\dfrac{35}{6}\sqrt6

\sin\dfrac X2\sin\dfrac Y2\sin\dfrac Z2=\dfrac4{35}

\sin^2\left(\dfrac{X+Y}{2}\right)=\dfrac35

Answer and worked solution

Official answer: A, C, D

Sides. Let $s-x=4k$ , $s-y=3k$ , $s-z=2k$ . Adding gives $s=9k$ , so $x=5k$ , $y=6k$ , $z=7k$ .

Area and inradius.
$\Delta=\sqrt{9k\cdot4k\cdot3k\cdot2k}=6\sqrt6k^2,\qquad r=\frac\Delta s=\frac{2\sqrt6}3k.$
From $\pi r^2=\frac{8\pi}3$ : $\frac{24}9k^2=\frac83$ , so $k=1$ .

Check each option.

(A) $\Delta=6\sqrt6$ . True.
(B) $R=\dfrac{xyz}{4\Delta}=\dfrac{210}{24\sqrt6}=\dfrac{35\sqrt6}{24}$ . False.
(C) $\sin\frac X2\sin\frac Y2\sin\frac Z2=\dfrac r{4R}=\dfrac{2\sqrt6/3}{35/\sqrt6}=\dfrac4{35}$ . True.
(D) $\sin^2\frac{X+Y}2=\cos^2\frac Z2=\dfrac{s(s-z)}{xy}=\dfrac{18}{30}=\dfrac35$ . True.

(A), (C), (D)

Topics:Properties of triangles

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JEE Advanced 2016 · Paper 1, Mathematics question 45