Question

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 $\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}$ and area of incircle of the triangle $XYZ$ is$\frac{8\pi }{3}$, then

• A. The radius of circumcircle of the triangle $XYZ$ is $\frac{35}{6}\sqrt{6}$
• B. ${\sin }^2\left(\frac{X+Y}{2}\right)=\frac{3}{5}$
• C. $\sin \frac{X}{2}\sin \frac{Y}{2}\sin \frac{Z}{2}=\frac{4}{35}$
• D. Area of the triangle $XYZ$ is $6\sqrt{6}$

Answer 1

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

Area of the triangle $XYZ$ is $6\sqrt{6}$


$\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=k$
$s–x=4k$ ... (i)
$s–y=3k$ ... (ii)
$s–z=2k$ ... (iii)
$3s–(x+y+z)=9k$
$3s–2s=9k$
$s=9k$
From (i), (ii), (iii)
$x=5k,y=6k,z=7k$
Area of incircle = $\pi r^2=\frac{8\pi }{3}$
$r^2=\frac{8}{3}$
$\Rightarrow \frac{{\Delta }^2}{s^2}=\frac{8}{3}$
$\Rightarrow \frac{s(s-x)(s-y)(s-z)}{s^2}=\frac{8}{3}$
$\frac{\left(9k\right)\left(4k\right)\left(3k\right)\left(2k\right)}{81k^2}=\frac{8}{3}$
$k^2=1$
$\Rightarrow k=\pm 1$
Now side length, $x=5,y=6,z=7$ and $s=9$
(1) Area of triangle $XYZ=\sqrt{s(s-x)(s-y)(s-z)}$
$=\sqrt{9\cdot 4\cdot 3\cdot 2}$
$=3\cdot 2\sqrt{6}$
$=6\sqrt{6}$
(2) Radius of circumcircle of $\Delta XYZ$
$R=\frac{xyz}{4\Delta }=\frac{5\cdot 6\cdot 7}{4\cdot 6\sqrt{6}}=\frac{35}{4\sqrt{6}}$
(3) $\sin \frac{X}{2}\cdot \sin \frac{Y}{2}\cdot \sin \frac{Z}{2}=\sqrt{\frac{(s-x)(s-z)}{xz}\frac{(s-y)(s-z)}{yz}\frac{(s-x)(s-y)}{xy}}$
$=\frac{(s-x)(s-y)(s-z)}{xyz}$
$=\frac{4\cdot 3\cdot 2}{5\cdot 6\cdot 7}$
$=\frac{4}{35}$
(4) ${\sin }^2\left(\frac{X+Y}{2}\right)={\cos }^2\frac{Z}{2}=\frac{s(s-z)}{xy}=\frac{9\cdot 2}{5\cdot 6}=\frac{3}{5}$.