Answer:
Option A,C,D
Explanation:
Given a $\triangle XYZ$ ,where 2x=x+y+z
and $\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}$
$\therefore$ $\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}$
= $\frac{3x-(x+y+z)}{4+3+2}=\frac{s}{9}$
or $\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=\frac{s}{9}=\lambda(let)$
$\Rightarrow$ $s=9\lambda,s=4\lambda+x, s=3\lambda+y$
and $s=2\lambda+z$
$\therefore$ $s=9\lambda,x=5\lambda, y=6\lambda, z=7\lambda$
Now, $\triangle= \sqrt{s(s-x)(s-y)(s-z)}$ [heron's formula]
$= \sqrt{9\lambda.4\lambda.3\lambda.2\lambda}=6\sqrt{6}\lambda^{2}$ .........(i)
Also, $\pi r^{2}=\frac{8\pi}{3}$
$\Rightarrow$ $ r^{3}=\frac{8}{3}$ .......(ii)
and $R=\frac{xyz}{4\triangle}$
$=\frac{(5\lambda)(6\lambda)(7\lambda)}{4.6.\sqrt{6}\lambda^{2}}=\frac{35\lambda}{4\sqrt{6}}$ .......(iii)
Now, $r^{2}=\frac{8}{3}=\frac{\triangle^{2}}{S^{2}}=\frac{216\lambda^{2}}{81\lambda^{2}}$
$\Rightarrow$ $\frac{8}{3}=\frac{8}{3}\lambda^{2}$ [from Eq. (ii)]
$\Rightarrow$ $\lambda=1$
(a) $\triangle XYZ=6\sqrt{6}\lambda^{2}=6\sqrt{6}$
$\therefore$ Option (a) is correct.
(b) Radius of circumcircle
$(R)=\frac{35}{4\sqrt{6}}\lambda=\frac{35}{4\sqrt{6}}$
$\therefore$ Option (b) is incorrect.
(c) Since, $r= 4R\sin\frac{X}{2}.\sin\frac{Y}{2}.\sin\frac{Z}{2}$
$\Rightarrow\frac{2\sqrt{2}}{\sqrt{3}}=4.\frac{35}{4\sqrt{6}}\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2}$
$\Rightarrow\frac{4}{35}=\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2}$
$\therefore$ Option (c) is correct.
(d) $\sin^{2}\left(\frac{X+Y}{2}\right)=\cos^{2}\left(\frac{Z}{2}\right)$ . as
$\frac{X+Y}{2}=90^{0}-\frac{Z}{2}=\frac{s(s-z)}{xy}=\frac{9\times 2}{5\times 6}=\frac{3}{5}$
$\therefore$ Option (d) is correct
