Question

A $\Delta ABC$ with sides $a,b,c$ opposite to the angles $A,B,C$ satisfies the inequality $\frac{a^2+b^2+c^2}{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}\le 9$.
Given $a\ge b\ge c$, and $a$ attends its maximum possible value, then

• A. Maximum area of the triangle is $\frac{9}{4}$
• B. Maximum area of the triangle is $\frac{3}{4}$
• C. Circumradius of the triangle is $3$
• D. Inradius of the triangle when its area is maximum is $\frac{3}{2}(\sqrt{2}-1)$

Answer 1

Answer: (A), (D)

The correct options are
A Maximum area of the triangle is $\frac{9}{4}$
D Inradius of the triangle when its area is maximum is $\frac{3}{2}(\sqrt{2}-1)$

If $R$ be the circumradius of $\Delta ABC$, then
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\Rightarrow a=2R\sin A,b=2R\sin B,c=2R\sin C$

$\frac{a^2+b^2+c^2}{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}\le 9$

$\Rightarrow \frac{(2R\sin A{)}^2+(2R\sin B{)}^2+(2R\sin C{)}^2}{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}\le 9$

$\Rightarrow (2R{)}^2\le 9$
$\Rightarrow 2R\le 3$

Now, $a=2R\sin A$
$\Rightarrow a\le 3\sin A$
$\therefore a_{max}=3$ at $\sin A=1$
$\Rightarrow A={90}^{\circ }\Rightarrow B+C={90}^{\circ }$

Area of $\Delta ABC$ $=\frac{1}{2}bc\sin A=\frac{1}{2}\left(2R\sin B\right)\left(2R\sin C\right)=\frac{9}{4}\left(2\sin B\cos B\right)(\because B+C={90}^{\circ })=\frac{9}{4}\sin 2B$
Area is maximum when $\sin 2B=1$ and maximum area $=\frac{9}{4}$
$\sin 2B=1\Rightarrow B={45}^{\circ }\Rightarrow C={45}^{\circ }$

Since, $\Delta ABC$ is a right angled triangle,
$\therefore$ Circumradius, $R=\frac{3}{2}$

Inradius, $r=4R\sin \left(A/2\right).\sin \left(B/2\right).\sin \left(C/2\right)$
$=2\times 3\times \frac{1}{\sqrt{2}}\times {\sin }^2\left(B/2\right)(\because B=C)$
$=\frac{6}{\sqrt{2}}\times \left(\frac{1-\cos B}{2}\right)=\frac{3}{2}(\sqrt{2}-1)$