QJEEM24Ma7e6e: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7e6e Solution: A circle is inscribed in an equilateral triangle of side of…

Inradius of $\triangle A B C=r=\frac{\Delta}{s}=\frac{\frac{\sqrt{3}}{4} \times(12)^2}{18}$

$r=2 \sqrt{3}$

Side length of square is $a$ , then $a^2=2 r^2$

$\Rightarrow a^2=24$

Area of square, $m=24$

Perimeter of square, $n=4 \sqrt{24}$

$\begin{aligned}& \Rightarrow m+n^2=24+384 \\& =408\end{aligned}$