JEE Main 2024 (Online) 6th April Morning ShiftQuestion 1
A circle is inscribed in an equilateral triangle of side length $12$. If the area and perimeter of any square inscribed in this circle are $m$ and $n$, respectively, then $m+n^{2}$ is equal to
Circle
Solution
Inradius $=\frac{12}{2\sqrt3}=2\sqrt3$. Inscribed square has diagonal $4\sqrt3$, side $2\sqrt6$. $m=24$, $n=8\sqrt6$, $n^2=384$. $m+n^2=408$. Option (A).