JEE Main 2024 — 6 April, Shift 1 — Question 15

$\because \mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{\sqrt{3} \mathrm{a}^{2}}{4 \cdot \frac{3 \mathrm{a}}{2}}=\frac{\mathrm{a}}{2 \sqrt{3}}=\frac{12}{2 \sqrt{3}}=2 \sqrt{3}$

$\therefore \mathbf{A}=\mathrm{r} \sqrt{2}=2 \sqrt{6}$

Area $=m=A^{2}=24$

Perimeter $=\mathrm{n}=4 \mathrm{~A}=8 \sqrt{6}$

$\therefore \mathrm{m}+\mathrm{n}^{2}=24+384$

$=408$