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A circle is inscribed in an equilateral triangle of side of 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
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408
$\because \mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{\sqrt{3} \mathrm{a}^2}{4 \cdot \frac{3 \mathrm{a}}{2}}=\frac{\mathrm{a}}{2 \sqrt{3}}=\frac{12}{2 \sqrt{3}}=2 \sqrt{3}$
$\begin{aligned} & \therefore A=\mathrm{r} \sqrt{2}=2 \sqrt{6} \\ & \text { Area }=\mathrm{m}=\mathrm{A}^2=24 \\ & \text { Perimeter }=\mathrm{n}=4 \mathrm{~A}=8 \sqrt{6} \\ & \therefore \mathrm{m}+\mathrm{n}^2=24+384 \\ & =408\end{aligned}$
$\begin{aligned} & \therefore A=\mathrm{r} \sqrt{2}=2 \sqrt{6} \\ & \text { Area }=\mathrm{m}=\mathrm{A}^2=24 \\ & \text { Perimeter }=\mathrm{n}=4 \mathrm{~A}=8 \sqrt{6} \\ & \therefore \mathrm{m}+\mathrm{n}^2=24+384 \\ & =408\end{aligned}$
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