JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 1)
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
408
414
312
396
Explanation
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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, $$n=4 \sqrt{24}$$
$$\begin{aligned} & \Rightarrow m+n^2=24+384 \\ & =408 \end{aligned}$$
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