A circle is inscribed in an equilateral triangle of side of length 12.

Question

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

Accepted Answer

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, $$n=4 \sqrt{24}$$

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

Solutions · 1