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:

• A. 396
• B. 408
• C. 312
• D. 414

Answer 1

Answer: (B)

408

Answer 2

The radius $r$ of the circle inscribed in an equilateral triangle is given by:

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

The side of the square inscribed in this circle is:

$\lambda = r\sqrt{2} = 2\sqrt{3} \cdot \sqrt{2} = 2\sqrt{6}.$

Area of the square:

$m = \lambda^2 = (2\sqrt{6})^2 = 24.$

Perimeter of the square:

$n = 4\lambda = 4(2\sqrt{6}) = 8\sqrt{6}.$

$m + n^2 = 24 + (8\sqrt{6})^2 = 24 + 384 = 408.$