A circle is inscribed in an equilateral trian | vedclass

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(B) The radius $r$ of the incircle of an equilateral triangle with side $a$ is given by $r = \frac{a}{2\sqrt{3}}$.Given $a = 12$,we have $r = \frac{12

Vedclass · Accepted Answer

$408$

Vedclass · Answer

(B) The radius $r$ of the incircle of an equilateral triangle with side $a$ is given by $r = \frac{a}{2\sqrt{3}}$.Given $a = 12$,we have $r = \frac{12}{2\sqrt{3}} = 2\sqrt{3}$.Let the side of the square inscribed in the circle be $A$. The diagonal of the square is equal to the diameter of the circle,so $\sqrt{2}A = 2r$.$\sqrt{2}A = 2(2\sqrt{3}) = 4\sqrt{3}$.$A = \frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$.Area $m = A^2 = (2\sqrt{6})^2 = 4 	imes 6 = 24$.Perimeter $n = 4A = 4(2\sqrt{6}) = 8\sqrt{6}$.We need to find $m + n^2$.$m + n^2 = 24 + (8\sqrt{6})^2 = 24 + 64 	imes 6 = 24 + 384 = 408$.

$A$ circle is inscribed in an equilateral triangle of side 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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