SAT Mathematics Practice Test 2

The regular hexagon can be divided into 6 equilateral triangles of side length a by drawing the six segments from the center of the regular hexagon to each of its 6 vertices. Since the area of the hexagon is $384\sqrt{3}$ square inches, the area of each equilateral triangle will be $\frac{384\sqrt{3}}{6}=64\sqrt{3}$ square inches.Drawing any altitude of an equilateral triangle divides it into two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. If the side length of the equilateral triangle is a, then the hypotenuse of each $30^{\circ}-60^{\circ}-90^{\circ}$ triangle is a, and the altitude of the equilateral triangle will be the side opposite the 60° angle in each of the $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. Thus, the altitude of the equilateral triangle is $\frac{\sqrt{3}}{2}a$ and the area of the equilateral triangle is $\left(\frac{1}{2}\right)(a)\left(\frac{\sqrt{3}}{2}a\right)=\frac{\sqrt{3}}{4}a^{2}$. Since the area of each equilateral triangle is $64\sqrt{3}$ square inches, it follows that $a^{2}=\frac{4}{\sqrt{3}}(64\sqrt{3})=256$ square inches. And since the area of the square with side length $a$ is $a^{2}$ , it follows that the square has area 256 square inches.Choices B, C, and D are incorrect and may result from calculation or conceptual errors.