Question 17·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

Let $s=BC$. The right angle is at $B$, and $BC$ is opposite the $30^\circ$ angle at $A$, so $BC$ is the short leg. The $30^\circ$-$60^\circ$-$90^\circ$ ratio is $s:s\sqrt3:2s$. So the long leg is $AB=s\sqrt3$, and the hypotenuse, across from the right angle, is $AC=2s$.

A perimeter adds the sides around the outside. Add $BC$, $AB$, and $AC$, and set that sum equal to the given perimeter:

Combine the three terms containing $s$ on the left:

Type $(3+\sqrt3)s_1\sim24+8\sqrt3$ in Desmos. The subscripted $s_1$ is the length Desmos solves for, and $\sim$ asks it to fit the two sides of the equation. Under PARAMETERS, Desmos shows $s_1=8$, so $BC=8$ centimeters.

The matching tick marks mean $AD$ and $BC$ are congruent, or equal in length. So the shaded triangle’s base is

$\triangle ADE$ has the same $30^\circ$ angle at $A$ and a right angle at $D$. Its $60^\circ$ angle is at $E$, across from $AD$. So $AD$ is now the long leg, and $DE$ is the short leg. Side roles come from the angles, not the tick marks. Use the ratio:

Divide by $\sqrt3$:

$AD$ and $DE$ meet at a right angle, so they are the base and perpendicular height. Type $\frac12(8)\frac8{\sqrt3}$ in Desmos; it shows about $18.475$. To match an exact radical choice, multiply the lengths:

Multiply the numerator and denominator by $\sqrt3$:

The shaded area is $\frac{32\sqrt3}{3}$ square centimeters. Choice D.