SAT Question 6ab30ce3: Right Triangles And Trigonometry answer and explanation with Desmos solution

Choice D is correct. Corresponding angles in similar triangles have equal measures. It's given that triangle $ABC$ is similar to triangle $DEF$, where $A$ corresponds to $D$, so the measure of angle $A$ is equal to the measure of angle $D$. Therefore, if $\tan \left(A\right)=\sqrt{3}$, then $\tan \left(D\right)=\sqrt{3}$. It's given that angles $C$ and $F$ are right angles, so triangles $ABC$ and $DEF$ are right triangles. The adjacent side of an acute angle in a right triangle is the side closest to the angle that is not the hypotenuse. It follows that the adjacent side of angle $D$ is side $DF$. The opposite side of an acute angle in a right triangle is the side across from the acute angle. It follows that the opposite side of angle $D$ is side $EF$. The tangent of an acute angle in a right triangle is the ratio of the length of the opposite side to the length of the adjacent side. Therefore, $\tan \left(D\right)=\frac{EF}{DF}$. If $DF=125$, the length of side $EF$ can be found by substituting $\sqrt{3}$ for $\tan \left(D\right)$ and $125$ for $DF$ in the equation $\tan \left(D\right)=\frac{EF}{DF}$, which yields $\sqrt{3}=\frac{EF}{125}$. Multiplying both sides of this equation by $125$ yields $125\sqrt{3}=EF$. Since the length of side $EF$ is $\sqrt{3}$ times the length of side $DF$, it follows that triangle $DEF$ is a special right triangle with angle measures $30^{\circ }$, $60^{\circ }$, and $90^{\circ }$. Therefore, the length of the hypotenuse, $\overline{DE}$, is $2$ times the length of side $DF$, or $DE=2\left(DF\right)$. Substituting $125$ for $DF$ in this equation yields $DE=2\left(125\right)$, or $DE=250$. Thus, if $\tan \left(A\right)=\sqrt{3}$ and $DF=125$, the length of $\overline{DE}$ is $250$.