SAT Question b5d62bba: Lines, Angles, and Triangles answer and explanation with Desmos solution

Choice D is correct. The figure shows that $\angle MRL$ and $\angle PRQ$ are vertical angles. Since vertical angles are congruent, $\angle MRL$ and $\angle PRQ$ are congruent. It’s given that $\overline{LM}$ is parallel to $\overline{PQ}$. The figure also shows that $\overline{LQ}$ intersects $\overline{LM}$ and $\overline{PQ}$. If two parallel segments are intersected by a third segment, alternate interior angles are congruent. Thus, $\angle MLR$ and $\angle PQR$ are congruent. Since $△LMR$ and $△PQR$ have two pairs of congruent angles, the triangles are similar, where $\overline{MR}$ in $△LMR$ corresponds to $\overline{PR}$ in $△PQR$. Since the areas of similar triangles are proportional to the squares of the lengths of corresponding sides, it follows that $\frac{\text{area of}△PQR}{\text{area of}△LMR}=(\frac{RP}{MR}{)}^2$. It’s given that the lengths of $\overline{MR}$ and $\overline{RP}$ are $8$ and $14$ units, respectively, and the area of $△LMR$ is $36$ square units. Substituting $14$ for $RP$, $8$ for $MR$, and $36$ for $\text{area of}△LMR$ in the equation $\frac{\text{area of}△PQR}{\text{area of}△LMR}=(\frac{RP}{MR}{)}^2$ yields $\frac{\text{area of}△PQR}{36}=(\frac{14}{8}{)}^2$, which is equivalent to $\frac{\text{area of}△PQR}{36}=(\frac{7}{4}{)}^2$, or $\frac{\text{area of}△PQR}{36}=\frac{49}{16}$. Multiplying each side of this equation by $36$ yields $\text{area of}△PQR=36(\frac{49}{16})$, or $\text{area of}△PQR=\frac{441}{4}$. Therefore, the area of $△PQR$ is $\frac{441}{4}$ square units.