source: #alcumus tags: #special-right-triangles #area course: [[Geometry]] status: #solved ### Problem In the diagram, square $A B C D$ has sides of length 4, and $\triangle A B E$ is equilateral. Line segments $B E$ and $A C$ intersect at $P$. Point $Q$ is on $B C$ so that $P Q$ is perpendicular to $B C$ and $P Q=x$. ![[alc-eq-tr-in-sq.svg]] Find the area of $\triangle A P E$ in simplest radical form. #### Solution $\triangle APE=\triangle ABE - \triangle APB$ ##### To find APB Notice that $\overline{PQ}$ is emphasized, even though its length is not given. Labeling a few more lengths in terms of $x$ will show that an altitude of $\triangle APB$ drawn from $P$ to $\overline{AB}$ has length $4-x$ since it is congruent to $\overline{BQ}$, which can be found with the [[◊ Segment Addition Postulate]]. $\overline{QC}=x$ since $PQC$ is an [[Isosceles Right Triangle]] Simultaneously, labeling the angles based on knowledge of the interior angles of the [[Equilateral Triangle]] and [[Isosceles Right Triangle]], as well as some [[Triangle Sum Theorem]] ![[alc-eq-tr-in-sq-1.svg]] Because $\triangle PQB$ is a [[Special Right Triangles#30-60-90]], the ratio of the legs is known. $x$ can be solved for, and the denominator should be [[‡ Rationalized Denominator|rationalized]] to make writing $4-x$ simple. $\begin{aligned} \frac{4-x}{x}&=\sqrt{3}\\ {4-x}&=x\sqrt{3} \\ 4 &= x\left(1+\sqrt{3}\right)\\ \frac{4}{1+\sqrt{3}}&=x\\ {2\sqrt{3}-2}&=x\\ 6-2\sqrt{3}&=4-x \end{aligned}$ the area of $\triangle APB$ is $\frac{4\cdot\left(6-2\sqrt{3}\right)}{2}=12-4\sqrt{3}$ ##### To find ABE [[Area of an equilateral triangle]] Area of $\triangle ABE$ is $4\sqrt{3}$ ##### Subtracting $\begin{aligned} \triangle APE&=&\triangle ABE &&-&& \triangle APB \\ \triangle APE&=& 4\sqrt{3} &&-&& \left(12-4\sqrt{3}\right)\\ \triangle APE&=&& &8\sqrt{3}-&12 \end{aligned}$ $\triangle APE=\boxed{8\sqrt{3}-12}$