In the diagram, $\triangle A B C$ is isosceles with $A B=A C$ and $B C=30 \mathrm{~cm}$. Square $E F G H$, which has a side length of $12 \mathrm{~cm}$, is inscribed in $\triangle A B C$, as shown. The area of $\triangle A E F$, in $\mathrm{cm}^{2}$, is
(A) 27
(D) 48
(B) 54
(C) 51
(E) 60

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-124.jpg?height=222&width=309&top_left_y=1553&top_left_x=1336)
Let $M$ be the midpoint of $E F$ and $N$ be the midpoint of $H G$. By symmetry, $N$ is also the midpoint of $B C$. Also, the line through $A$ and $M$ will also pass through $N$, and will be perpendicular to both $E F$ and $B C$.

Since the side length of the square is 12 , then

$E M=H N=6$ and $E H=12$.

Since we are told that $B C=30$, then $B N=15$ and so $B H=9$.

Since $E F G H$ is a square, then $E F$ is parallel to $H G$, and so $\angle A E M=\angle E B H$, ie. $\triangle A M E$ is similar to $\triangle E H B$.

Therefore, $\frac{A M}{6}=\frac{12}{9}$ or $A M=8$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-225.jpg?height=317&width=398&top_left_y=2067&top_left_x=1362)

Thus, the area of $\triangle A E F$ is $\frac{1}{2}(12)(8)=48 \mathrm{~cm}^{2}$.

ANSWER: (D)