In the diagram, $\triangle A B C$ is a right-angled isosceles triangle. $D$ is the midpoint of $B C$ and $E$ is the midpoint of $A B$. If $A B=B C=24 \mathrm{~cm}$, what is the area of $\triangle A E D$ ?
(A) $48 \mathrm{~cm}^{2}$
(B) $36 \mathrm{~cm}^{2}$
(C) $72 \mathrm{~cm}^{2}$
(D) $9 \mathrm{~cm}^{2}$
(E) $54 \mathrm{~cm}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-003.jpg?height=309&width=314&top_left_y=2250&top_left_x=1339)
Solution 1

The area of $\triangle A E D$ is equal to one-half its base times its height.

Suppose the base of $\triangle A E D$ is $A E$, then its height is $B D$ ( $A E$ is perpendicular to $B D$ ).

Since $A B=B C=24 \mathrm{~cm}$ and $E$ and $D$ are the midpoints of their respective sides, then $A E=12 \mathrm{~cm}$ and $B D=12 \mathrm{~cm}$.

Thus, the area of $\triangle A E D$ is $\frac{1}{2} \times 12 \mathrm{~cm} \times 12 \mathrm{~cm}=72 \mathrm{~cm}^{2}$.

## Solution 2

The area of $\triangle A E D$ is equal to the area of $\triangle A B D$ minus the area of $\triangle E B D$.

Suppose the base of $\triangle E B D$ is $B D$, then its height is $E B$.

Since $A B=B C=24 \mathrm{~cm}$ and $E$ and $D$ are the midpoints of their respective sides, then $E B=12 \mathrm{~cm}$ and $B D=12 \mathrm{~cm}$.

Thus, the area of $\triangle E B D$ is $\frac{1}{2} \times 12 \mathrm{~cm} \times 12 \mathrm{~cm}=72 \mathrm{~cm}^{2}$.

The area of $\triangle A B D$ is equal to $\frac{1}{2} \times B D \times A B=\frac{1}{2} \times 12 \mathrm{~cm} \times 24 \mathrm{~cm}=144 \mathrm{~cm}^{2}$.

Thus, the area of $\triangle A E D$ is $144 \mathrm{~cm}^{2}-72 \mathrm{~cm}^{2}=72 \mathrm{~cm}^{2}$.

ANSWER: (C)