In the diagram, $A B C D$ is a square with side length $8 \mathrm{~cm}$. Point $E$ is on $A B$ and point $F$ is on $D C$ so that $\triangle A E F$ is right-angled at $E$. If the area of $\triangle A E F$ is $30 \%$ of the area of $A B C D$, what is the length of $A E$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_ff0a25f37b4b6ded9838g-2.jpg?height=366&width=333&top_left_y=1725&top_left_x=1362)
Solution 1

Since $\triangle A E F$ is right-angled at $E$, its area equals $\frac{1}{2}(A E)(E F)$.

Since $\triangle A E F$ is right-angled at $E$ and $A B C D$ is a square, then $E F=A D=8 \mathrm{~cm}$.

We are told that the area of $\triangle A E F$ is $30 \%$ of the area of square $A B C D$, and so the area of $\triangle A E F$ equals $0.3(8 \mathrm{~cm})^{2}$.

Therefore, $\frac{1}{2}(A E)(8 \mathrm{~cm})=0.3(8 \mathrm{~cm})^{2}$ and so $A E=2(0.3)(8 \mathrm{~cm})=0.6(8 \mathrm{~cm})=4.8 \mathrm{~cm}$.

## Solution 2

Since $\triangle A E F$ is right-angled at $E$ and $A B C D$ is a square, then $E F$ is parallel to $A D$ which makes $A E F D$ a rectangle.

Since $A F$ is a diagonal of rectangle $A E F D$, then the area of rectangle $A E F D$ is twice the area of $\triangle A E F$, or $60 \%$ of the area of the square.

Since rectangle $A E F D$ and square $A B C D$ share the same height, then it must be that $A E$ equals $60 \%$ of the length of $A B$.

Therefore, $A E=0.6 \cdot A B=0.6 \cdot 8 \mathrm{~cm}=4.8 \mathrm{~cm}$.

ANSWER: $4.8 \mathrm{~cm}$