Geometry Right triangles in a rectangle, find the area. 
Please help, I've been struggling to figure out this problem for too long... 
Given the area of rectangle $ABCD = 1200 \text{ unit}^2$, find the area of right triangle $ABE$

We have $[ABCD]=1200$, therefore the area of $\Delta{ABD}=\dfrac{1}{2}[ABCD]=600$. Now, calculate length $AD$ and $BD$.
$$
\begin{align}
[ABD]&=600\\
\dfrac{1}{2}AB\cdot AD&=600\\
\dfrac{1}{2}\cdot40\cdot AD&=600\\
20\cdot AD&=600\\
AD&=30
\end{align}
$$
Using Phytagoras' formula, we get
$$
BD^2=AB^2+AD^2\quad\Rightarrow\quad BD=\sqrt{40^2+30^2}=50.
$$
Now, calculate length $AE$.
$$
\begin{align}
[ABD]&=600\\
\dfrac{1}{2}BD\cdot AE&=600\\
\dfrac{1}{2}\cdot50\cdot AE&=600\\
25\cdot AE&=600\\
AE&=24.
\end{align}
$$
Again we use Phytagoras' formula to obtain $BE$.
$$
AB^2=AE^2+BE^2\quad\Rightarrow\quad BE=\sqrt{AB^2-BE^2}=32.
$$
Thus, the area of $\Delta{ABE}$ is
$$
\begin{align}
[ABE]&=\dfrac{1}{2}AE\cdot BE\\
&=\dfrac{1}{2}\cdot24\cdot 32\\
&=384.
\end{align}
$$
