
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 $ABE$

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 $ABE$
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As an alternative way to do this:
You can use similarity of the triangles $ BCD$ and $AEB$ which we can show are similar since all their angles are the same. $$\angle AEC \equiv \angle BCD=90^\circ \;\;\text{ and }\;\; \angle EAB \equiv \angle DBC \;\; \text{ and } \;\; \angle ABE \equiv \angle BDC$$ Thus all we need is to find the ratio of two matching sides of the triangles to be able to work out the others.
$$1200=BC \times AB=BC\times 40\;\; \therefore \;\; BC=\frac{1200}{40}=30$$
$BC=30$ and $AB=40$ therefore from the Pythagorean theorem we know that: $$BD=\sqrt{BC^2+AB^2}=\sqrt{30^2+40^2}=50$$ Side $BD$ is similar to $AB$ therefore allowing us to calculate the ratio between the two triangle sides as: $$ \frac{AB}{BD}=\frac{40}{50}=\frac{4}{5} $$ thus now we can find the length of $AE$ simply by multiplying $BC$ (the side of the smaller triangle that corresponded with $AE$) by $\frac{4}{5}$.
$$AE=\frac{4}{5}\times 30=24 \;\;\; \text{ and }\;\;\; EB=\frac{4}{5}\times DC=\frac{4}{5}\times40=32$$ Thus allowing you find the area of the triangle: $$\text{Area}=\frac{1}{2}\times 32\times 24=384$$
Indicdently you could also use the fact that the areas of two similar triangles are proportional to the square of the ratio of their sides. That is, we know the ratio of the sides is $\frac{4}{5}$ thus the square of this ratio is $ \frac{16}{25}$ so all we need to do is multiply the area of the larger triangle (which we already know is 600) by $\frac{16}{25}$ to gives the same answer: $$ \text{Area}=\frac{16}{25}\times 600=384 $$
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 $$ \begin{align} [ABE]&=\dfrac{1}{2}AE\cdot BE\\ &=\dfrac{1}{2}\cdot24\cdot 32\\ &=384. \end{align} $$