Like the title suggests, the figure below is two overlapping right triangles. The overlapped area forms a quadrilateral and the goal is to find the area of that region given some lengths:
Here, $AB=2, BC=6, CD=8$ and $DE=4$.
This problem is essentially the same problem posted by Professor Michael Penn on YouTube in a video that asks "can you solve it in a different way?" And that's essentially why I'm sharing it here. I want to see several different approaches to this problem. Michael Penn attempted it using a coordinate system and setting up an integral, I will share my geometric approach as an answer below. Please share your own approaches!
Here's my approach to this puzzle:
1.) Draw $CO$ from $C$ to $O$ such that it bifurcates quadrilateral $BODC$ into $\triangle COD$ and $\triangle COB$. As done in an earlier post, we are going to use the lemma that the ratio of area of two triangles is the same as the ratio of their base lengths if they share a common height. This means that $A(\triangle COD)=2A(\triangle ODE)$ and $A(\triangle COB)=3A(\triangle OBA)$. We can label them as:
$$A(\triangle COD)=2a$$ $$A(\triangle ODE)=a$$ $$A(\triangle COB)=3b$$ $$A(\triangle OBA)=b$$
2.) We know that the area of $\triangle BEC=3a+3b$. We can also calculate the area of $\triangle BEC$, which you comes out to be $36$. Thus:
$$3a+3b=36$$ $$a+b=12$$
For $\triangle ACD$, we can see that its area is $2a+4b$, we can also see that its area comes out to $32$, thus:
$$2a+4b=32$$ $$a+2b=16$$
We now have a system of equations with two variables:
$$a+2b=16$$
$$a+b=12$$
Subtracting the second one from the first one, we get $b=4$, and thus $a=8$
The area of quadrilateral $BODC =3(4)+2(8)=12+16=28$