Finding area problem

Question

There is this simple geometry question that seems so easy but I think the question lacks some information (does it?). Or maybe there are other ways to solve the problem.

So the problem says, there are two squares $ABCD$ and $FCHG$ with a side of length $8$ and $10$, respectively. We are asked to find the area of the shaded region.

What makes this question a bit tricky is because the small triangle $DEF$ is not shaded. And if we are given the length of AD, then the question will be easy. Is there a way to find AD or is there any other methods to solve the problem?

By the way, the answer is $48.4$.

And also, I am pretty sure we don't need to use advance methods such as trigs, etc.

I really appreciate any helps!

Answer 1

Triangles $BCF$ and $DEF$ are similar triangles, which means that their sides are proportional. You know side $BC=8,$ $CF=10$ and side $EF=2$. Can you proceed?

Answer 2

Hint: Look for similar triangles. What is $\Delta DEF$ similar to?

Answer 3

Let say lines $BG$ and $FC$ intersects at $K$,

So $CK+KF=10$ and $FKG$ and $BCK$ are similar triangles.

So $\frac{KF}{CK}=10/8$ therefore $CK=40/9$ and $KF=50/9$

The area of FKG triangle = $0.5*10*50/9=27.7$

Also $AD+DE=8$ and $ADB$ and $DEF$ are similar triangles. Similarly it gives

$AD=32/5$ and $DE=8/5$

The area of $BDEK$ = area of $BFC$- area of $DEF$ - area of $BCK$ = $0.5*8*10-0.5*2*8/5-0.5*8*40/9$= 20.7

Total area = The area of $BDEK$ + The area of FKG triangle =$27.7+20.7=48.4$