# Finding area problem

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!

## 4 Answers

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?

• yes I can! thanks! – user71346 Nov 11 '13 at 10:50

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

• ah yes! so careless... thanks! – user71346 Nov 11 '13 at 10:49

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 =$27.7+20.7=48.4$

"Is there a way to find AD or is there any other methods to solve the problem?"

AD is proportioan to AE as BD is proportional to BF as EC is proportional to FC.

So AD ~ 8 as 8 ~ 10. Or AD/8 = 8/10.