What's the area of the region $F$ in this diagram? I have exams coming up. So I would appreciate your help a lot. 
The region named "$F$" is defined in the diagram below. With Pythagoras, I found out that $AB$ is 8 cm and $IG$ is 5cm. I also thought about splitting $F$ into a triangle and a rectangle. Can anyone please help me to solve this task?

 A: It looks as if you want the area of the shaded region. The whole big triangle has area $(1/2)(6)(6)$.
Now find the areas of the unshaded bits, and subtract from the whole. 
The area of the top unshaded triangle is very easy. Now let's work on the area of the unshaded triangle on the left. It is an isosceles right-angled triangle with hypotenuse $4$. The nice way to find its area is to observe that $4$ copies of it can be put together to make a $4\times 4$ square. Or else you can find the legs using the Pythagorean theorem. 
A: Hint: The area of the region $F$ is the area of the whole triangle $\triangle ABC$, minus the areas of the three non-shaded triangles.
A: Call $\,M,N\,$ resp. the two points where those perpendiculars from $\,I,G\,$ intersect $\,AB\,$ .Since all the triangles in the diagram are straight-angled and isosceles (why?) , we have that 
$$MI=AM=\frac{AI}{\sqrt 2}=2\sqrt 2\;\;,\;\;NG=NB=\frac{GN}{\sqrt 2}=\sqrt 2$$
Thus, the basis of the trapezoid $\,MNIG\,$ are $\,MI, NG\,$ , and its height is 
$$h=AB-AM-NB=6\sqrt 2-2\sqrt 2-\sqrt 2=3\sqrt 2$$ 
so
$$F=A_{MNIG}=\frac{1}{2}\left(2\sqrt 2+\sqrt 2\right)3\sqrt 2=\frac{18}{2}=9$$
