Area of the grey region in the following triangle

Question

$ABC$ is a triangle having area of $240$.

Given that $BD:DF:FC=3:4:3$; $BE:EA=5:7$ ; $CG:GA=2:7$ then what is the area of the grey region?

I am not getting any way to start this question, any hint on the theorem will be helpful.

Answer 1

$$\frac{S_\Delta{BED}}{S_\Delta ABC}=\frac{5\cdot3}{12\cdot10}=\frac{1}{8};$$

$$\frac{S_\Delta{GFC}}{S_\Delta ABC}=\frac{2\cdot3}{9\cdot10}=\frac{1}{15}$$ Thus, $$S_{AEDFG}=240\left(1-\frac{1}{8}-\frac{1}{15}\right)=194.$$

Answer 2

Let H be the point of (BC) s.t. (AH) // (ED)

You can find the ratio of BD/BH, hence the area of BAH (which has the same height from A than ABC).

Then you can calculate the area of BED which is a triangle similar to BAH.

You can find the area of GFC in the same way.

The answer is Area ABC - Area BED - Area GFC.