issues with geometry triangle $4$ line drawn parallel to base of triangle such that they are equidistant.if the area of the most bottom part is 4 sq cm. find area of triangle?
MY THOUGHTS :
being weak in geometry i couldn't make a single step forward.i am giving what i could analyzed after drawing the diagram

with out these two analysis i couldnt make a single step forward though i gave a lot time behind it
please help.thanks in advance
 A: Hint: The area of the large triangle is equal to the sum of area of the triangle you get by removing the bottom trapezium and the area of the bottom trapezium.
Hint: The height and base of the smaller triangle which you get by removing the bottom trapezium is exactly four fifths the height and base of the full triangle (show this).
A: Hint: The area of the entire triangle is the area of the second largest triangle plus $4$ on the bottom. Since the lines are parallel all the triangles are similar to each other,  all the heights and bases are similar to each other. If we call the area of the entire triangle $a$, the area of the second largest triangle is $\frac{4}{5}  \cdot \frac{4}{5}a$, since the base and height are both $\frac{4}{5}$ of the entire triangle. 
$a - \frac{16a}{25} = 4$
A: Such triangles can be broken up into equal pieces. Counting those pieces will tell you what the areas you need are.

If the bottom part has an area of $4$, then $7A=4$. So $A = \dfrac 47$
So the area of the triangle is $16A = \dfrac{64}{7} = 9\frac 17$
