How to solve this geometry assigment?

Question

The three elements all have the same area (64cm2). How can I calculate the h? The result is 4cm, but I don't understand how to calculate it.

EDIT: Guys, thanks a lot for the quick help! Trust me I've spent quite some time trying to solve it but I didn't realize that I can use the triangle area formula in reverse to calculate the height of the triangle (16cm). That was most helpful.

Answer 1

Assuming the rectangle is a square: you know the side, so you know the common area $A$. Of the triangle, you know the area and one side, so you also know the height perpendicular to it. Which turns out to be the difference between the two basis of the trapezium. Finally, you have a trapezium of which you know the smaller basis, the difference between the larger basis and the smaller basis, and the area: therefore you know its height.

Answer 2

HINT

Try and figure out the value of ? based on the top right triangle.

Answer 3

I think what you wanted to say ( but did n't) is like:

The top two (square,obtuse triangle) have same area = 64.

Bottom two (half square , right triangle) have same area =32.

Use the fact that area between parallels of same base have same area. And, area does not change if height is doubled when base is halved.

Under this assumption $h=4$ as marked.

Answer 4

Alternatively, labeling the larger base of trapezium by $b$, make up a system of two equations for areas of small and large trapeziums: $$\begin{cases} \frac{8+b}{2}\cdot h=64 \\ \frac{8+b}{2}\cdot (8+h)=3\cdot 64 \end{cases} \stackrel{(2):(1)}{\Rightarrow} \frac{8+h}{h}=3 \Rightarrow h=4.$$