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I think it's supposed to be a trapezium/trapezoid. The reference claims that the area is 92. It looks to be right since I guess

Area ABCE = 8*7=56

And

Area CDE = 8*9/2 = 36

However, if we have the height as 8cm (given?), and the two bases as 7cm (given?) and 16cm = 7cm (given?) + 9cm, then by Pythagorean theorem,

$$8^2+\text{mCF}^2 = 9^2$$

And hence

$$9 = 2\text{mCF} = 2\sqrt{17}$$

?

I guess I'm assuming that DF is a bisector of CE. Is it not? Actually I think CDE is equilateral. Is it not? I lost my notes, but I think I was able to show those two.

If I'm wrong please explain why.

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    Nope, $DF$ isn't a bissector (it is the height of the triangle, you partly hid the right angle) and $CDE$ isn't equilateral. – 2017-02-05

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Yes, the shape is possible. First consider the parallelogram $ABCD$, with sides $7$ and $9$. It still has a degree of freedom (the angle at $B$), which allows it to have a height of $8$ ($<9$).

Then by extending the lower side and intersecting with a parallel to the vertical side at distance $8$, you find the vertex $D$ (projecting to $F$).

As no other constrain is expressed, the construction is possible.

Note that an equilateral triangle cannot have $\dfrac89$ as the ratio of height to side.