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In the parallelogram it is given that $AD=20$ and that the perimeter is $140$. (Ok so therefore $AB=DC=50$).
It is also given that $AS=AM+18$.
We need to find $AM$, therefore $AS$ and then to calculate the area of the Parallelogram which derives from finding $AM$.
Now, I managed to find the right answer but not the way to prove it, I just played around with Pythagoras.
If I could prove that $SC=CB$ then the rest follows and $AM=12$, but I don't know how.
Please help, thanks!

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    The picture is blank. – 2017-02-27
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    @fleablood it now isn't – 2017-02-27
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    Also, could someone rewrite it in Math Jax? I cannot do it on handheld. – 2017-02-27
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    @Nicholas You should be able to borrow some ideas from [this answer](http://math.stackexchange.com/a/2163130/291201). – 2017-02-27

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The area $|ABCD| = AS\cdot AD = AM\cdot AB$ (by regarding different sides as the base)

So
$(AM+18)\cdot 20 = AM\cdot 50$
$360 = 30\cdot AM$
$AM=12$

Area $|ABCD| = 12\cdot 50 = 600$

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    Thank you very much . My basic geometry is very lousy so I did not fully understand. The area is Base multiplied by height, so for example AM x AB. but why is it also AS x AD? – 2017-02-27
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    @Nicholas - take $AD$ as the base, then $AS$ is the height. Be careful not to hurt your neck as you look at it that way. (for less contortion, take $BC$ as the base, but it's the same value) – 2017-02-27
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    But once the height "exits" the Parallelogram, aren't we talking about a whole new Parallelogram. Thanks again and sorry for being so thick. – 2017-02-27
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    No, the height is the height; just the distance between the parallel lines (extended if necessary). If you want to prove this to yourself, add on the two (congruent) triangles necessary to fill the parallelogram into a rectangle, push them together then subtract them off again. – 2017-02-27