My Math Forum Simple middle school Parallelogram problem

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 February 27th, 2017, 12:19 PM #1 Newbie   Joined: Feb 2017 From: Israel Posts: 1 Thanks: 0 Simple middle school Parallelogram problem In the bottom 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!
 February 27th, 2017, 01:51 PM #2 Senior Member     Joined: Feb 2010 Posts: 633 Thanks: 104 Since the opposite angles of a parallelogram are equal, we have that $\displaystyle \angle B \cong \angle D$. Along with the right angles at $\displaystyle M$ and $\displaystyle S$ this makes $\displaystyle \triangle AMD \sim \triangle ASB$ by AA. Now by ratios of corresponding sides we have $\displaystyle \dfrac{AM}{20} = \dfrac{AS}{50}= \dfrac{AM+18}{50}$ Cross multiplying the first with the last we get that $\displaystyle AM=12$ and so $\displaystyle AS = 30$.

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