
Geometry Geometry Math Forum 
 LinkBack  Thread Tools  Display Modes 
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 $\displaystyle AS = 30$. 

Tags 
middle, parallelogram, problem, school, simple 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Problem for middle school students  agentredlum  Algebra  11  June 14th, 2013 03:42 PM 
Interesting problem for middle school students  agentredlum  Elementary Math  2  May 25th, 2013 02:52 PM 
Hard middle school number theory problem  julien  Number Theory  43  August 13th, 2011 05:41 PM 
Pretty Complex Middle School problem  DQ  Elementary Math  1  February 11th, 2008 08:19 PM 