My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum

LinkBack Thread Tools Display Modes
February 27th, 2017, 11:19 AM   #1
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!
SELBY8 is offline  
February 27th, 2017, 12:51 PM   #2
Senior Member
mrtwhs's Avatar
Joined: Feb 2010

Posts: 674
Thanks: 127

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 $\displaystyle AS = 30$.
mrtwhs is offline  

  My Math Forum > High School Math Forum > Geometry

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 02:42 PM
Interesting problem for middle school students agentredlum Elementary Math 2 May 25th, 2013 01:52 PM
Hard middle school number theory problem julien Number Theory 43 August 13th, 2011 04:41 PM
Pretty Complex Middle School problem -DQ- Elementary Math 1 February 11th, 2008 07:19 PM

Copyright © 2018 My Math Forum. All rights reserved.