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Given Perimeters, Find Sides

Date: 8/5/96 at 12:35:16
From: tstorer
Subject: Rectangle and triangle

I'm in grade 9 math, and have a problem with one of these algebra 

As a description of the question, there are two diagrams side by side. 
One is a rectangle, and one is a right triangle. 

The bottom edge of the triangle is marked "x+1", the perpendicular 
side is marked "x", and the hypoteneuse is marked "x+2". 

The rectangle beside it has two of its parallel lines marked "x+1" and 
the opposite two parallel lines are marked "x-1". The direction above 
reads: "These figures have equal perimeters.  Find the sides."    


Date: 8/6/96 at 10:44:26
From: Doctor Mike
Subject: Re: Rectangle and triangle

Hello Pat,  I think I can imagine what the diagrams look like.
The perimeter of any polygon is the distance completely around
it.  For the rectangle it is : 
   R  =  (x+1) + (x+1) + (x-1) + (x-1)  =  4x.
For the triangle it is : 
   T  =  (x+1) + x + (x+2)  =  3x + 3.
The first thing to do is to find out what x is.  In order for the 
rectangle perimeter R to equal the triangle perimeter T you have to 
set R equal to T and solve for x.  Let's do it :
   4x  =  3x + 3.
This is pretty easy to solve.  Just subtract 3x from both sides
to get x = 3.
Now we know what x is but we have not answered the question. 
We have to tell what the side lengths are for the 2 polygons.
The side lengths for the triangle are 3, 4, and 5.  For the 
rectangle they are .... WAIT .... You do that part, okay?  
I hope this helps.  Write back again if you have more questions.
-Doctor Mike,  The Math Forum
Associated Topics:
Middle School Algebra

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