Given Perimeters, Find SidesDate: 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 questions. 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." Pat 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 Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/