# Homework Help: Finding dimensions of a rectangle.

1. Dec 8, 2012

### rcmango

1. The problem statement, all variables and given/known data

the perimeter of a rectangle is 24 ft. the length is 4 ft longer than the width find the dimensions

width x
length x+4

however, I should be doing it like this:

a first equation should start like: 2x+2y=?

and the second should start like x=y+?

so whats the length and the width?

2. Relevant equations

3. The attempt at a solution

x + x + (x+4)+(x+4) = 24
4x+8=24
-8 -8

4x = 16
4x/4 = 16/4

x = 4

so width 4 ft
and length 4+4, so 8 ft.
makes sense, since length should be greater than width.

2. Dec 8, 2012

### phinds

x + x + (x+4)+(x+4) = 24

x = 4

exactly

3. Dec 8, 2012

### Ray Vickson

Why do you say you "should be doing it like this:"? There is no *should*---any correct method is OK. Doing it two ways might be helpful towards learning goals, but it is not necessary for correct presentation of the solution.

4. Dec 8, 2012

### Michael Redei

If you've learned one way of solving problems of this kind, and you're expected to show that you have, in fact, understood the technique, then you probably should be doing just what you are, namely translating the facts bit by bit into equations, i.e.
• "rectangle" suggests the equation 2x + 2y = 24
• "width versus height" remark means y = x + 4
If you're just interested in the answer, any method will do, for instance this one, which I'll call "guess and revise":

Start with any width you like, say 7. Then the length will be 4 more, so 11. This gives you a perimeter of 7+7+11+11 = 36.

Oops, that's 12 too much. Let's subtract this "extra" length from the four sides, taking 12/4 = 3 away from each: width now is 7-3 = 4, and length is 11-3 = 8.

5. Dec 9, 2012

### rcmango

thankyou phinds for confirming.

Thanks for completing those equations Michael Redei, i seen them done that way and I didn't understand it, its much more clear now.

Last edited: Dec 9, 2012