Dimensions of a Rectangle Word Problem

by Teri
(England)

A rectangle has area 12a squared and a perimeter 14a. What are the dimensions of the rectangle and how do you solve this?

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Dimensions of a Rectangle Word Problem

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Feb 21, 2010
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by: Anonymous

Hi Terri,

This is quite an interesting problem and it does seem confusing at first. Let's break it down.
You know that you have a rectangle and you are given the Area of 12a^2 with a perimeter of 14a.

The first thing that you should notice is that the dimensions of the rectangle will have the variable a. (When you multiply the length times the width to find the area, you will end up with a^2 and when you add all four sides, you will end up with a.)

So, for now, let's ignore the a. I know that I have a rectangle whose area is 12 and perimeter is 14. So, when you multiply the length times width you must end up with 12 and when you add all four sides for the perimeter, you will end up with 14.

Let's write the equations:
lw = 12 (length times width = area)
2l +2w = 14 (2 times the length + 2 times the width = perimeter)

From here, I would have to assume that you know strategies for solving a system of equations and how to solve quadratic equations.

I would use substitution. Let's solve 2l + 2w = 14 for w.

2l -2l + 2w = 14 - 2l - Subtract 2l from both sides.

2w/2 = (14 - 2L)/2 - Divide both sides by 2

w = 7 - L

Now substitute 7-L for w into the area equation.

L(w) = 12
L(7-L) = 12
7L - L^2 = 12
or
-L^2 +7L = 12
-1(-L^2 + 7L = 12) Multiply everything by -1 to make -L^2 positive.
L^2 - 7L = -12

Now let's set this equation equal to 0 so that we can factor.

L^2 - 7L + 12 = 0
(L -4) (L - 3) = 0
L = 4, L = 3

Therefore, the dimensions are L = 4a and W = 3a

Check:
A = LW
A = (4a)(3a)
A = 12a^2

P = 2L + 2W
P = 2(4a) + 2(3a)
P = 8a + 6a
P = 14a

I hope this helps. There are a lot of skills involved if you solve it algebraically.

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