PDA

mook11
05-11-2006, 05:28 PM
I'm a homeschooled student and this problem is the last on a test. I don't know what's not clicking in my brain but I just can't figure out a formula. Can someone please help?

A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the dimensions of the original rectangle?

I understand the first and second rectangles would look like this:

4w 4w+5
w [ ] w + 2 [ ]

but that's as far as I get! What would the formula look like?

Thanks in advance for any help!

skeeter
05-11-2006, 06:02 PM
A rectangle is 4 times as long as it is wide
length = L, width = W

L = 4W

A second rectangle is 5 centimeters longer and 2 centimeters wider than the first.
length = L', width = W'

L' = 4W+5
W' = W+2

The area of the second rectangle is 270 square centimeters greater than the first.

L'W' = 270 + LW

now ... do the substitutions that will get the last equation in terms of a single variable

(4W+5)(W+2) = 270 + (4W)(W)

solve for W ... the width of the original rectangle, then you can find all else.

mook11
05-11-2006, 06:56 PM
Okay!

So the dimensions of the original rectangle are 20 and 80, which makes it's square footage 1,600, right?

Awesome, thank you very much, Skeeter!