geometry

[jerzey]

the length of a rectangle is is 1 cm longer than its width if the diagonal is 4cm , what are the dimensions (the length and width) of the rectangle

[Matt]

Let L and W be the length and width of the rectangle respectively. Then, since "the length of a rectangle is is 1 cm longer than its width," we have L = W + 1.

We know the diagonal is 4cm long. Thus, by Pythagorean's Theorem:

. . . . L² + W² = 4²

but we also know that L = W + 1, and so:

. . . . (W+1)² + W² = 4²

Now, solve for W.

edit: squared

[Denis]

Ya forgot a "^2" Matt; should be:
(W+1)^2 + W^2 = 4^2

After some juggling/simplifying, jerzey, you'll be at:
2W^2 + 2W - 15 = 0
You will not get a "nice integer" solution; just so you know!

[Matt]

Ya forgot a "^2" Matt; should be:
(W+1)^2 + W^2 = 4^2
Thanks Denis. It's fixed now.