jerzey

08-22-2005, 08:04 PM

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

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jerzey

08-22-2005, 08:04 PM

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

08-22-2005, 08:07 PM

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:

. . . . L<sup>2</sup> + W<sup>2</sup> = 4<sup>2</sup>

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

. . . . (W+1)<sup>2</sup> + W<sup>2</sup> = 4<sup>2</sup>

Now, solve for W.

edit: squared

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

. . . . L<sup>2</sup> + W<sup>2</sup> = 4<sup>2</sup>

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

. . . . (W+1)<sup>2</sup> + W<sup>2</sup> = 4<sup>2</sup>

Now, solve for W.

edit: squared

Denis

08-22-2005, 11:47 PM

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!

(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

08-23-2005, 03:02 AM

Ya forgot a "^2" Matt; should be:

(W+1)^2 + W^2 = 4^2

Thanks Denis. It's fixed now.

(W+1)^2 + W^2 = 4^2

Thanks Denis. It's fixed now.

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