SOLUTION: Could someone help me with this please. Word problems. Geometry Geometry: The length of a rectangle is 1 cm longer than it's width. If the diagonal of the rectangle is 4 cm, wha

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Question 49205: Could someone help me with this please. Word problems. Geometry
Geometry: The length of a rectangle is 1 cm longer than it's width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let the length of the rectangle be L and its width be W.
The length, L, is 1 cm longer than the width, W, so L = W+1
The sides of the rectangle can be written in terms of the width: L=(W+1) and W=(W).
The diagonal of the rectangle forms the hypotenuse (c) of a right triangle whose sides are (W+1) and (W).
Using the Pythagorean theorem: c%5E2+=+a%5E2%2Bb%5E2, you can find W:
4%5E2+=+%28W%2B1%29%5E2+%2B+W%5E2 Simplify.
16+=+W%5E2%2B2W%2B1+%2B+W%5E2 Collect like-terms.
2W%5E2%2B2W%2B1+=+16 Subtract 16 from both sides of the equation.
2W%5E2%2B2W-15+=+0 Use the quadratic formula to solve: W=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
W+=+%28-2%2B-sqrt%282%5E2-4%282%29%28-15%29%29%29%2F2%282%29
W+=+%28-2%2B-sqrt%284%2B120%29%29%2F4
W+=+%28-2%2B-sqrt%28124%29%29%2F4
W+=+%28-2%2B-2sqrt%2831%29%29%2F4
W+=+-1%2F2-%281%2F2%29sqrt%2831%29 Discard this solution as the width must be a positive number.
W+=+-1%2F2+%2B+%281%2F2%29sqrt%2831%29
For an approximate answer:
W+=+-1%2F2%2B%281%2F2%29%285.57%29
W+=+2.28cm
L+=+W%2B1
L+=+3.28cm