SOLUTION: The length of a rectangle is 1cm longer than its width. If the diagonal of the rectangle is 4cm, what are the dimensions (length and width) of the rectangle? Talk about confusin

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Question 65712: The length of a rectangle is 1cm longer than its width. If the diagonal of the rectangle is 4cm, what are the dimensions (length and width) of the rectangle?
Talk about confusing to me! Thank you for the help!
Found 2 solutions by funmath, Nate:
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 1cm longer than its width. If the diagonal of the rectangle is 4cm, what are the dimensions (length and width) of the rectangle?
We can use the pythagorean theorem to solve this:
highlight%28c%5E2=a%5E2%2Bb%5E2%29, where c=hypotenuse, a and b are both legs of a right triangle
In this case the diagonal is c=4
Let the width, a=x
Then the length, b=x+1
4%5E2=%28x%29%5E2%2B%28x%2B1%29%5E2
16=x%5E2%2Bx%5E2%2B2x%2B1
16=2x%5E2%2B2x%2B1
16-16=2x%5E2%2B2x%2B1-16
0=2x%5E2%2B2x-15
This is in the form: 0=ax%5E2%2Bbx%2Bc
Now use the quadratic formula: highlight%28x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a%29
a=2, b=2, and c=-15
x=%28-2%2B-sqrt%282%5E2-4%282%29%28-15%29%29%29%2F%282%282%29%29
x=%28-2%2B-sqrt%284%2B120%29%29%2F4
x=%28-2%2B-sqrt%28124%29%29%2F4
x=%28-2%2B-sqrt%284%29%2Asqrt%2831%29%29%2F4
x=%28-2%2B-2%2Asqrt%2831%29%29%2F4
x=2%28-1%2B-sqrt%2831%29%29%2F4
x=%28-1%2B-sqrt%2831%29%29%2F2
Ignore the negative answer because dimensions are positive.
x=%28-1%2Bsqrt%2831%29%29%2F2
x~~2.28388
The width is: x~~2.28 cm
The length is x+1=2.28+1=3.28 cm
Happy Calculating!!!

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
width = w
length = w + 1
A width and a length can make legs of a right triangle with the diagonal being its hypotenuse. *Draw a picture to better understand.
a^2 + b^2 = c^2
(w)^2 + (w + 1)^2 = 4^2
w^2 + w^2 + 2w + 1 = 16
2w^2 + 2w = 15
w^2 + w = 7.5
(w + 0.5)^2 = 7.75
w + 0.5 = +- sqrt(7.75)
w = -0.5 +- sqrt(7.75)
width about 2.2839
length about 3.2839
*You can use other ways to solve: w^2 + w - 7.5 = 0