SOLUTION: The width of a rectangle is 8 less than twice its length. If the area of the rectangle is 200 cm^2, what is the length of the diagonal?

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Question 341864: The width of a rectangle is 8 less than twice its length. If the area of the rectangle is 200 cm^2, what is the length of the diagonal?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The width of a rectangle is 8 less than twice its length.
If the area of the rectangle is 200 cm^2, what is the length of the diagonal?
:
Let x = the length
then
(2x-8) = the width
and
x(2x-8) = 200; the area
A quadratic equation
2x^2 - 8x - 200 = 0
Simplify, divide by 2
x^2 - 4x - 100 = 0
Find x using the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this equation: a=1; b=-4; c=-100
x+=+%28-%28-4%29+%2B-+sqrt%28-4%5E2-4%2A1%2A-100+%29%29%2F%282%2A1%29+
:
x+=+%284+%2B-+sqrt%2816+%2B+400+%29%29%2F2+
:
x+=+%284+%2B-+sqrt%28416+%29%29%2F2+
The positive solution is what we want here
x+=+%284+%2B+20.4%29%2F2+
x = 24.4%2F2
x = 12.2 is the length
then
2(12.2) - 8 = 16.4 is the width
:
Check solution by finding the area: 12.2 * 16.4 = 200.1 ~ 200, close enough
:
Find the diagonal (d)
d = sqrt%2812.2%5E2+%2B+16.4%5E2%29
c = 20.44 cm is the diagonal