SOLUTION: The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 68 cm^2, find the dimensions of the rectangle to the nearest thousandth.

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Question 118352: The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 68 cm^2, find the dimensions of the rectangle to the nearest thousandth.
Found 2 solutions by MathLover1, stanbon:
Answer by MathLover1(20850) (Show Source):
You can put this solution on YOUR website!
The length:
The width:
the area:
given:

��.substitute and

��use quadratic formula to solve for

�.

Since we are looking for the width, we need only positive root:

Now we can find the length;
�.substitute

Check:
��.
��.
��.which can be rounded to

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 68 cm^2, find the dimensions of the rectangle to the nearest thousandth.
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Let width be "x" cm ; Then length is "5x+2" cm.
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EQUATION:
x(5x+2) = 68
5x^2+2x-68=0
x = [-2 +- sqrt(4-4*5*-68)]/10
x = [-2 +- sqrt(4+1360)]/10
x = [-2 +- sqrt(1364)]/10
x = [-2 +- 36.9324]/10
Positive answer:
x = 3.4932 cm (width)
5x+2 = 19.4662 cm (length)
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Cheers,
Stan H.