SOLUTION: The length of a rectangle is five times its width. If the area of the rectangle is 500in^2, find its perimeter

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Question 896644: The length of a rectangle is five times its width. If the area of the rectangle is 500in^2, find its perimeter
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
m=5.
L for length, w for width.
p=unknown.
A=500

L=mw and wL=A.
w%2A%28mw%29=A
mw%5E2=A
w%5E2=A%2Fm
w=sqrt%28A%2Fm%29, L=m%2Asqrt%28A%2Fm%29.
Rationalizing their denominators,
highlight%28w=sqrt%28Am%29%2Fm%29 and highlight%28L=sqrt%28Am%29%29.

Question is to find p.
p=2w%2B2L
highlight%28p=2sqrt%28Am%29%2Fm%2B2%2Asqrt%28Am%29%29
And substitute the given values occurring from the description.
Compute p.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

The length of a rectangle is five times its width. If the area of the rectangle is 500in^2, find its perimeter 


Let width be W

Then length is: 5W

Thus, W(5W) = 500

5W%5E2+=+500

W%5E2+=+500%2F5

W%5E2+=+100

W, or width = sqrt%28100%29, or 10

Length: 5(10), or 50

Perimeter: 2(L + W), or 2(10 + 50), or 2(60), or highlight_green%28120%29 inches