SOLUTION: The length a rectangle is twice that of its width. If the area of the rectangle is 72 square inches, then find the length and width. How do I do this please include details and e

Algebra ->  Square-cubic-other-roots -> SOLUTION: The length a rectangle is twice that of its width. If the area of the rectangle is 72 square inches, then find the length and width. How do I do this please include details and e      Log On


   

Question 939757: The length a rectangle is twice that of its width. If the area of the rectangle is 72 square inches, then find the length and width.
How do I do this please include details and each step thank you
Someone else answered the question with the answer as w=6
L=12. But I don't understand how they got it if someone can give me the steps or explain it to me please.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
The very same example problem has been asked and answered a few times already.

w for width, L for length;
Given is L=2w, and using A for area, A=72.

Formula for area using the assigned variables, wL=A.

Summarizing the list of equations, purely in symbols:
L=2w
wL=A
-
KNOWN: A=72
UNKNOWN: w and L

Solution Process:
wL=A
w%282w%29=A
w%2A2%2Aw=A
w%5E2%2A2=A
w%5E2=A%2F2
highlight%28w=sqrt%28A%2F2%29%29

Your choice if you want to rationalize the denominator, but knowing that A%2F2 will be a whole number here, rationalizing does not matter much.

L=2w means now L=2%2Asqrt%28A%2F2%29.

Plug-in the given value and evaluate w and L.

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!

The length a rectangle is twice that of its width. If the area of the rectangle is 72 square inches, then find the length and width.
How do I do this please include details and each step thank you
Someone else answered the question with the answer as w=6
L=12. But I don't understand how they got it if someone can give me the steps or explain it to me please.


Let width be W

Then length = 2W

Area = LW, and since area = 72 sq in, we get:

2W(W) = 72

2W%5E2+=+72

W%5E2+=+72%2F2

W%5E2+=+36

W, or width = sqrt%2836%29, or highlight_green%286%29 inches

Length: 2(6), or highlight_green%2812%29 inches