SOLUTION: what is the side length and width of a rectangle if the area is 200ft and the perimeter is 100ft

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Question 986480: what is the side length and width of a rectangle if the area is 200ft and the perimeter is 100ft
Found 2 solutions by macston, solver91311:
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
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L=length; W=width; P=perimeter=100ft; A=area=200ft^2
P=2(L+W)
100ft=2(L+W)
50ft=L+W
50ft-L=W
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A=LW}
200ft%5E2=LW Substitute for W
200ft%5E2=L%2850ft-L%29}
200ft%5E2=-L%5E2%2B50L
0=-L%5E2%2B50L-200
0=L%5E2-50L%2B200
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aL%5E2%2BbL%2Bc=0 (in our case 1L%5E2%2B-50L%2B200+=+0) has the following solutons:

L%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-50%29%5E2-4%2A1%2A200=1700.

Discriminant d=1700 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--50%2B-sqrt%28+1700+%29%29%2F2%5Ca.

L%5B1%5D+=+%28-%28-50%29%2Bsqrt%28+1700+%29%29%2F2%5C1+=+45.6155281280883
L%5B2%5D+=+%28-%28-50%29-sqrt%28+1700+%29%29%2F2%5C1+=+4.3844718719117

Quadratic expression 1L%5E2%2B-50L%2B200 can be factored:
1L%5E2%2B-50L%2B200+=+1%28L-45.6155281280883%29%2A%28L-4.3844718719117%29
Again, the answer is: 45.6155281280883, 4.3844718719117. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-50%2Ax%2B200+%29

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ANSWER: The rectangle is 45.62 feet long by 4.38 feet wide
(or 45.62 feet wide by 4.38 feet long).

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


This problem cannot be solved as written because area is not measured in feet, in other words the statement "the area is 200 ft" is nonsense. I'm going to go out on a limb and make the assumption that you meant "the area is 200 square feet". Don't roll your eyes and say "Whatever"; details like this do make a difference.

The length plus the width of a rectangle is equal to one-half of the perimeter. Since the perimeter is 100 feet, the sum of the measures of the sides of the rectangle must be 50 feet. Hence:



The area is the length times the width, so



But we can substitute from Equation (1) into Equation (2) to get an expression for the area as a function of the width:



Since we know that the area is equal to , we can write:



Solve for and then calculate

Hint: The quadratic does not factor, so you need to use the quadratic formula.

John

My calculator said it, I believe it, that settles it