SOLUTION: 2508 is the area of a rectangle and 202 is the perimeter what is the length; width?

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Question 949086: 2508 is the area of a rectangle and 202 is the
perimeter what is the length; width?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website!
w, width
L, lengh
p, perimeter
A, area

FORMULAS
2w+2L=p;
wL=A.

KNOWN VARIABLES:
p=202
A=2508
UNKNOWN VARIABLES:
w, L.

SOLVE FOR UNKNOWN VARIABLES

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Just know that sometimes, the actual values can make the equation here easier to work with.
You may want to choose the substitute the given values NOW into this quadratic equation before
continuing. That is what will do here:

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OR

Finding corresponding L values from earlier found formula
OR

Dimensions, however you want to assign them, or 44 and 57.

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

2508 is the area of a rectangle and 202 is the
perimeter what is the length; width?

Let length and width be L and W, respectively
Then: LW = 2,508 ------- eq (i)
Also, 2(L + W) = 202______2(L + W) = 2(101)______L + W = 101______W = 101 - L ------ eq (ii)
L(101 - L) = 2,508 ------- Substituting 101 - L for W in eq (i)

(L - 44)(L - 57) = 0
L - 44 = 0 OR L - 57 = 0
L = 44 OR L = 57
This means that if length = 44, then width = 57
However, if length = 57, then width = 44
Thus, dimensions are: