SOLUTION: the perimeter of a rectangle is 32m and the area of the rectangle is 55m squared. Find the dimensions of the rectangle

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Question 29496: the perimeter of a rectangle is 32m and the area of the rectangle is 55m squared. Find the dimensions of the rectangle
Found 2 solutions by sdmmadam@yahoo.com, cleomenius:
Answer by sdmmadam@yahoo.com(530) (Show Source):
You can put this solution on YOUR website!
the perimeter of a rectangle is 32m and the area of the rectangle is 55m squared. Find the dimensions of the rectangle
Let the length of therectangle be L meters
and let the width be B meter.
Given that the perimeter of a rectangle is 32m
That is 2(L+B)= 32 ----(1)
which on dividing by 2 gives (L+B)=16 ----(2)
Given that the area of the rectangle is 55m
That is LXB = 55----(3)
By formula (L-B)^2 = (L+B)^2-4LB
=(16)^2-4X(55) [using (2) and (3)]
=256-220
=36
(L-B)^2 = 36
taking the positive sqroot
(since length >width by convention, therefore L>B implying (L-B) >0 }
Therefore L-B = 6 ----(4)
We have L+B= 16----(2)
Adding (4) and (2)
(L+L) = (6+16)
2L= 22
L=22/2 = 11
L=11 in (2) gives B = 5
Therefore Length of the rectangle = 11m
and width of the rectangle = 5 ms
Verification: Area = length X width = 11X5=55 sq meters which is correct
Therefore our values are correct.

Answer by cleomenius(959) (Show Source):
You can put this solution on YOUR website!
I set up 2 equations
2L + 2W = 32 or L + W = 16 (equation 1)
L * W = 55
Next, I set the second equation to L
L = 55 / W
I substituted this for L in equation 1.
55/W + W = 16
Multiply all by W
55 + W^2 = 16W or W^2 -16W + 55 = 0
Which factors to ( W - 5) ( W - 11)
So, we obtain W = 5, W = 11.
Checking 5 in first equation, 2L + 10 = 32
2L = 22
L = 11.
Substituting 5 * 11 into the second equation, we obtain 55.
Therefore, we check.
Cleomenius.