Question 30040: Please help me solve this equation:
The perimeter of a rectangle is 41 inches and its area is 91 square inches. Find the lenght of the shortest side.
Thanks:)
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! The perimeter of a rectangle is 41 inches and its area is 91 square inches. Find the lenght of the shortest side.
Let the length of the rectangle be L inches and
let the width of the rectangle be B inches
The perimeter of a rectangle is 41 inches
That is 2(L+B) =41 which implies (L+B) = 20.5----(1)
And area is 91 square inches
That is (LB) = 91 ----(2)
By formula (L-B) ^2 = (L+B)^2 - 4LB
(L-B)^2 = (20.5)^2 - (4X91) (using (1) and (2)
(L-B)^2 =(420.25) -364 = 56.25
Taking the postitive sqrt
(as length by convention larger than width,L>B,that is L-B >0)
L-B = 7.5 ----(3)
(L+B) = 20.5----(1)
(3)+ (1) gives
2L= 28
L = 28/2 = 14
L=14 in (1) gives B = 20.5-L = 20.5 - 14 = 6.5
Answer:Length of the rectangle= 14 inches and
Width of the rectangle = 6.5 inches
Verification: Area = Length X Width = 14 X 6.5 = 91 which is correct
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