SOLUTION: what shape has a perimeter of 14 inches and the area of 8sq inches

Question 503224: what shape has a perimeter of 14 inches and the area of 8sq inches

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
For a regular polygon with n sides of length s:
Perimeter = n*s = 14
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The area of a triangle (n=3) is:
s = 14/3
Area = 3*(14/3)^2*cot(60)
Area = 37.72
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Area of a square (n=4) is:
s = 14/4
Area = 4*(14/4)^2*cot(45)
Area = 49
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The area increases with the number of sides, so it's not a regular polygon.
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Try a parallelogram
Area = b*h/2 = 8
2b + 2h = 14
b+h = 7
h = 7-b
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b*h = 16
b*(7-b) = 16
7b - b^2 - 16 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -15 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

In the field of imaginary numbers, the square root of -15 is + or - .

The solution is , or
Here's your graph:

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No real solutions.
I think there's no such plane figure.
Maybe a typo.
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