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July 5th, 2016, 08:42 PM   #11
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Originally Posted by skimanner View Post
By the way do either of you know why all of the regular pentagon area formulas seem overly complicated? From what I have found the area is side squared x ~1.72048 which I'm sure could be contained in a better way such as 1 + sqrt(2) that is used for a regular octagon.
The problem is that your formula is an approximation. As usual, Denis found the exact formula, but it is far from intuitive with that 54 degrees in there.

The sum of the interior angles of a pentagon is 180(5 - 2) = 540 degrees. So each of the 5 interior angles is 108 degrees. We can decompose a regular pentagon into 10 congruent right triangles thereby splitting each angle of 108 into two equal angles of 54 degrees.
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July 6th, 2016, 05:30 AM   #12
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Ski, something a bit different; try this:

an isosceles triangle, equal sides = 1, vertex = 1 degree.
Calculate area.
Multiply area by 360.
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July 8th, 2016, 07:55 AM   #13
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I see thanks Jeff and Denis
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