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Calculating the surface area of a two and half sided regular polygon

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Part 2 of this post. I struggled for a day over this.

First, write down what you know…

1. Any regular polygon can be split up into equal isosceles triangles, one for each side. The angle in the middle is 360 divided by number of sides and the other two are just (180 deg minus the middle angle) divided by 2. So you can divide a square into four triangles, the angle in the middle is 90 deg and the other two angles are 45 deg. The area of a triangle = height (known as apothem is this context) x base/2.

2. A pentagram with side length 10 has surface area of 31.027 (handy calculator here).

3. A two and a half sided regular polygon can divided into two and a half triangles with an angle in the middle of 144 deg (360/2.5), the outer angles are 18 deg (half of 36 deg, the total angle inside each point). 144 + 18 + 18 = 180.

4. Using cosines and tangents, the height/apothem of each triangle is 1.625, so the area of each triangle is 1.625 x 10/2 = 8.125.

5. It has two and a half sides, so the total area = 20.3125.

6. The other way of doing it is using the formula for calculating the surface area, “The area of any regular polygon is given by the formula: Area = (a x p)/2, where a is the length of the apothem (the line joining the half way point of a side and the centre) and p is the perimeter of the polygon”. The surface area using this method is the same = 1.625 x 25/2 = 20.3125.

7. The surface area of a three-sided regular polygon (= an equilateral triangle) with side length 10 is 43.3, and the surface area of a regular two sided polygon (= a straight line) is zero, so 20.3 looks about right to me.

(If I had to take a wild guess, I would say that the difference (i.e. 31 – 20.3 = 10.7) is because what appears to be the pentagon in the middle of the two and a half sided polygon does not actually exist, it has no real corners, but something went wrong in my workings, the area of the pentagon in the middle of a regular five pointed star = about 9.9. That’s the bit I struggled over and failed to reconcile.)


Source: http://markwadsworth.blogspot.com/2018/12/calculating-surface-area-of-two-and.html


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