h will be opposite leg and R hypotenuse so use sine: h = R sine (180/N). That makes the length of a side 2h or 2R sin(180/N)

a will be adjacent leg so use cosine: a = R cos(180/N)

This makes the area of each isosceles triangle

be 1/2 (2R sin(180/n)(R cos 180/N), or

R^2 sin(180/N)(cos 180/N) and there are N triangles so the area is

N R^2 sin(180/N)(cos 180/N)

Area = 1/2 * apothem * perimeter

The apothem of any N-gon is a straight line drawn from the center of the figure to the middle part of any side (NOT the vertex).

We also have to know one more equation:

[180 (n-2)] / n, which gives a single interior angle of each N-gon.

Ok, so draw a picture out, let’s say of a triangle.

Draw the radius out to one of the angles (bisect it) and then draw the apothem. You now have a small right triangle.

Now solve for the other leg using sine and cosine and the previous equation.

cos ( 1/2 (180) (n-2) / n) = side / radius

side = radius cos (90 (n-2) / n)

you can simplify this to

side = radius * sin (180/n)

Now double this number: 2 * radius * sin (180/n)

Now multiply it by n to get the perimeter =

2N * radius * sin(180 / n)

—–

Now solve for the apothem:

sin ( 1/2 * 180(n-2)/n) = apothem / radius

apothem = sin (90 * (n-2) / n) * radius

apothem = cos (180 / n) * radius

FINALLY…

plug it back in to the equation:

Area = 1/2 apothem * perimeter

A = 1/2 * 2 * sin (180/N) * cos (180/n) * NR^2

You can clean this up a bit

A = 1/2 * sin (360/N) * NR^2

do you have a book? look in there they should have a formula and examples as to how to do it. or go to mathhelp.com type in you book, author and stuff and they will have solved problems look for the chapter and such.

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