what's the area of a regular octagon?

Answer 1

Join each vertex to the centre of the octagon and you will have 8 identical isosceles triangles. Calculate the area of one triangle and then multiply by 8.

Answer 2

Hi. Try this : http://www.answers.com/main/ntquery?s=area+of+a+regular+octagon&gwp=13

Answer 3

The area of any n-sided figure of side s is equal to

A = (n/4) (s^2) cot(pi/n)

In this case, since n = 8, we have

A = (8/4) (s^2) cot(pi/8)
A = 2s^2 cot(pi/8)

But cot(pi/8) is equal to cos(pi/8) / sin(pi/8), and we can solve this using the half angle identities. These identities go as follows:

cos^2(x) = (1/2) (1 + cos(2x))
sin^2(x) = (1/2) (1 - cos(2x))

Therefore,

cos^2(pi/8) = (1/2) (1 + cos(2*pi/8))
cos^2(pi/8) = (1/2) (1 + cos(pi/4))

And we know what cos(pi/4) is, since it is one of our known values on the unit circle. cos(pi/4) is equal to sqrt(2)/2.

cos^2(pi/8) = (1/2) (1 + sqrt(2)/2)
cos^2(pi/8) = [2 + sqrt(2)] / 4
Normally when taking the square root of both sides, we have to insert a "plus or minus". In this case though, we only take the positive value because we know what quadrant pi/8 is in.

cos(pi/8) = sqrt[2 + sqrt(2)] / 2

Now we solve this one:
sin^2(pi/8) = (1/2) (1 - cos(2*pi/8))
sin^2(pi/8) = (1/2) (1 - cos(pi/4))
sin^2(pi/8) = [1 - sqrt(2)/2] / 2
sin^2(pi/8) = [2 - sqrt(2)]/4, so
sin(pi/8) = sqrt[2 - sqrt(2)] / 2

It follows that

cos(pi/8) / sin(pi/8) = { sqrt[2 + sqrt(2)] / 2 } / {sqrt[2 - sqrt(2)] / 2}

This can be reduced to

sqrt[2 + sqrt(2)] / sqrt[2 - sqrt(2)]

We can rationalize the denominator,

sqrt(4 - 2) / [2 - sqrt(2)]
sqrt(2) / [2 - sqrt(2)]

We can further rationalize the denominator,

sqrt(2) (2 + sqrt(2)) / [4 - 2]
sqrt(2) (2 + sqrt(2)) / 2

Therefore, our simplified area is:

A = 2s^2 [sqrt(2) (2 + sqrt(2)) / 2]
A = (s^2) [sqrt(2) (2 + sqrt(2))]