Check out some similar questions!

augustcab · Sep 5, 2008, 01:37 PM

Formula for calculating the voulume of an Octogon

galactus · Sep 5, 2008, 02:13 PM

An octagon has no volume. Though, It has area.

You can derive a general formula.

Let r equal the distance from the center to a vertex.

We can break it up into 16 right triangles by drawing another line from the center to the middle of a side and then find the area of each.

The angle of each triangle will be 22.5 degrees because 360/16=22.5.

The area of a triangle is $\frac{bh}{2}$

The area of each can be gotten from $\frac{\overbrace{rsin(22.5)}^{\text{b}}\overbrace{ rcos(22.5)}^{\text{h}}}{2}$

There are 16 such triangles, so we have

So, the area of the octagon is $8r^{2}sin(22.5)cos(22.5)=2\sqrt{2}r^{2}$

If you need volume, multiply by the depth as well.

Just enter your value of r and the depth, call it d, to find the volume.

For that matter, you could Google octagon volume and find many things. I just came up with this one.

Unknown008 · Sep 6, 2008, 10:30 AM

Galactus? You sure? An octagon has eight sides i think? So, 8, triangles instead of 16?

Then, your final equation should be :

$Area = \sqrt{2} r^2$

EDIT: Oh, i didn't check the numbers... sorry, i need to check more carefully next time.

galactus · Sep 6, 2008, 10:44 AM

I broke it up into 16 right triangles. Instead of the 8 isosceles. See what I mean?

Since there are 16 of these little right triangles, we have $8r^{2}cos(22.5)(sin(22.5)=2\sqrt{2}r^{2}$

If we use the 8 isosceles triangles each with area $\frac{1}{2}r^{2}sin(45)$ , we get $4r^{2}sin(45)=2\sqrt{2}r^{2}$

Same thing. I should've done that before. Easier. A good check though.

galactus · Sep 6, 2008, 12:45 PM

That's OK Unknown. I doubt if the poster responds back anyway.