Area of a regular octagon with a side-length of 10 km I am asked to calculate the area of a regular octagon given the side-length of 10 km.  
I saw some examples saying that I should start by splitting the octagon into eight isosceles triangles, and that the length of the base would be 10 km, since we're given that the sides of the octagon are all 10km.  What I don't know is what to do next.  
 A: I think that a different approach is easier: try cutting it up as in the rough sketch below. Once you work out the lengths of the legs of the right triangles, you should be able to calculate the area pretty easily.

A: I'd like to suggest an alternate strategy.  A regular octagon has angles of 135$^o$.  I don't know if it's true where you're from, but in the US, stop signs are octagon shaped.  So picture a stop sign, or look at the drawing in Brian's answer.  Now extend the top, bottom, left, and right sides until they meet to form a square with a 45-45-90 right triangle at each corner.  The hypoteneuse of each triangle is a side of the octagon.
Can you finish it from here, calculating the area of the square and subtracting the area of the 4 triangles?
A: The following trigonometric approach is less elegant (and somewhat harder) than the ones that have been suggested. Its main advantage is that something like it works for any regular polygon. 
Let $P$ be the point at the centre of the octagon. Join $P$ to all the vertices. We have divided the octagon into $8$ triangles. Now we find the area of one of these triangles, and multiply by $8$.
Let $PAB$ be one of our $8$ triangles. Draw a perpendicular from $P$ to $AB$. Suppose that this perpendicular meets $AB$ at $Q$. Then the area of $\triangle PAB$ is $\frac{1}{2}(AB)(PQ)$.  We know that $AB=10$, and we need to find $PQ$.
Angle $APQ$ is $360^\circ/8$, that is, $45^\circ$. So the two equal angles  $PAQ$ and $PBQ$ add up to $180^\circ-45^\circ$, that is, $135^\circ$. So each of them is $67.5^\circ$.
Note that
$$\frac{PQ}{AQ}=\tan(\angle PAQ)=\tan(67.5^\circ).$$
But $AQ=5$, and therefore
$$PQ=5\tan(67.5^\circ).$$
So the area of $\triangle PAB$ is $\frac{1}{2}(10)(5\tan(67.5^\circ))$. Multiply by $8$, and simplify. We get that the octagon has area $200\tan(67.5^\circ))$. If you want a decimal approximation, the calculator will give you one. 
A: Area of a $n$-sided regular polygon, $$A=\frac{na^2\cot(\pi/n)}{4}$$ where $a$ is the side length. For your case, $n=8$ and $a=10$ which gives $A=200\cot(\pi/8) km^2$. 
A: Use The equation :
A = .5asn

This works with all regular polygons.
a is the length of the apothem(Perpendicular bisector of one side to the center point of the polygon)
s is the length of each side
n is the number of sides
This is really is splitting the polygon into triangles where the apothem is the height of said triangle and each side of the polygon is the base of that triangle. n is just the number of sides/number of triangles
