Using trigonometric ratios to express area of regular polygons I am very confused.  My book just asked me to use trigonometric ratios to express the area of a regular polygon with 9 sides and lengths of 8.  
I don't even know what this means.
So far I have learned how to use the sin cos and tan in right triangles and have no idea how this applies to all polygons.  Can someone please explain this to me :)
 A: If the polygon has 9 sides, the angle of each of the 9 sectors is $$\frac{2\pi}{9}$$ If you cut this angle in half and draw an altitude to one of the sides, you get a right triangle with base 4 (half of 8) and an angle of $$\frac{2\pi}{18}$$ (the angle is also cut in half). So the area of that piece is base*height/2 or $${1\over 2}(4)(4)\cot\left(\frac{2\pi}{18}\right)$$ The area of the whole polygon is 18 times that, so $$A=8\cdot 18\cot\left(\frac{2\pi}{18}\right)$$ and I'm bad at mental math so I'll let you finish simplifying.
A: It is difficult for me to insert a picture in the answer, so you will need to read this with pencil in hand.
Sketch a regular $9$-gon. Join the centre of the $9$-gon to the $9$ vertices. This divides the $9$-gon into $9$ congruent triangles. To find the area of the $9$-gon, we will find the area of one of these triangles, and multiply the result by $9$.
Concentrate now on one of these triangles $AOB$, where $O$ is the centre and $A$ and $B$ are two consecutive vertices. We will find the area of $\triangle AOB$.
We have $AB=8$, and $\angle AOB=\frac{360^\circ}{9}=40^\circ$.
Drop a perpendicular from $O$ to $AB$, meeting $AB$ at $M$. Note that $\angle AOM=20^\circ$. 
Also, $\frac{AM}{OM}=\tan(20^\circ)$. But $AM=4$, so $OM=\frac{4}{\tan(20^\circ)}$.
Now we know the base $AB$ of $\triangle AOB$, and the height $OM$. The area of $\triangle AOM$ is therefore $\frac{1}{2}\cdot 8\cdot \frac{4}{\tan(20^\circ)}$. 
Finally, multiply by $9$ and simplify a bit. We get $\frac{144}{\tan(20^\circ)}$, or if you prefer, $144\tan(70^\circ)$. 
