Area of a Regular Polygon

**Quick** · May 31st 2006, 04:35 PM

I was trying to find an equation to find the area of any regular polygon and I came up with this:
$A=\frac{tan(90-\frac{180}{s})d^2s}{4}$

$s$ =# of sides
$d$ =length of each side
and $90-\frac{180}{s}$ comes out in degrees

I think this equation works but I'm not sure, does it? Also, is there an easier equation out there?

**ThePerfectHacker** · May 31st 2006, 05:23 PM

$s=\mbox{ length of sides }$
$n=\mbox{ # 0f sides }$

1)Consider your regular polygon to be inscribed in a circle.

2)Now from the vertices draw lines to the center.

3)You form $n$ congruent triangles.

4)The central angle is $\frac{360}{n}$ .

5)Let $r$ we the radius of the circle.

6)Then you need to find the area of a triange having sides $r,r$ with angle included of $\frac{360}{n}$ . Apply the area formula* you have, $\frac{1}{2}r^2\sin \left(\frac{360}{n} \right)$ .

7)In total you have, $\frac{1}{2}nr^2\sin \left( \frac{360}{n} \right)$ .

8)Now we find what $r$ is.

9)Using law of cosine we have, $s^2=r^2+r^2-2r^2\cos \left( \frac{360}{n} \right)$

10)Thus, $s^2=2r^2\left(1-\cos \left( \frac{360}{n} \right) \right)$

11)Apply half angle formula thus,
$s^2=4r^2\sin^2 \left(\frac{180}{n} \right)$

12)Thus, $r^2=\frac{s^2}{4\sin^2 \left( \frac{180}{n} \right) }$

13)Substitute into 7: $\frac{1}{2}\frac{s^2\sin \left( \frac{360}{n} \right)}{4\sin^2 \left( \frac{180}{n} \right) }$

14)Use double angle, $\frac{1}{2}\frac{2s^2\sin \left( \frac{180}{n} \right)\cos \left( \frac{180}{n} \right)}{4\sin^2 \left( \frac{180}{n} \right) }$

15)Thus we have, $\frac{1}{4}s^2\cot \left( \frac{180}{n} \right)$

16)But $\cot x=\tan (90-x)$ thus,

17)Thus, $\frac{1}{4}s^2\tan \left(90- \frac{180}{n} \right)$

*)If a triangle has sides A and B with included angle C then is area is (1/2)A*B*sinC

**Soroban** · June 2nd 2006, 09:27 AM

Hello, Quick!

I was trying to find an equation to find the area of any regular polygon
and I came up with this: $A \:=\:\frac{s^2n}{4}\tan\left(90 -\frac{180}{n}\right)$

$n$ = number of sides
$s$ = length of each side
and $90 - \frac{180}{s}$ is degrees

I think this equation works but I'm not sure . . . does it? . . . Yes!

I think I know how you reasoned this out . . . Nice work!

Code:

            A
            *
           /:\
          / : \
         / θ:θ \
        /   :   \
       /    :h   \
      /     :     \
     /      :      \ 
    * - - - * - - - *
    B       D  s/2  C

We have $n$ isosceles triangle as shown above.

$\angle BAC = \frac{360^o}{n}\quad\Rightarrow\quad \theta = \frac{180^o}{n}$

The height $h$ of the triangle is: $h \;= \;\frac{s}{2}\cdot\tan C \;= \;\frac{s}{2}\cdot\tan(90^o - \theta)$

The area of the triangle is: $A_{\Delta}\;=\;\frac{1}{2}bh\;=\;\frac{1}{2}(s)[\frac{s^2}{2}\tan(90^o - \theta)] \;=\;\frac{s^2}{4}\tan(90 - \theta)$

And $n$ triangles have an area of: $A\;=\;n \times \frac{s^2}{4}\tan(90^o - \theta) \;= \;\frac{s^2n}{4}\tan\left(90^o - \frac{180^o}{n}\right)<br />$

**Quick** · August 2nd 2006, 06:48 PM

You know what I find ironic about this thread, both of your explanations go way beyond my understanding of geometry at that time (now I know everything but theta), but because of this thread I think you guys thought I was somewhere in the middle to the ending of my year of geometry.

but getting to my question: what is theta?

**ThePerfectHacker** · August 2nd 2006, 07:08 PM

Originally Posted by Quick

but getting to my question: what is theta?

Would do you mean? What it means for in this problem? I did not use it, Soroban did. It is the angle of that triangle he drew. Or why is it used? Because mathematicians for some reason decided to use theta when it comes to measuring angles. I usually do not use it. One reason it takes to long in LaTex. Theta is charachter from the divine language (Greek).

**c_323_h** · August 2nd 2006, 08:31 PM

$\theta$ (theta) is often used as a symbol for an unknown angle in trig/geometry.

For example the sine of theta, which is the same as saying "the sine of an angle":

$sin\theta=\frac{opposite}{hypotenuse}$

This gives you $sin\theta$ . If you wanted to find $\theta$ , that is, the angle, you would have to take the inverse sine.

$\theta=sin^{-1}\bigg(\frac{y}{r}\bigg)$

where $y$ is the opposite side and $r$ is the hypotenuse.

If you were referring to theta in the problem, then forget this post. I just wanted to show off my superior trig skills so I can compensate for my crappy calculus skills

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