Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was trying to find the area of a regular polygon in terms of n, the side length and s, the number of sides.

Because there are $s$ sides number of isosceles triangles in a regular polygon, I decided to work out the area of an isosceles triangle in terms of $A$, the unique angle and $a$, the unique side: $$Area=\frac{1}{2}ab\sin{C}$$ $$b=\frac{\sin{C}\times a}{\sin{A}}$$ (Sine rule, $C=B$) $$b=\frac{\sin{\frac{180-A}{2}}\times a}{\sin{A}}$$ $$Area=\frac{1}{2}a\times\frac{\sin{C}\times a}{\sin{A}}\times\sin{C}=\frac{{(\sin{C}\times{a})}^{2}}{\sin{A}}$$ $$C=\frac{180-A}{2}=90-\frac{A}{2}$$ $$Area=\frac{{(\sin({90-\frac{A}{2}})\times{a})}^{2}}{\sin{A}}$$ And that was where I got to in finding the area of an isosceles triangle. Then I tried to find the area of the whole regular polygon: $$Area=s\times\frac{{(\sin({90-\frac{A}{2})}\times{n})}^{2}}{\sin{A}}$$ Where $s$ is the number of sides and $n$ replaces $a$ $$A=\frac{360}{s}$$ $$Area=s\times\frac{{(\sin({90-\frac{\frac{360}{s}}{2}})\times{n})}^{2}}{\sin{\frac{360}{s}}}=s\times\frac{\sin^{2}({90-\frac{180}{s}})\times{n}^{2}}{\sin{\frac{360}{s}}}$$ $$Area=\frac{s{n}^{2}\sin^{2}({90-\frac{180}{s}})}{\sin{\frac{360}{s}}}$$ However, when I tested this formula it was wrong. Can someone tell me where I've gone wrong?

share|improve this question

2 Answers 2

$$Area=\frac{1}{2}a\times\frac{\sin{C}\times a}{\sin{A}}\times\sin{C}=\frac{{(\sin{C}\times{a})}^{2}}{\sin{A}}$$

It seems that you dropped $\frac{1}{2}$ here. :-)

share|improve this answer

enter image description here

Here's a kind of different approach:

Let $n$ be number of sides.

$\angle DAC= \dfrac{180}{n}$

$\tan \angle DAC=\dfrac{x/2}{AD} \implies AD= \dfrac{\frac{x}{2}}{\tan \frac{180}{n}} \implies AD=\dfrac{x}{2 \tan \frac{180}{n}}$

Call $\dfrac{180}{n}=p$

Area of $\triangle ABC= \dfrac{x^2}{4 \tan p}$, you will have $n$ triangles in $n$-sided regular polygon. Area of polygon=$\dfrac{x^2 \times n}{4 \tan p}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.