# Area of outter hexagon in circle

• May 15th 2012, 06:49 PM
johnsy123
Area of outter hexagon in circle
Consider a circle with radius r. How do i work out the area of the circumscribe hexagon? I have managed to work out the area of the inner to be 3asqrt{r^2-a^2}.
• May 15th 2012, 07:21 PM
Prove It
Re: Area of outter hexagon in circle
Quote:

Originally Posted by johnsy123
Consider a circle with radius r. How do i work out the area of the circumscribe hexagon? I have managed to work out the area of the inner to be 3asqrt{r^2-a^2}.

Inner hexagon:

http://i22.photobucket.com/albums/b3...n/hexagon3.jpg

You can see that the radius of the circle is the length of each equilateral triangle, so evaluating the area of each equilateral triangle using Heron's Formula:

\displaystyle \displaystyle \begin{align*} s &= \frac{r + r + r}{2} \\ &= \frac{3r}{2} \\ \\ A &= \sqrt{s(s - r)(s - r)(s - r)} \\ &= \sqrt{\frac{3r}{2}\left(\frac{3r}{2} - r\right)\left(\frac{3r}{2} - r\right)\left(\frac{3r}{2} - r\right)} \\ &= \sqrt{\frac{3r}{2} \cdot \frac{r}{2} \cdot \frac{r}{2} \cdot \frac{r}{2}} \\ &= \sqrt{\frac{3r^4}{16}} \\ &= \frac{\sqrt{3}\,r^2}{4}\end{align*}

So the area of the hexagon is

\displaystyle \displaystyle \begin{align*} 6\cdot \frac{\sqrt{3}\,r^2}{4} &= \frac{3\sqrt{3}\,r^2}{2}\textrm{ units}^2 \end{align*}

Outer hexagon:

http://i22.photobucket.com/albums/b3...n/hexagon2.jpg

You can see that the radius of the circle is the height of each equilateral triangle, and splits each equilateral triangle into two triangles with the sides in the ratio \displaystyle \displaystyle \begin{align*} 1, \sqrt{3}, 2 \end{align*}. So the if the height of each triangle is \displaystyle \displaystyle \begin{align*} r \end{align*}, then the base is \displaystyle \displaystyle \begin{align*} \frac{2r}{\sqrt{3}} \end{align*}.

Therefore the area of each triangle is \displaystyle \displaystyle \begin{align*} \frac{1}{2}\cdot r \cdot \frac{2r}{\sqrt{3}} = \frac{r^2}{\sqrt{3}}\end{align*}, and therefore the area of the hexagon is

\displaystyle \displaystyle \begin{align*} 6\cdot \frac{r^2}{\sqrt{3}} &= \frac{6r^2}{\sqrt{3}} \\ &= \frac{6\sqrt{3}\, r^2}{3} \\ &= 2\sqrt{3}\,r^2 \textrm{ units}^2\end{align*}