# Thread: Area of outter hexagon in circle

1. ## 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}.

2. ## Re: Area of outter hexagon in circle

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:

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

Outer hexagon:

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 \begin{align*} 1, \sqrt{3}, 2 \end{align*}. So the if the height of each triangle is \displaystyle \begin{align*} r \end{align*}, then the base is \displaystyle \begin{align*} \frac{2r}{\sqrt{3}} \end{align*}.

Therefore the area of each triangle is \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 \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*}

### Area of an isosceles triangle by heron's formula

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