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Thread: Area of outter hexagon in circle

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    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}.
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    Re: Area of outter hexagon in circle

    Quote Originally Posted by johnsy123 View Post
    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:

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

    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:



    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

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