Results 1 to 2 of 2

Math Help - Area of outter hexagon in circle

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    108

    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}.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1393

    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 \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*}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Circle and hexagon problem.
    Posted in the Geometry Forum
    Replies: 2
    Last Post: April 17th 2012, 08:35 PM
  2. Hexagon inscribed in circle gr.10
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 30th 2011, 03:41 AM
  3. Hexagon around circle
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: February 7th 2010, 11:18 AM
  4. A Circle Circumscribed about a hexagon
    Posted in the Geometry Forum
    Replies: 3
    Last Post: January 12th 2010, 06:57 PM
  5. A Circle Circumscribed about a hexagon
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: January 12th 2010, 05:30 AM

Search Tags


/mathhelpforum @mathhelpforum