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Circumscribed and inscribed circles of a regular hexagon?

  1. Sep 18, 2004 #1
    Hey ppl,
    Could anyone help me with this: what is the ratio of the areas of the circumscribed and inscribed circles of a regular hexagon? how do I go about working it out from first principles?
    Cheers, joe
  2. jcsd
  3. Sep 18, 2004 #2


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    Well for any regular polygon its "apotema" (apothem perhaps?) is

    [tex] a = \frac{L}{2}tan \frac{\alpha}{2} [/tex]

    where [tex] L [/tex] is the length of a side and [tex] \alpha [/tex] is the inner angle of the polygon.

    To calculate this inner angle just use the formula

    [tex] \alpha = \frac{(n-2)180}{n} [/tex]

    where n is the number of sides

    and its Area is

    [tex] A = \frac{1}{2}pa [/tex]

    where p is the perimeter or nL

    Now if i remember circumscribed correctly means a circle inside the hexagone and inscribed means a hexagone inside the circle, right?

    maybe this could be calculated with right triangles...anyhow the formulas above could help you for a regular hexagone

    Edit: sorry for so many edits, seems i need a break.
    Last edited: Sep 18, 2004
  4. Sep 18, 2004 #3
    The regular hexagon consists of six equilateral triangles, the radius of the inscribed circle is equal to the height (i.e. distance from top to middle of base) of one such triangle and the radius of the circumscribed circle is equal to the length of a side of one such triangle.
  5. Sep 18, 2004 #4


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    In other words, the ratio of radii is clearly equal to the tangent of 30.
  6. Sep 18, 2004 #5
    Or you just use phytagoras to see that:

    heigth^2 + (1/2 * side)^2 = side^2
    heigth^2 = side^2 - 1/4 * side^2
    height^2 = 3/4 * side^2

    and then:

    area of circumscibed circle:
    2Pi * side^2
    area of inscribed circle:
    2Pi * 3/4 * side^2

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