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Area of a hexagon

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data

    A circle of radius r is impressed in a hexagon. Find the area of the hexagon.

    2. Relevant equations

    Area of a triangle = (1/2)bh

    3. The attempt at a solution

    The hexagon can be split up into six triangles, and with the formula for the area of a triangle, becomes (6)(1/2*bh).

    Does the circle in the problem even matter? It makes it seem like there is more to the problem than there really is. Is there something I overlooked?
  2. jcsd
  3. Mar 31, 2008 #2
    what do you mean with : a circle of radius r is impressed in a hexagon? I think you have to use the radius of the circle on your answer. But if you could explain this part a lill bit more, what do you mean by that.
  4. Mar 31, 2008 #3
  5. Mar 31, 2008 #4


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    Do you mean "inscribed" instead of "impressed"? This question (post #1) is not really a Calculus problem; did you intend it to be for Calculus? Without using Calculus, this is a very simple exercise needing only intuitive knowledge of pattern in a shape and the pythagorean theorem.
  6. Mar 31, 2008 #5


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    It's just the sum of the areas of six identical triangles, each with two legs of length r and a central angle of 360 / 6 = 60 degrees. It's basic plane geometry.

    - Warren
  7. Mar 31, 2008 #6
    well, i think what they are asking you to do is express the area of the hexagon in terms of the radius of the circle. Your formula is right, just try to express the height of the triangles in terms of the radius, and use the letter r. Also notice that all three sides of the triangles are of size r.
  8. Apr 1, 2008 #7
    It's meant to be a Calculus problem, or at least it's on my Calculus homework.

    Alright, I went back and re-did this problem and got (3/2)r^2(sqrt(3)). I solved for the area of one equilateral triangle using r as one of the sides, multiplied it by six, and then reduced it to what is above.
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