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Math Help - Reuleaux Pent. Area

  1. #1
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    Reuleaux Pent. Area

    Hi, I having trouble trying to find the area of a regular, pointed Reuleaux pent. Like this in the picture: http://www.squidzone.ca/photos/uncat...uxpentagon.jpg Thanks in advance
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  2. #2
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    A Reuleaux triangle consists of an inscribed equilateral trinagle and 3 circular segments. The area of an equilateral triangle is given by \frac{\sqrt{3}}{4}s^{2}. Where s is the length of a side.

    The area of a circular segment is \frac{1}{2}r^{2}({\theta}-sin{\theta}). There are three of those and the angle is, of course, Pi/3 or 60 degrees.

    Add them up and you have your area. Now, you can use the same reasoning for a pentagon. Only you have 5 circular segments and use the area of an inscribed pentagon. The angles for the polygon will be 360/5=72.
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    Hi,
    Thank you for your help so far.

    I have to keep the final answer in terms of h, the constant width
    In a triangle, the constant width is a side of a the triangle so for the area of a triangle I used h^2 instead of  s^2

    I was able to find the area of a Reuleaux Triangle by:
    sector + 2 segments
    sector + 2(sector - Triangle)
    sector + 2 sector - 2 Triangle
    3 sectors - 2 Triangles
    3(\frac{{60}}{360}{\pi}h^2)- 2(\frac{\sqrt{3}}{4}h^{2})
    3(\frac{{1}}{6}{\pi}h^2)- 2(\frac{\sqrt{3}}{4}h^{2})
    \frac{{1}}{2}{\pi}h^2 - \frac{{1}}{2}h^2\sqrt{3}
    \frac{{1}}{2}h^2( \pi - \sqrt{3})

    However I am stuck on the pentagon. I have to find the area in terms of h, the constant width. AD=h
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  4. #4
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    One way to find the area of a pentagon is to count up the area of the 5 isosceles triangles that make up the pentagon.

    The area of one of those is \frac{1}{2}r^{2}sin{\theta}

    But there are 5 of them, so we have \frac{5}{2}r^{2}sin(\frac{2\pi}{5})

    Now, all you need is the area of the circular segments.
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    I still think I'm doing something wrong, I am getting the wrong answer.
    With h=5 the answer should be 18.96 rounded.

    5 Iso. Tri's + 5 segments
    [\frac{5}{2}h^{2}sin\frac{2\pi}{5}] + [\frac{5}{2}h^{2}(\frac{\pi}{5}-sin\frac{\pi}{5})]

    I am getting 6.02 rounded

    What am I doing wrong?

    thanks
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  7. #7
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    The center of each arc of a Reuleaux is at the vertex of the pentagon...not in the center. See what I mean?. If the center of the pentagon were the center of the arc you would just have a circle. With the triangle it works OK, but with the pentagon, the circular arc is a little less than it would be if it were centered at the center of the pentagon. Here is a hackneyed diagram so you can see what I mean. Pardon my lop-sided pentagon.
    Last edited by galactus; November 24th 2008 at 06:38 AM.
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