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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- May 30th 2008, 10:55 PMbulldog106Reuleaux 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

- May 31st 2008, 06:10 AMgalactus
A Reuleaux triangle consists of an inscribed equilateral trinagle and 3 circular segments. The area of an equilateral triangle is given by $\displaystyle \frac{\sqrt{3}}{4}s^{2}$. Where s is the length of a side.

The area of a circular segment is $\displaystyle \frac{1}{2}r^{2}({\theta}-sin{\theta})$. 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. - May 31st 2008, 09:31 AMbulldog106
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 $\displaystyle h^2 $ instead of $\displaystyle s^2$

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

$\displaystyle 3(\frac{{60}}{360}{\pi}h^2)- 2(\frac{\sqrt{3}}{4}h^{2})$

$\displaystyle 3(\frac{{1}}{6}{\pi}h^2)- 2(\frac{\sqrt{3}}{4}h^{2})$

$\displaystyle \frac{{1}}{2}{\pi}h^2 - \frac{{1}}{2}h^2\sqrt{3}$

$\displaystyle \frac{{1}}{2}h^2( \pi - \sqrt{3})$

However I am stuck on the pentagon. I have to find the area in terms of h, the constant width. AD=h

http://img164.imageshack.us/img164/1701/pentnw8.jpg - May 31st 2008, 10:09 AMgalactus
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 $\displaystyle \frac{1}{2}r^{2}sin{\theta}$

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

Now, all you need is the area of the circular segments. - May 31st 2008, 10:14 AMbulldog106
Thanks

- May 31st 2008, 11:02 AMbulldog106
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

$\displaystyle [\frac{5}{2}h^{2}sin\frac{2\pi}{5}] + [\frac{5}{2}h^{2}(\frac{\pi}{5}-sin\frac{\pi}{5})]$

I am getting 6.02 rounded

What am I doing wrong?

thanks - May 31st 2008, 12:11 PMgalactus
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.(Doh)