Expressing Area of polygon using trigonometry

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meesh_was_here

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May 23, 2006
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I have absolutely no idea how to do the following:

"In the diagram, ABCDEF....is a regular polygon with n sides and apothem OM. Let OD=r.

a. Express m<(angle)COM in terms of n. (I got .5n=<COM)

b. Use your answer to part (a) to express the area of the polygon in terms of n and r, using trigonometry."

So just let me know if I got part (a) right and please help to figure out what part (b) is. Thanks a bunch!
 
I was given it because it's one of many problems that I have to do for a project in my geometry class. I think I got it after I had posted it, but now I just want to check myself. Why? Do you know how to do it?
 
meesh_was_here said:
I think I got it after I had posted it, but now I just want to check myself. Why? Do you know how to do it?
Click to expand...
So post what you think is the answer.
Let us check it.
 
Area of any polygon (A=1/2ap) therefore a=r cos 360/2n and p=2n x sin 360/2n area of polygon=1/2(r cos 360/2n)(2n x sin 360/2n). If you don't understand anything that I wrote just let me know, but I think I did it right, sooo, yeah.
 
Are you missing a factor of r in the last term?
I get \(\displaystyle \L
\quad\frac{{nr^2 }}{2}\sin \left( {\frac{{2\pi }}{n}} \right)\).
 
meesh_was_here said:
I have absolutely no idea how to do the following:

"In the diagram, ABCDEF....is a regular polygon with n sides and apothem OM. Let OD=r.

a. Express m<(angle)COM in terms of n. (I got .5n=<COM)

b. Use your answer to part (a) to express the area of the polygon in terms of n and r, using trigonometry."

So just let me know if I got part (a) right and please help to figure out what part (b) is. Thanks a bunch!
Click to expand...

If you draw the radii from the center of the regular polygon O to each vertex of the n-sided polygon, you'll get "n" isosceles triangles, each with its vertex angle at O.

Since the angles about O must total 360 degrees, and you have "n" of them, then the measure of each of those central angles should be 360/n degrees.

Now, if you draw an apothem OM in one of those isosceles triangles, it will divide the vertex angle of that triangle in half, and m<COM will be (1/2)(360/n), or 180/n.

Since OD is a radius of the polygon, and OM is an apothem, OM/OD = cos (180/n).
So, OM = OD * cos (180/n)

I'll let you work on this for a while.....
 
Can you guys try to explain it to me in simpler terms? I really have a lot of trouble understanding all of this stuff when someones sitting right in front of me. Thanks for all of your help! This is helping me so much, really! :p
 
meesh_was_here said:
I really have a lot of trouble understanding all of this stuff when someones sitting right in front of me.
Click to expand...
Well, tell that "someone" to move :roll:
 
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