December 31st, 2018, 07:44 AM  #1 
Member Joined: May 2018 From: Idaho, USA Posts: 65 Thanks: 7  Geometry Question
Hello, So, first thing you have to understand is that in my spare time I like to experiment with math problems. What experiment I have been thinking about for a while, is if you had an irregular polygon, how would you find the area? Yesterday, I was on my calculator just while thinking about this and I did something that actually worked for finding the area. Truth be told, I have no idea how I got it to work, so I was wondering if you guys could help me figure out how this worked. I love geometry, and I love studying and experimenting with it. Anyway, here are the specs on the polygon I experimented with. Side A = 5 Side B = 5 Side C = 3 Side D = 3 Side E = 2 Angle AB = 60 Angle BC = 120 Angle CD = 60 Angle DE = 120 Angle EA = 60 Basically made up of two equilateral triangles. Here's what I did first to find the answer, then I started experimenting. I took one part of the polygon, and found its area. I did it with the next part, then I added them together. According to what I did on paper, the area equals 14.72243186. Then, I started experimenting, thinking there was another way to find the area. I did something interesting, and I don't know how it worked. Here is what I did. A sin (AB) + B sin (BC) + C sin (CD) + D Sin (DE) + E sin (EA) This equals 15.58845727 according to my calculator. Not the answer I was looking for. I then looked at the polygon, and noticed angle DE went in toward the center of the polygon. I then thought to myself, " What would happen if I added the sine of 360 minus Angle DE?" On a hunch, I added 1sin (240) to 15.58845727. 1sin (240) equals 0.866025404. After doing this, the answer I got was the same as before, 14.72243186. My hunch worked. I have no idea how it worked. Does anyone know how this worked? I am puzzled by this. Jared 
December 31st, 2018, 10:05 AM  #2 
Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147 
Well I'm confused. You have described a five sided figure (a pentagon) whose five interior angles 60, 120, 60, 120, and 60 add up to 420 degrees. You're 120 degrees short.

December 31st, 2018, 10:22 AM  #3 
Member Joined: May 2018 From: Idaho, USA Posts: 65 Thanks: 7 
Sorry, I put down one of the numbers wrong. This might help explain it. Imagine a triangle that has three side lengths all equaling five, and another with all three side lengths equaling three. Connect them at their bases so it is a Pentagon and the side lengths are 5, 5, 3, 3, and 2. Angle DE on the OUTSIDE is 120. Little mistake there on my part. I apologize. Would it help if I showed a picture on paper? Jared 
December 31st, 2018, 10:28 AM  #4 
Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147 
OK I think I've got it. The area is $\displaystyle \frac{17 \sqrt{3}}{2}$ or about $\displaystyle 14.72243186$

December 31st, 2018, 10:43 AM  #5 
Member Joined: May 2018 From: Idaho, USA Posts: 65 Thanks: 7 
That's it. Thank you. What I did was this: 5 sin (60) + 5 sin (120) + 3 sin (60) + 3 sin the OUTER ANGLE DE, which is 120, in other words 3 in (120) + 2 sin (60) I got what I said before, 15.58845727. I added 1 sine the INNER angle DE, which is 240. In other words, 1 sin (240) After adding this, I got the same answer as you. 14.72243186. Well, it looks like there is more than one way to find the area! Thanks for the help. Jared 
January 1st, 2019, 05:06 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,968 Thanks: 2216 
Try using lengths of, say, 7, 5 and 2 instead of 5, 3 and 2. The area is then (7² + 5²)√3/4 instead of (5² + 3²)√3/4. 
January 1st, 2019, 07:36 AM  #7 
Member Joined: May 2018 From: Idaho, USA Posts: 65 Thanks: 7 
Just tried what you said. The area should equal 32.04293994, correct? You were saying to have the same number of sides and the same angles, but different side lengths, correct? When I did what I did before with all the sines, it didn't work. I don't know how it worked when I had the side length of 5, 3, and 2. It doesn't work with these side lengths now. I have no idea how I made it work before. Anyway, I appreciate the help with this. Jared 

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