My Math Forum Geometry Question

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 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
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OK I think I've got it. The area is $\displaystyle \frac{17 \sqrt{3}}{2}$ or about $\displaystyle 14.72243186$
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 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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