anonymous
  • anonymous
A hexagon is inscribed in a circle. If the difference between the area of the circle and the area of the hexagon is 36 m2, use the formula for the area of a sector to approximate the radius r of the circle. (Round your answer to three decimal places.)
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
A buddy is in the same class and we both had trig in high school, so we are wondering how everyone else is doing since we're having so much trouble. I managed to finally figure out how to solve using sectors only it requires knowing the relationship between the sides of a 30-60-90 triangle, which is a couple chapters down the road. This is my work: \[A=\frac{1}{2}r^{2}θ\] \[A=12rh\] \[\frac{1}{2}r^{2}θ−\frac{1}{2}rh=36\] \[\theta = \frac{\pi}{3}\] \[\frac{1}{2}r^{2}(\frac{\Pi}{3})−\frac{1}{2}rh=36\] \[h=\frac{\sqrt{3}}{2}r\] \[\frac{1}{2}r^{2}(\frac {\Pi}{3})−(\frac{1}{2})r(\frac{\sqrt{3}}{2})r=36\] \[\frac{1}{2}r^{2}(\frac{\Pi}{3})−\frac{1}{2}(\frac{\sqrt{3}}{2})r^{2}=36\] \[\frac{1}{2}r^{2}[(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})]=36\] \[6(\frac{1}{2}r^{2}[(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})])=36\] \[3r^{2}[(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})])=36\] \[r^{2}[(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})])=12\] \[r=\sqrt{12 \div [(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})])}\] \[r \approx 8.139\] While this works, is there a way to do it with trigonometric functions, but still apply sector theory?
goformit100
  • goformit100
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AkashdeepDeb
  • AkashdeepDeb
It is a regular hexagon right?

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anonymous
  • anonymous
Yeah, that's assumed from the problem given, so each of the six triangles formed from the hexagon would be equilateral triangles with 60 degree angle measures.
AkashdeepDeb
  • AkashdeepDeb
Yes excatly! :D
AkashdeepDeb
  • AkashdeepDeb
And then the area of the circle would be pi*r*r - area of hexagon pi*r*r - [3 sqrt 3*r^2/2] = 36 Just find this and you'll get radius!! :D
anonymous
  • anonymous
That's what I did with my work shown above. I was wondering if there was a way to do it without involving any roots and by using trig functions because our book technically hasn't covered 30-60-90 triangles yet.
AkashdeepDeb
  • AkashdeepDeb
Yes you can do it but this method is the EASIEST and the fastest. And where would you require 30-60-90 triangles? :D
anonymous
  • anonymous
I did it when I split the equilateral triangle in half to find h in terms of r and got h = (sqrt3/2)r
AkashdeepDeb
  • AkashdeepDeb
Hahah...But why? :') You already have a direct formula for equilateral triangle = sqrt 3*a^2/4 Why do you want to slpit it? XD
anonymous
  • anonymous
. . . I like right triangles :D I usually end up breaking things down too far and over-analyze them. It was making a lot of sense to break it that way even if its the long way.
AkashdeepDeb
  • AkashdeepDeb
Hahahaha!! LOL But you understood this now right? :D
anonymous
  • anonymous
Not really, I remember getting that equation before and I was wondering how the pi*r*^2 part worked because the formula for the area of a sector has a measure of theta involved.
dan815
  • dan815
|dw:1378103963122:dw|
anonymous
  • anonymous
Man, my algebra skills aren't that sharp to be trying to solve that . . . Lol.
AkashdeepDeb
  • AkashdeepDeb
|dw:1378104178575:dw|
anonymous
  • anonymous
Yeah, I understand how the shapes break down, how to get to that point, and eventually to the answer, but that's not exactly what I'm struggling with. What I want to know is if there is a way to figure this out using sines and cosines if you break down one equilateral triangle for example.
AkashdeepDeb
  • AkashdeepDeb
The thing is you obviously can break it down to two 30-60-90 triangles and then find their areas [NO ISSUE] but you don't really NEED to do that. as you already have a direct formula for finding the area of an equilateral triangle. Getting me @jlangley95 ?
AkashdeepDeb
  • AkashdeepDeb
If you want to do it with trigonometry I can show it to you! Do you want me to do it with sines and cosines?
anonymous
  • anonymous
That might be helpful. I see how the way you've already shown will work, but the only thing is, I'm not good at doing the algebra to solve it.
anonymous
  • anonymous
Because I've got: \[2\Pi r^{2}=\sqrt{3}r ^{2}=62\] and don't know where to go from there
anonymous
  • anonymous
That first = should be a -
AkashdeepDeb
  • AkashdeepDeb
Haha..Chill look at this now. :D |dw:1378105164136:dw| BC = r AD = sqrt 3 * r/2 Area = sqrt 3 * r^2/4 Understood this? :)
anonymous
  • anonymous
Yeah, it makes sense. The only thing is the book has not talked about any of this in the chapter we're on. We literally are not supposed to know that the short leg on a 30-60-90 is 1/2 the hypotenuse or that the longer leg is sqrt3 times the shorter leg. I realize that this is a very essential relationship between the sides of this type of triangle, but is it possible at all to do this problem without knowing at least that much, and if not, then how could you derive those values?
AkashdeepDeb
  • AkashdeepDeb
Oh cool! It is good that you know it now!! :D Well see there are 3 ways of solving this. 1) By using the equilateral triangle area formula [I already showed you] 2) Pythagorus theorem [To find the height after dividing it] 3) And by Trigonometry/30-60-90 triangle method which I just showed you. :D That is pretty much it! :)
anonymous
  • anonymous
Alright! Thanks! Hopefully I can explain this to my friend
AkashdeepDeb
  • AkashdeepDeb
:)

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