anonymous
  • anonymous
Geometry question w/ a Circle. find area of a shaded region. (Harder problem)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1345779977023:dw|
anonymous
  • anonymous
I think what you have to do is to bisect the triangle into 60 degree up top and then the bottom left and right are still 30 degrees. then you do the 30 60 90 triangle thing.
anonymous
  • anonymous
i am stuck from there

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alexwee123
  • alexwee123
i did it a different way...
anonymous
  • anonymous
well explain.
alexwee123
  • alexwee123
with integrals...
anonymous
  • anonymous
well. it shouldn't be that complicated im only doing geometry.
Mimi_x3
  • Mimi_x3
Area of minor segment \[A = \frac{1}{2} r^2 \left(\theta-\sin\theta\right) \] But the only problem is \(r\) is not given..
Hero
  • Hero
I think I meant to post the area of a sector of a circle
anonymous
  • anonymous
you guys are thinking too hard its only like subtracting a sector
Mimi_x3
  • Mimi_x3
Area of sector - area of triangle
Hero
  • Hero
Yeah that
Mimi_x3
  • Mimi_x3
But \(r\) is not given!!! Im thinking of using the cosine rule and sine rule to find \(r\) would it work?
anonymous
  • anonymous
here ill take it a bit further. |dw:1345780578744:dw| does that help?
Hero
  • Hero
It will not be hard to find \(r\)
Hero
  • Hero
Is the vertex of angle 120 is the center of the circle?
anonymous
  • anonymous
yep
Mimi_x3
  • Mimi_x3
|dw:1345780720301:dw| looks like you can use the sine rule..
Hero
  • Hero
Okay, so yeah just do Area of Sector - Area of Triangle like what @Mimi_x3 suggested
anonymous
  • anonymous
soo all we need is the area of the circle and then we can use the 120/360 * pi r2 or something and subtract the area from that.
Mimi_x3
  • Mimi_x3
lets find \(r\) first
anonymous
  • anonymous
|dw:1345780842471:dw|
Mimi_x3
  • Mimi_x3
I got 8 as well using the sine rule
Mimi_x3
  • Mimi_x3
Now its easy..you can do it yourself
anonymous
  • anonymous
ok so 64 pi is the area of the circle
anonymous
  • anonymous
and the area of the triangle left side is 1/2 * 4sqroot3 * 4
anonymous
  • anonymous
*2 to make up for the other side.
Mimi_x3
  • Mimi_x3
I will just use this formula\[A = \frac{1}{2} r^2 \left(\theta-\sin\theta\right) \] No need to derive it
anonymous
  • anonymous
ok
Mimi_x3
  • Mimi_x3
and it must be in radians..so convert it..
anonymous
  • anonymous
oh heck nah jose
alexwee123
  • alexwee123
you guys are thinking too hard o.0
anonymous
  • anonymous
oh wait i got the answer. just do 120/360 which is 1/3 and then i did 1/3 * 64pi which is the area of the circle and it is 64pi/3 and now subtract it by the area of the triangle which is
anonymous
  • anonymous
1/2 4sqrt3 * 4 or 8sqrt3 * 2 for the other side
alexwee123
  • alexwee123
|dw:1345781211630:dw| |dw:1345781273195:dw|
alexwee123
  • alexwee123
|dw:1345781325118:dw|
anonymous
  • anonymous
so the final answer is \[64\pi/3 - 16\sqrt{3}\]
alexwee123
  • alexwee123
yes good job @Lethal
anonymous
  • anonymous
yep, i would best response both of u but i can only do one. thx alex.

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