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Help! Lots of points

Help! Lots of points-example-1
User Dkova
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2 Answers

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Answer: i belive the answer is 3

User Philnext
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We first try to find the area of the circle. Easy, 6^2* pi = 36pi


That means half the circle would be 18pi


We try focus on the sector with angle 120 degrees to the right of the circle. We want to find the area of that. We can do 120/360 * 36pi = 12pi


We subtract that from the half circle to get 6pi. Now we are left with the triangle to the left of that sector (the one with 6m) on it.


We know however that it is an equilateral triangle since the vertex angle is 60 degrees and since two sides of the triangle are radii, its isoceles. Either way its equilateral so we can find the area.


The formula for finding that is sqrt(3)/4 * s^2 where s is a side


sqrt(3)/4 * 6^2 = 9sqrt(3)


We now subtract this from 6pi


That would be the area of the shaded area


6pi - 9sqrt(3) = 3.26 m^2


The shaded segment is closest to 3m^2


Hope this helped!!!!! Sorry for the long answer though :p


User Shiffon
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