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Dallasb22
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A regular hexagon with sides of 3" is inscribed in a circle. Find the area of a segment formed by a side of the hexagon and the circle.
(Hint: remember Corollary 1the area of an equilateral triangle is 1/4 s2 √3.)
 one year ago
 one year ago
Dallasb22 Group Title
A regular hexagon with sides of 3" is inscribed in a circle. Find the area of a segment formed by a side of the hexagon and the circle. (Hint: remember Corollary 1the area of an equilateral triangle is 1/4 s2 √3.)
 one year ago
 one year ago

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JakeV8 Group TitleBest ResponseYou've already chosen the best response.2
maybe you can find the radius of the circle by looking at one of the triangles that make up the inscribed hexagon. That will allow you to get the whole circle area, and 1/6 of that area falls in the "pie piece" that includes the triangle. Then, if you have the area of that triangle, subtract it from the pie piece to get the segment in question.
 one year ago

isratalo Group TitleBest ResponseYou've already chosen the best response.0
the radius of that circle will be 3. so are=pi r^2=28.27 by joining the centre with each corner of the hexagon u'll get the area of each part. So Area of each part=28.27/6=4.545.............(i) Area of the triangle= (3)^2*sq. root of 3/4=3.9.............(ii) Now subtracting the area of equation (ii) from equation (i) u will get the result.
 one year ago

Dallasb22 Group TitleBest ResponseYou've already chosen the best response.0
@isratalo
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.2
Area of circle with radius 3 = 3^2 pi = 9 pi. 1/6 of the circle = (9/6) pi = (3/2) pi The triangle has hypotenuse 3, base 3, so the height is (3/2)sqrt(3). The area of that triangle is (1/2)bh = (1/2)(3)(3/2)sqt(3) = (9/4)sqrt(3) Area of 1/6 circle  Area of triangle = (3/2)pi  (9/4)sqrt(3) = 4.712  3.897 = 0.815 That's what I got... but you need to double check the geometry and the math.
 one year ago

isratalo Group TitleBest ResponseYou've already chosen the best response.0
Ooops, i'm really sorry, JakeV8 is right. I've checked it right now. Actually, previously i made a calculation mistake..Sorry again..
 one year ago
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