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See the attached figure
radius of the circle = r = 6
the length of one side of the equilateral triangle = L
the height if the triangle = h
The shaded area is the required area.
∴The area = (1/3) (area of circle - area of triangle )
area of circle = π r² = π * 6² = 36 π
An equilateral triangle is inscribed in a circle of radius r
∴ The length of one side of the triangle = √3 * r
= 6√3
and the height of the triangle = (√3 /2) * 6√3 = 9
∴ Area of the triangle = 0.5 * 9 * 6√3 = 27√3
The required area = (1/3) (area of circle - area of triangle )
= (1/3) * ( 36π - 27√3 )
= 12π - 9√3
radius of the circle = r = 6
the length of one side of the equilateral triangle = L
the height if the triangle = h
The shaded area is the required area.
∴The area = (1/3) (area of circle - area of triangle )
area of circle = π r² = π * 6² = 36 π
An equilateral triangle is inscribed in a circle of radius r
∴ The length of one side of the triangle = √3 * r
= 6√3
and the height of the triangle = (√3 /2) * 6√3 = 9
∴ Area of the triangle = 0.5 * 9 * 6√3 = 27√3
The required area = (1/3) (area of circle - area of triangle )
= (1/3) * ( 36π - 27√3 )
= 12π - 9√3
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Each point is then equal distance from the center, that distance being the radius of the circle.
Using the distance formula, we can find the center of the circle (x,y):
Plugging in points A and B into distance formula, then setting them equal to each other gives:
Right away we can cancel out the x terms leaving:
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