An equivalent triangle is inscribed inside a circle of radius 4. What is the area of each of the three wedges outside the triangle but inside the circle?
If all three vertices of an equilateral triangle touches the circle, then the three wedges each have the same area. The triangle also shares the same center point with the circle.umaa wrote:An equivalent triangle is inscribed inside a circle of radius 4. What is the area of each of the three wedges outside the triangle but inside the circle?
Area of each wedge = (Area of circle - area of triangle)/3
Lets say the vertices of the triangle are A, B and C, and the center point is X.
AX = BX = CX=4
If you draw a line from X to the midpoint of AB (lets call this midpoint N), you will form a 30-60-90 triangle. We know that for a 30-60-90 triangle, we have the following ratios for the lengths:
1:√3:2
Since AX forms the hypotenuse of the triangle, we have XN = 2 and AN=
So the area of the equilateral triangle = 0.5*(2*AN)*(4+XN)
=(2√3)*(6)
=12√3
Area of the circle = pi*(4^2) = 16pi
So the area of each wedge = (16pi - 12√3)/3
-BM-
