I divided the triangle into two parts such that one part remains an equilateral triangle and the other becomes a trapezium. I think the area of the triangle will be the area of the trapezium plus the area of the equilateral triangle which is root 3/4 a^2. I'm having trouble finding the sides of the Trapezium.
-
Forget the trapezoid. Draw the triangle inscribed in the circle. Now draw a line from the top of the triangle to the base that is perpendicular to the base. Then draw two more radii from the center of the circle to the lower angles of the triangle.
http://oi44.tinypic.com/2uxtg86.jpg
Those radii bisect the lower angles to give you two 30/60/60 right triangles. The leg opposite your 30 degree angle is 0.5r. So now you know the length the triangle's height, which is r + 0.5r = 1.5r
You also can figure out the length of your base, which is r√3.
So the area of the circle equals:
A = 1/2bh
A = 1/2 (r√3)(1.5r)
A = 3/4(√3)r^2
A ~ 1.29903r^2
http://oi44.tinypic.com/2uxtg86.jpg
Those radii bisect the lower angles to give you two 30/60/60 right triangles. The leg opposite your 30 degree angle is 0.5r. So now you know the length the triangle's height, which is r + 0.5r = 1.5r
You also can figure out the length of your base, which is r√3.
So the area of the circle equals:
A = 1/2bh
A = 1/2 (r√3)(1.5r)
A = 3/4(√3)r^2
A ~ 1.29903r^2
-
I suggest that you finish the question!