File comment: circle and equilateral
gmatcircle.jpg [ 17.44 KiB | Viewed 1938 times ]
Above image shows a pictorial view of the problem.
Perimeter of equilateral triangle = z
side of the equilateral triangle = z/3
A line from center to the Vertex A of the triangle will make 30 degree angle from base.
A perpendicular from center will bisect the base. Hence AP = z/6
Now , Cos 30 = sqrt(3)/2 = AP/OA
or , OA = 2* AP/sqrt(3)
Radius of circle = OA = 2*(z/6)/sqrt(3) = z/3 *sqrt(3)
Area of circly = y = pi* radius^2
y = pi (z/3*sqrt(3))^2
y = pi*z^2/27
27y = pi* z^2
pi*z^2 - 27y = 0
Hence answer is E
Help me with Kudos if it helped you "
Mathematics is a thought process.