Attachment:

**File comment:** circle and equilateral
gmatcircle.jpg [ 17.44 KiB | Viewed 1176 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

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