lan583 wrote:

. An equilateral triangle that has an area of 9root3 is inscribed in a circle. What is the area of the circle?

A. 6 Pi B. 9 Pi C. 12 Pi D. 9 Pi Root3 E. 18 Pi Root3

c. 12pi

aet= area equ tria with side a

hence height h= sqrt3*a/2

aet = 9sqrt3 = (1/2)* (a*sqrt3*a/2)

= (1/2)*(a^2sqrt3/2) = a^2sqrt3/4

hence 9 = a^2/4

a = 6

2 ways to proceed from here -

1-

If an equilateral triangle is inscribed in a circle, then the square on the side of the triangle is triple the square on the radius of the circle.

so, a^2 = 3r^2

r^2 = a^2/3 = 36/3 = 12

area circle = 12pi

2-

since the equilateral triangle is inscribed in a circle, the centre of the circle bisects the height from vertex:opposite base in a ratio of 2:1

hence radius r = (2/3)*sqrt3*a/2 = 2sqrt3

area circle = pi(2sqrt3)^2 = 12pi