video explanation

text explanation

we are told the circle has an area of 9. from archimedes' famous formula a = r², we know that the radius of the circle must equal r = 3. this means ao = od = 3, which means ad = 6. now, we have the length of the side of triangle abd.

abd is the kind of triangle we get when we bisect an equilateral. the angle b is still 60°, the angle at d is a right angle, and the angle at a has been bisected, so that angle dab = 30°. this is a 30-60-90 triangle, which has special properties. you can read about these special properties at this gmat blog.

suppose you don't remember all those properties. look at side db --- that's exactly half the side of the equilateral triangle. if we call db = x, then a full side, like bc or ab, must equal 2x. given that ad = 6, bd = x, and ab = 2x, we can use the pythagorean theorem in right triangle abd.

this means that a full side of the equilateral is twice this:

incidentally, if you remembered your 30-60-90 triangle properties, you could have found:

if you are unfamiliar with roots, and particular with the procedure of rationalizing a denominator (used here to divide 12 by root 3), then see this gmat blog where more is explained, or the "rationalizing" video below.

either way, now we have a height, ad = 6, and a base,

well, now we can find the area.

that's the area of the equilateral triangle abc.

answer = c

faq: i solved the problem for 36/√3. is this wrong?

a: your answer is correct—just not in its final form. the explanation above describes "fixing the denominator," and links to a video, which you should watch if you think this can be the final answer. ultimately, never leave roots on the bottom of a fraction.

faq: why isn't 18/√3 an answer? i used the area formula but didn't get the right answer.

a: you are so close to the right answer. you found the area for half of the triangle. multiply 18/√3 by 2 to find the total area of the triangle.

faq:  if i draw a line from c to o, won't that bisect angle c and give me a 30-60-90 triangle?  then i can proceed from there to find the length of the base.

no.  if you join those points to form a line oc, that would not split angle c equally into two 30 degree angles.

in an equilateral triangle, the altitude, angle bisector, and perpendicular bisector are all the same line.  thus, to bisect angle c, we would actually need to draw a line from angle c to the midpoint of side ab.

however, that line would not pass through the center of the circle (o) in this problem.  therefore, oc is not a line that would bisect angle c.