<p>Hi guys, I need help:
An equilateral triangle with an area of 3/4 rad3 is inscribed within a circle. What is the area of the circle? </p>
<p>I know that each side is then rad 3 because s^2 rad3 /4 is the area of an equilateral triangle do each side is rad3. But how do I use This to find the area?</p>
<p>The center of the circle is the center of the triangle. Let r be the radius, C be the center, and A, B be 2 adjacent vertices of the triangle. ABC is an obtuse triangle with central angle 120 degrees. To solve for r, use the Law of Cosines. (sqrt3)^2=r^2+r^2-2(r)(r)cos120==>3=2r^2-2r^2(-0.5)==>3=3r^2==>r=1. The area of the circle is then pir^2=pi(1)^2=pi. pi is your answer.</p>
<p>I was thinking of doing that, but how do you know that the center of the circle is the center of the triangle?</p>
<p>@thenerdyjew, the center of the triangle is equidistant from its three vertices, at a distance R (where R is the radius of the circumscribed circle).</p>
<p>Another, possibly more intuitive way, of thinking about this problem is to start by drawing lines from the center of the circle to each of the equilateral triangle’s vertices. You end up with 3 congruent isosceles triangles. They are congruent because each has two sides equal to the radius and the other side equal to the length of the original triangle.</p>
<p>Work out the area of one of the 3 congruent triangles. Easiest is to drop a perpendicular from the center of the circle to the side. You end up with a 30-60-90 triangle. You should be able to proceed from here. The area of each of the three triangles is (1/4)sqrt(3)r^2. You’ll get r=1. And the area is therefore pi.</p>
<p>Another solution would be to construct 30-60-90 triangles in the same way that fogcity did, but work out the lengths instead (particularly if you already know that the side length is sqrt(3)).</p>
<p>Note that the sides are in the ratio R/2 : (R/2)sqrt(3) : R, where R is the circumradius. The side corresponding to (R/2)sqrt(3) is simply sqrt(3)/2 (half the side length of the equilateral triangle). Hence, (R/2)sqrt(3) = sqrt(3)/2, R = 1, area = pi.</p>
<p>The law of cosines would probably be my second choice method, but too much algebra. Always better to save time.</p>