<p>Now, I usually consider these types of problems to be rather simple (whole area minus the nonshaded region), however I encountered one that has stumped me:</p>
<p>An equilateral triangle with side 12 is inscribed in a circle. Find the shaded area. (The figure shows the shaded area to be the area outside the triangle.) I can find the nonshaded region fairly easily, but I need the radius to find the area of the whole circle, and I don't know how I'd go about doing that.</p>
<p>The problem is from Barron's SAT 2400, Page 217.</p>
<p>The correct answer is 48pi - 36sqrt(3).</p>
<p>To find the radius, you could split the triangle into three smaller congruent triangles. When you do this, you are bisecting each of the vertex angles. The lines converge at the exact center of the triangle (and the circle).</p>
<p>So now you have three isosceles tri’s with angle measures of 30, 120, 30.</p>
<p>You can then divvy up one these isosceles triangles into two right tri’s, which happen to be 30/60/90.</p>
<p>The middle leg of this tri (the x√3 side) is 6. We know this because it’s half the length of a side of the original equilateral tri.</p>
<p>Now we can figure out the hypotenuse, which is also the radius of the circle. Setting x√3=6 and solving for x we get x=2√3. Hypotenuse is 2x, or 4√3.</p>
<p>So there you have it, radius=4√3.</p>
<p>Kinda hard to explain without a sketch but I hope that helps. BTw that seems a little bit complicated for an SAT question, but I guess the 2400 book is meant to challenge you’s.</p>