This morning I was lying in bed waiting for my alarm to go off, and a question just popped into my head. This happens to me fairly often, but usually I’m too lazy to write it down, then I forget it, then I spend the rest of the day trying to convince myself that it wasn’t a very good question anyway so I shouldn’t worry about forgetting it. This morning, though, I wrote it down. Which means someone’s going to win a Math Guide.
As always, the first correct response in the comments will win the book. If you submit your comment and it doesn’t show up right away, don’t freak out and submit a million more. There’s sometimes a delay because of comment moderation, but rest assured that I receive all comments in the order they’re submitted. If you’re the first one who got it right, I’ll know. (Full contest rules.)
Let’s do this.
An equilateral triangle is inscribed in a circle with area 9π. What is the area of the triangle?
UPDATE: Rob got it first! Solution below the cut.
If the area of the circle is 9π, then its radius must be 3 (A = 9π = πr2; 3 = r):
Since you’re dealing with an equilateral triangle, you know each angle in the triangle measures 60º. Since we just made a bunch of smaller, isosceles triangles in there, each of the smaller angles must measure 30º. Which means the central angles must each be 120º.
We want the area of the equilateral triangle, so we need to find its dimensions. To do this, we’re going to use the properties of 30º-60º-90º triangles.
The 30º-60º-90º triangle on the bottom right has a hypotenuse of 3, which means its short leg must be half of that (1.5), and its long leg must be 1.5√3.
Of course, that means the height of the equilateral triangle is 4.5, and its base is 3√3.
Now that we have the base and height, we can calculate the area of the triangle.