A step-up SAT/GRE problem -- Circles (Rhombus edition)

In an earlier post, we talked about "step-back problems." The idea is that, wherever possible, each problem should be associated with at least one problem that uses similar format and relies on similar concepts but which "steps up" (is more difficult) or "steps down" (is easier).

In that previous post we talked about problems where you had to find the shaded area of a circle. This problem covers similar territory but takes things up a notch.

Circle 1 and Circle 2 both have radius 2. Each passes through the center of the other. Find the area of the rhombus formed by the two points of intersection (A and B) and the centers of each circle (C1 and C2).

Solution after the break.






Getting started:

If you have no idea where to begin with a problem, sometimes it’s a good idea to just label everything you can think of and then ask yourself what you know about each part.

In this case, take a close look at the five line segments that start with the center of either circle. Remember that both circles have radius 2. Knowing that, what can you figure out about the length  of each side of the two triangles?

Remember, every line segment that connects the center of one of these circles has length two.

Now, all we need to do is find the area of two equilateral triangles, each with sides equal to two.

If you've gotten this far you can probably handle this by yourself, but in case you need a hint, try bisecting the triangle and using the Pythagorean Theorem.

Take a minute, we'll wait.

You should have gotten a height of square root of three for the equilateral triangle, which means that, since the base is two the area is also the square root of three. Therefore, the area of the rhombus is two times the square root of three.