## Wednesday, March 5, 2014

### A challenging standardized test geometry problem

[If you thought the last two problems in the series were too easy, this might be more your speed.]

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).

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.
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Knowing that, what can you figure out about the length  of AC1
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and C1C2 ?

You might want to pause for a minute here to make sure you see where this comes from.

So, our rhombus breaks down to two equilateral triangles with all sides equal to the radius (2). Figuring out the area of each of these takes a bit of work but it’s a pretty straightforward problem.

Next we need to find the area  of an equilateral triangle.

All sides have length 2.

The formula for area of a triangle is

We have the base (2). Now we need the height.

If we draw a vertical bisector, we get two right triangles where the length of the vertical leg is the height of our original equilateral triangle.

Now we can use the Pythagorean Theorem to find the height.

So, our rhombus breaks down to two equilateral triangles with all sides equal to the radius (2).

The area of each triangle is √3. That makes the area of the rhombus 2√3.