# Problem of the Day: 1/8/13

In how many distinct ways can the letters of the word ALGEBRA be arranged?

Solution to yesterday’s problem:
Let’s show the diagram again, with a few small changes:

As you can see, I’ve created two 30-60-90 right triangles (the angle between two sides of a regular hexagon is always 120°). Along with this, you should remember that if the short side is $a$, the hypotenuse is $2a$ and the long side is $a\sqrt{2}$.

Knowing these facts about 30-60-90 triangles, let’s find a. The hypotenuse is 6 inches at 2a, so the short side must be 3 inches. Then, the long side is $3\sqrt{2}$ inches long. With two of these long sides put together, the total side length of the square is $6\sqrt{2}$ inches.

To find the area of the square, we square this side length to get $36 \times 2$, or 72 square inches.