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 , the hypotenuse is and the long side is .

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 inches long. With two of these long sides put together, the total side length of the square is inches.

To find the area of the square, we square this side length to get , or *72 square inches.*

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