At supper, I asked the kids if they had any jokes and was rewarded with:

Why are math books depressed?

They have *so many* problems.

Volunteered by fifth grader. Fortunately, eighth grader later in the evening reported being over challenged by one such problem, which spared me the agony of watching either election returns or a dull World Series game, the outcome of which (I'm talking about the ballgame only) did not interest me in the least. She was supposed to calculate the area of a regular hexagon that's 20 centimeters on a side.

I hemmed and hawed embarrassingly for a couple of minutes. In my defense, she had the advantage of stewing about it for 15 minutes, or more likely 15 seconds, before submitting the problem to dad and expecting instant insight. My first idea was to think of the hexagon as being inscribed within a rectangle, then subtracting the areas of four identical right triangles at the corners of that rectangle. But what is known about the triangles? Only that their hypotenuse is 20 cm and their angles 30º--60º--90º. That's enough, but it's a trigonometry problem, and her class is geometry. I asked whether the phrase "cosine of thirty degrees" meant anything to her and got the gaze I associate with a prizefighter regarding his opponent while the referee gives instructions before the fight.

I could see from the other problems that the lesson seemed to be "applications of the Pythagorean Theorem," and what ended up working was to carve up the hexagon into triangles. She knew that the interior angles were all 120º—goes up by 180º for every side added, so 540º for a pentagon (108 per angle), 720º for a hexagon (120 per angle), &c. The "spokes" bisect these angles, so the resulting triangles have to be equilateral:

We don't have right triangles yet, but each equilateral triangle can be divided into two identical right triangles that look like this:

The area of these right triangles, each one-twelfth of the hexagon, is given by the product of half the base and the height. Half the base is 5. She knew how to use the Pythagorean Theorem to find h:

10^{2} + h^{2} = 20^{2}------->h = √300 = 10√3.

So the area of each right triangle is 50√3 and the area of the whole hexagon is (12)(50)(√3), or 600√3 sq cm. She endured me pointing out that the same strategy can be used to derive a general formula for the area of a regular hexagon with the length of a side equal to s, for the triangles then look like:

And, by the Pythagorean Theorem:

(s/2)^{2} + h^{2} = s^{2}------>h^{2} = s^{2} - (s/2)^{2} = s^{2} - s^{2}/4 = 3s^{2}/4----->h = s√3/2.

The area of each right triangle is then half the base (s/4) times the height (s√3/2), which equals s^{2}√3/8. Since the regular hexagon is made of 12 of these right triangles, the general formula for its area is:

12 s^{2} √3/8 = **(3/2) (s ^{2}) (√3)**.

As to the so-called election "results," I now sleep in camo and have loaded my weapons into the minivan. I assume we'll be marching on Richmond but am awaiting specific instructions from Soros and Hillary.

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