I’ve spruced it up a tad, but an *extremely* similar question was #17 (not even #20!) on an SAT in 2006. The prize this week (pardon my proselytizing): Someday you will be as good at something as Mike Miller is at songwriting.

UPDATE: A couple folks nailed this one, and doing so represents, in my opinion, a promising level of nimbleness. Color me impressed. Solution below the cut. The first thing you need to recognize here is that the pink shaded region is a quadrilateral.

The angles in a quadrilateral will always add up to 360°, just like the angles in a triangle will always add up to 180°. Since *v* + *w *= 72, the two unlabeled angles must add up to 360° – 72° = 288°. Since the question states that all the angles in the polygon are the same, we can conclude than each of the unlabeled angles is 144°.

From here, there are a few ways to get the solution. Since this is a rare bird of a question, I’m going to solve it for you first in a way that I think might actually have some utility when applied to other questions, and then I’ll give you the quick solution.

##### Triangles, friend-o

The SAT friggin’ loves triangles. So even though it’s *highly* unlikely that you’ll come across a question like this on the SAT since I’ve only seen it once, practice with breaking confusing diagrams into triangles might pay off for you on game day. Since this is a regular polygon (that is, it has equal sides and equal angles), it can be broken down into a bunch of triangles that meet at its center. Those triangles will all be isosceles, and their legs will bisect the interior angles of the polygon. This is easier shown than said:

So the question becomes “how many 36° angles fit in the center of this shape before it goes all the way around (or, how many times does 36° go into 360°? And your answer, of course, is 10. 10 of these triangles will form a 10-sided polygon.

##### The other way

I’m sure some of you are scratching your head, wondering why I haven’t used the (*n* – 2)×180° formula. Quite simply, not everyone knows it and it’s extremely rare that you’d actually need that formula on the SAT. I don’t want to fill anybody’s head with unnecessary formulas, and I think the triangle solution is elegant. But I’d be remiss not to mention that the sum of the degree measure of all the angles in a polygon can be calculated using (*n* – 2)×180°, where *n* is the number of sides of the polygon. For example, the sum of the degree measures of the angles in a 6-sided figure is (6 – 2)×180° = 720°. If you’re clever, you can use this formula once you know the degree measure of one angle in a *regular* polygon to calculate the number of sides. If the polygon is regular, that means all the angles will be the same, so the total measure divided by the number of sides will give the measure of one of the angles. In other words:

We can plug 144° in for the measure of the single angle, and solve for *n*:

*n*= (

*n*– 2)×180

*n*= 180

*n*– 360

*n*= -360

*n*= 10

Bangarang.