Put your answers in the comments; I'll post the solution here Monday. Good luck!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.
Triangles, friend-oThe 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:
The other wayI'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:
144n = (n - 2)×180
144n = 180n - 360
-36n = -360
n = 10Bangarang.