## transientframes asked:

In the circle above, the two diameters shown intersect. If the sum of the areas of the unshaded regions is equal to the average of the areas of the shaded regions, what is the value of x? (The circle looks like a railroad crossing sign with the top and bottom quarters shaded, the left and right quarters not shaded, and x in the angle of the left quarter (where one of the Rs is on a railroad crossing sign). It's not drawn to scale. The x refers to the angle.)Hah! I just did this question with a student *today*! It’s from the practice PSAT your counselor probably gave out, right?

It’s really important to understand the bit about the sum of the areas of the unshaded regions equalling the average of the areas of the shaded regions. And of course, it’s SUPER important that you recognize that since the diameters intersect and create vertical angles, the shaded regions are the same, and the unshaded regions are the same.

So call an unshaded region *u* and a shaded region *s*. You can write the 2nd sentence of the question like this:

*u* + *u* = (*s* + *s*)/2.

2*u* = (2*s*)/2

2*u* = *s*

So that’s helpful. A shaded region is twice as big as an unshaded one.

Now, you need to recognize that what you’ve just said about the regions is also true about the central angles that create them, since they’re all part of the circle. So not only is a shaded region twice as big as an unshaded one, but the angle that sets off the shaded region (call it *y* just so we have a name for it) is twice as big as the one that sets off an unshaded one (*x*).

You’re almost home free. Note that *x* + *y* = 180. If *x* is half as big as *y*, then *x* has to be 60. (And *y* has to be 120.)

I hope I didn’t make this seem overcomplicated by using all those variables…this is one that’s way easier to explain in person. :/

Good luck on that PSAT!