Area of Intersection of Three Circles 
I guess I'm just too dumb to solve this PSAT problem -- so what I tried for this is to break down the shaded area into a difference of sectors and triangles. I want to utilize the area of the 20 degree sector, and I have a hunch that all of this can be achieved without trigonometry.  I don't see a way to utilize that Q is the midpoint of PR. I've basically gone nowhere on this problem, and all help is appreciated.
 A: The area of the common wedge between the two horizontal semicircles = the area of the wedge between the left horizontal circle and the tilted circle (unfortunately I can't draw a diagram). Now delete the area common to both these regions. 
Think: if you rotate back the tilted circle by 20∘, it will superimpose itself on the right horizontal circle. 
A: 
Let $A$ be the area of the region bounded by arc $QS$, arc $SU$ and line segment $QU$.
The shaded area is equal to $A$ minus the sum of the area of the minor segment with centre angle $60^\circ$ and the area of the sector with centre angle $40^\circ$.
But $A$ minus the sum of the area of the minor segment with centre angle $60^\circ$ is equal to the area of the sector with centre angle $60^\circ$.
So the shaded area is equal to  the area of the sector with centre angle $20^\circ$, which is equal to $\displaystyle 72\times\frac{20}{180}=8$.
A: For variety, we can apply a variation on Cavalieri's principle.
The slanted semicircle is obtained by rotating the other semicircle 20 degrees around $Q$. The variation we need is:

If all of the circular slices of two regions have the same length, then the regions have the same volume

By circular slice, I mean the intersections of the region with circles centered on $Q$. 
Because of the rotation, it's easy to see slice of the shaded region has the same length as the corresponding slice of the $20^\circ$ wedge — each region is obtained by rotating a figure $20$ degrees — and thus they have the same area.
For the rigorously inclined, the aforementioned principle can be fairly easily proven via area integrals in polar coordinates.
A: Let $S$ is an area of the common part of two semicircles (one of them with diameter PR).
Thus, the needed area is $$S+\frac{\pi R^2}{18}-S=8.$$
A: A simple way to think through the answer:


*

*Think of rotating the semicircle in 20° increments to make a complete circle.

*This will take 18 steps for a full circle (20*18=360).

*Therefore, the small area is 1/18th of a circle or 1/9th of a semicircle.

*If the area of the whole semicircle is 72, then the area of the shaded part is 1/9*72=8

