Test 7 Section 3 - #16 (page 773)

Note: Pretty much any time you have a shaded region problem, you should be thinking about it in 3 parts, arranged into this equation:

In this particular question, we're going to need to do a little work before we can use that, and then a teensie bit more once we have.  We already know the shaded region's area is 64π.  We know the smaller circle has a radius of 6, so we can use A = πr² to find the area of that inner circle (the UNSHADED part) to be 36π.

Put those into the equation above and you get:

64π + 36π = 100π

So the area of the WHOLE THING is 100π.  Use A = πr² one more time now to find the radius of the large circle:

100π = πr²

100 = r²

10 = r