4/19/10

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

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