Question 1025035
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The area of the non-overlapped part of the larger rug is *[tex \Large 15\pi] square feet: square the radius (16) times *[tex \Large \pi], minus the *[tex \Large \pi] square feet overlap.  Similarly, the area of the non-overlapped part of the smaller rug is *[tex \Large 8\pi] square feet.


So *[tex \Large 15\pi\ +\ 8\pi\ +\ \pi\ =\ 24\pi] square feet.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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