XP Math - Forums - View Single Post - Word Problem
 Thread: Word Problem View Single Post
 06-16-2008 #2 Sillysidley   Join Date: Oct 2006 Posts: 822 Well, let the radius of the smaller circle be r, so the radius of the larger circle is 2r-1. We then have $(2r-1)^2-r^2=33$, since we just divided out the pi. Now, I just eyed it out and found r=4, so our answer is 7. If you can't see that, you can expand to get $4r^2-4r+1-r^2=33\rightarrow\3r^2-4r+1=33\rightarrow\3r^2-4r-32=0\rightarrow\(3r+8)(r-4)=0$, so we know r must be 4. Or, after getting $(2r-1)^2-r^2=33$, you can apply difference of squares to get $(3r-1)(r-1)=33$ and you can either make it a quadratic again, or in this case it's easy to examine factors and find r=4. __________________ .