SOLUTION: The width of the rectangle is 9 inches shorter than it's length and twice the radius of a circle. The difference in perimeters of the two shapes is 450 inches. Find the area of the

Question 103276: The width of the rectangle is 9 inches shorter than it's length and twice the radius of a circle. The difference in perimeters of the two shapes is 450 inches. Find the area of the circle.
Answer by alisttutors(8) (Show Source): You can put this solution on YOUR website!
W=L-9
W=2R
(2W+2L)-2PiR=450
You want to find Pi*R^2 which is the same as Pi*W^2/4
2W+2(W+9)-W*Pi=450
4W+18-Pi*W=450
(4-Pi)*W=432
W=432/(4-Pi)
Pi(432/(4-Pi))^2/4
Plug that puppy into a calculator and you get:
198,916.36920088...
or 1.98*10^5
Hmm, that is ugly - but I've checked it - let me know if it confers with your thoughts.
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