Question 509424
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Let *[tex \Large r] represent the radius of the pool.  Then the radius of the entire circle including the walkway is *[tex \Large r\ +\ 3].  Then the area of the entire circle, including both the pool and the walkway is *[tex \Large A_T\ =\ \pi(r\ +\ 3)^2], whereas the area of just the pool is *[tex \Large A_P\ =\ \pi{r^2}].  Subtract the area of the pool from the area of the entire circle to get the area of the walkway:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi(r\ +\ 3)^2\ -\ \pi{r^2}\ =\ 198]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi\left[r^2\ +\ 6r\ +\ 9\ -\ r^2\right]\ =\ 198]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6r\ +\ 9\ =\ \frac{198}{\pi}]


You can take it from here.  Leave your answer in terms of *[tex \Large \pi] for an exact answer or round your numeric approximation to the nearest whole meter since the precision of your given measurement was to the nearest square meter.  Or, if you use the old standby *[tex \Large \frac{22}{7}] as an approximation of *[tex \Large \pi] then your answer comes out to be a whole number anyway.


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