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(B) Let the radius of the circular pool be $r \ m$.The width of the concrete wall is $4 \ m$.Therefore,the radius of the outer circle (pool + wall) is

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$20$

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(B) Let the radius of the circular pool be $r \ m$.The width of the concrete wall is $4 \ m$.Therefore,the radius of the outer circle (pool + wall) is $(r + 4) \ m$.The area of the pool is $A_{pool} = \pi r^2$.The area of the wall is $A_{wall} = \pi(r + 4)^2 - \pi r^2$.According to the problem,$A_{wall} = \frac{11}{25} A_{pool}$.Substituting the expressions: $\pi(r + 4)^2 - \pi r^2 = \frac{11}{25} \pi r^2$.Dividing by $\pi$: $(r^2 + 8r + 16) - r^2 = \frac{11}{25} r^2$.$8r + 16 = \frac{11}{25} r^2$.Multiplying by $25$: $200r + 400 = 11r^2$.Rearranging into a quadratic equation: $11r^2 - 200r - 400 = 0$.Factoring the equation: $(r - 20)(11r + 20) = 0$.Since the radius cannot be negative,$r = 20 \ m$.

$A$ circular swimming pool is surrounded by a concrete wall $4 \ m$ wide. If the area of the concrete wall surrounding the pool is $11/25$ that of the pool,then the radius (in $m$) of the pool is

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