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1. A circular swimming pool is surrounded by a concrete wall 4 ft. wide. If the area of the concrete wall surrounding the pool is $\frac{11}{25}$ that of the pool, then the radius of the pool is :

A. 8 ft

B. 16 ft

C. 20 ft

D. 30 ft

Solution:

Let the radius of the pool be R ft.
Radius of the pool including the wall = (R + 4)ft.
Area of the concrete wall :

\begin{aligned} = π [{(R + 4)}^{2} - R^{2}] s q . f t \\ = π [(R + 4 + R) (R + 4 - R)] s q . f t \\ = 8 π (R + 2) s q . f t \\ 8 π (R + 2) = \frac{11}{25} π R^{2} \\ \Rightarrow 11 R^{2} = 200 (R + 2) \\ \Rightarrow 11 R^{2} - 200 R - 400 = 0 \\ \Rightarrow 11 R^{2} - 220 R + 20 R - 400 = 0 \\ \Rightarrow 11 R (R - 20) + 20 (R - 20) = 0 \\ \Rightarrow (R - 20) (11 R + 20) = 0 \\ \Rightarrow R = 20 \end{aligned}

∴ Radius of the pool = 20 ft.

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