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(D) Let the diameters of the concentric circles be $k, 2k$,and $3k$.Therefore,the radii of the concentric circles are $\frac{k}{2}, k$,and $\frac{3k}{

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$1: 3: 5$

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(D) Let the diameters of the concentric circles be $k, 2k$,and $3k$.Therefore,the radii of the concentric circles are $\frac{k}{2}, k$,and $\frac{3k}{2}$.Area of the inner circle,$A_1 = \pi \left(\frac{k}{2}\right)^2 = \frac{\pi k^2}{4}$.Area of the middle region,$A_2 = \pi(k)^2 - \frac{\pi k^2}{4} = \frac{3\pi k^2}{4}$.Area of the outer region,$A_3 = \pi \left(\frac{3k}{2}\right)^2 - \pi(k)^2 = \frac{9\pi k^2}{4} - \pi k^2 = \frac{5\pi k^2}{4}$.The required ratio of the areas is $A_1 : A_2 : A_3 = \frac{\pi k^2}{4} : \frac{3\pi k^2}{4} : \frac{5\pi k^2}{4} = 1 : 3 : 5$.

An archery target has three regions formed by three concentric circles as shown in the figure. If the diameters of the concentric circles are in the ratio $1: 2: 3$,then find the ratio of the areas of the three regions.

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