An archery target has three regions formed


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

then find the ratio of the areas of three regions.


Let the diameters of concentric circles be k, 2k and 3k.

$\therefore$ Radius of concentric circles are $\frac{k}{2}, k$ and $\frac{3 k}{2}$.

$\therefore \quad$ Area of inner circle, $A_{1}=\pi\left(\frac{k}{2}\right)^{2}=\frac{k^{2} \pi}{4}$

$\therefore$ Area of middle region, $A_{2}=\pi(k)^{2}-\frac{k^{2} \pi}{4}=\frac{3 k^{2} \pi}{4}$

$\left[\because\right.$ area of ring $=\pi\left(R^{2}-r^{2}\right)$, where $R$ is radius of outer ring and $r$ is radius of inner ring]

and area of outer region, $A_{3}=\pi\left(\frac{3 k}{2}\right)^{2}-\pi k^{2}$

$=\frac{9 \pi k^{2}}{4}-\pi k^{2}=\frac{5 \pi k^{2}}{4}$

$\therefore \quad$ Required ratio $=A_{1}: A_{2}: A_{3}$

$=\frac{k^{2} \pi}{4}: \frac{3 k^{2} \pi}{4}: \frac{5 \pi k^{2}}{4}=1: 3: 5$

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