An archery target has three regions formed by | vedclass

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Question

Let the diameters of concentric circles be $k , 2 k$ and $3 k$.

$	herefore$ Radius of concentric circles are $\frac{k}{2}, k$ and $\frac{3 k}{2}$

Vedclass · Accepted Answer

$1: 3: 5$

Vedclass · Answer

Let the diameters of concentric circles be $k , 2 k$ and $3 k$.

$	herefore$ Radius of concentric circles are $\frac{k}{2}, k$ and $\frac{3 k}{2}$

$	herefore \quad$ Area of inner circle, $A_{1}=\pi\left(\frac{k}{2}ight)^{2}=\frac{k^{2} \pi}{4}$

$	herefore$ Area of middle region, $A_{2}=\pi(k)^{2}-\frac{k^{2} \pi}{4}=\frac{3 k^{2} \pi}{4}$

$\left[\becauseight.$ area of ring $=\pi\left(R^{2}-r^{2}ight)$, where $R$ is radius of outer ring and $r$ is radius of inner ring]

and area of outer region, $A_{3}=\pi\left(\frac{3 k}{2}ight)^{2}-\pi k^{2}$

$=\frac{9 \pi k^{2}}{4}-\pi k^{2}=\frac{5 \pi k^{2}}{4}$

Required ratio $= A _{1}: A_{2}: A_{3}$

$=\frac{k^{2} \pi}{4}: \frac{3 k^{2} \pi}{4}: \frac{5 \pi k^{2}}{4}=1: 3: 5$

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

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