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  • 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

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.

Answers (1)

 

d1:d2:d3   =   1: 2 : 3 [multiplying by s]

=   s : 2s  :  3s       

 The radius of inner circle (r1)=\frac{s}{2}

The radius of the middle circle (r2)=\frac{2s}{2}=s

The radius of the outer circle (r3)=\frac{3s}{2}

Area of the region enclosed between second and first circle   

\\=\pi r_{2}^{2}-\pi r_{1}^{2}\\ =\pi s^{2}-\frac{\pi s^{2}}{4}\\ =\frac{3 \pi s^{2}}{4}

 Area of the region enclosed between third and second circle   

\\=\pi r_{3}^{2}-=\pi r_{2}^{2}\\ =\frac{\pi 9s^{2}}{4}-\pi s^{2}\\ =\frac{ 5\pi s^{2}}{4}

Area of the first circle =\pi r_{1}^{2}=\frac{\pi s^{2}}{4}

The ratio of area of the three regions

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

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