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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)

[1 : 3 : 5]

Solution

 

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

  =   s : 2s  :  3s       

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

Radius of middle circle (r2)=\frac{2s}{2}=s

Radius of outer circle (r3)=\frac{3s}{2}

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

Ratio of area of 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

           

Posted by

infoexpert24

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