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.

Diameters are in the ratio 1:2:3


So, let the diameters of the concentric circles be 2r, 4r and 6r.


∴ Radius of the circles be r, 2r, 3r respectively.


Now, Area of the outermost circle = π (Radius)2 = π (3r)2 = 9πr2


Area of the middle circle = π (Radius)2 = π (2r)2 = 4πr2


Area of the innermost circle = π (Radius)2 = π (r)2 = πr2


Now, Area of the middle region = Area of middle circle – Area of the innermost circle


= 4πr2 - πr2 = 3πr2


Now, Area of the outer region = Area of outermost circle – Area of the middle circle


= 9πr2 - 4πr2 = 5πr2


Required ratio = Area of inner circle: Area of the middle region: Area of the outer region


= πr2 : 3πr2 : 5πr2


⇒ Required Ratio is 1:3:5


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