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