An archery target has three regions formed by three concentric circles

1.	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.
Answer» 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

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.

Answer»

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