Question

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 ratios 1 : 2 : 3, then find the ratio of the areas of three regions.

Answer 1

Let the three regions be A, B and C.

The diameters are in the ratio 1 : 2 : 3.

Let the diameters be 1x, 2x and 3x

Then the radius will be `x/2, (2x)/2` and `(3x)/2`

Area of region A = `pi"r"_"A"^2`

= `pi(x/2)^2`

= `pix^2/4` 

Area of region B = `pi"r" _"B"^2-pi"r" _"A"^2`

= `pi(x)^2-pi(x/2)^2`

= `(3pi(x)^2)/4`

Area of region C = `pi"r"_"C"^2-pi"r"_"B"^2-pi"r"_"A"^2`

= `pi((3x)/2)^2-pi(x)^2-pi(x/2)^2`

= `pi((3x)/2)^2-(3pix^2)/4`

= `(5pix^2)/4` 

Thus, ratio of the areas of regions A, B and C  will be 

`pix^2/4 : (3pi(x)^2)/4 : (5pix^2)/4`

⇒ 1 : 3 : 5