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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.
Solution
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
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