find of the An archery target has three regions formed by three concentric circles as shown in Fig 15.8. If the diameters of the concentric circles are in the ratio `1:2:3`. , then find the ratio of the areas of three regions Fig. 15.8

Explanation

Let the diameters of concentric circles be k, 2k and 3k.
`:.` Radius of concentric circles are`(k)/(2), k and (3k)/(2)`
`:.` Area of innir circle, `A_(1)= pi((k)/(2))^(2) = (k^(2)pi)/(4)`
`:.` Area of middle region, `A_(2)= pi(k)^(2) - (k^(2)pi)/(4)=(3k^(2)pi)/(4)`
[`because` area of ring = `pi(R^(2) - r^(2))` , where R is radius of outer ring and r is radius of inner ring]
and area of outer region, `A_(3) = pi ((3k)/(2))^(2) - pik^(2)`
= `(9pik^(2))/(4) - pik^(2) = (5pik^(2))/(4)`
`:.` Required ratio = `A_(1) : A_(2) : A_(3)`
= `(k^(2)pi)/(4) : (3k^(2)pi)/(4): (5pik^(2))/(4) = 1 : 3 : 5`