Question

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

Answer 1

The diameter of three concentric circles are in the ratio of $1:2:3$

Let the diameter of the three circles be $1x,2x$ and $3x$ respectively.

So, their radii will also be in the ratio of $\frac{1x}{2}:\frac{2x}{2}:\frac{3x}{2}\Rightarrow 1:2:3$

Let, radius of the three concentric circles be $r_1,r_2,r_3$ respectively where $r_1=1y,r_2=2y,r_3=3y$

Therefore,

Area of the first region ie. a circle = $\pi r_1^2=y^2\pi$

Area of the second region ie. the middle ring = $\pi r_2^2-\pi r_1^2=\pi (4y^2-y^2)=3y^2\pi$

Area of the third region ie. the outer ring = $\pi r_3^2-\pi r_3^2=\pi (9y^2-4y^2)=5y^2\pi$

Ratio of their areas = $\pi r_1^2:(\pi r_2^2-\pi r_1^2):(\pi r_3^2-\pi r_2^2)$

= $y^2\pi ,3y^2\pi ,5y^2\pi$

= $1y^2:3y^2:5y^2$

= $1:3:5$

So, the ratio of their respective ratios is $1:3:5$