Question

An archery target has three regions formed by three concentric circles, If the diameters of the concentric circles are in the ratio $1:2:3$, then the ratio of the areas of three regions is

• A. $1:4:9$
• B. $9:5:4$
• C. $1:3:5$
• D. $1:2:3$

Answer 1

The correct option is C $1:3:5$

Given the diameters of concentric circles are in ratio 1:2:3 respectively.

Let the diameter of the concentric circles be $s,2s,3s$.
Radius of inner circle= $r_1=\frac{s}{2}$
Radius of middle circle = $r_2=\frac{2s}{2}=s$
Radius of outer circle = $r_3=\frac{3s}{2}$
Area of first circle (first region ) = $\pi r_1^2=\pi \frac{s^2}{4}$

Area of second circle $=\pi r_2^2=\pi s^2$

Area of third circle $=\pi r_3^2=\pi \frac{9s^2}{4}$
So Area of region enclosed between second and firstcircle $=\pi r_2^2-\pi r_1^2=\pi s^2-\pi \frac{s^2}{4}\Rightarrow \frac{3\pi s^2}{4}$

Area of region enclosed between third and second circle $=\pi r_3^2-\pi r_2^2=\pi \frac{9s^2}{4}-\pi s^2\Rightarrow \frac{5\pi s^2}{4}$

Ratio of area of three regions $=\pi \frac{s^2}{4}:\frac{3\pi s^2}{4}:\frac{5\pi s^2}{4}$
Ratio $=1:3:5$