Question

Question 13
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 ratio 1:2:3, then find the ratio of the areas of three regions.

Answer 1

Let the diameters of concentric circles be k, 2k and 3k.

$\therefore$ Radius of concentric are $\frac{k}{2},k$ and $\frac{3k}{2}$.

$\therefore$ Area of inner circle $A_1=\pi {\left(\frac{k}{2}\right)}^2=\frac{k^2\pi }{4}$

$\therefore$ Area of middle region, $A_2=\pi (k{)}^2-\frac{k^2\pi }{4}=\frac{3k^2\pi }{4}$

[$\because$ 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 $A_3=\pi {\left(\frac{3k}{2}\right)}^2-\pi k^2$

$=\frac{9\pi k^2}{4}-\pi k^2=\frac{5\pi k^2}{4}$

$\therefore$ Required ratio $=A_1:A_2:A_3$

$=\frac{k^2\pi }{4}:\frac{3k^2\pi }{4}:\frac{5\pi k^2}{4}=1:3:5$