Question

An archery target has three regions formed by three concentric circles as shown in figure15.8. If the diameters of the concentric circles are in the ratios 1 : 2 : 3 , then find the ratio of the areas of three regions .

Answer 1

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 $\frac{x}{2},\frac{2x}{2}and\frac{3x}{2}$
Area of region A = ${\mathrm{\pi r}}_{\mathrm{A}}^2=\mathrm{\pi }{\left(\frac{\mathrm{x}}{2}\right)}^2=\frac{{\mathrm{\pi x}}^2}{4}$
Area of region B = ${\mathrm{\pi r}}_{\mathrm{B}}^2-{\mathrm{\pi r}}_A^2=\mathrm{\pi }{\left(\mathrm{x}\right)}^2-\mathrm{\pi }{\left(\frac{\mathrm{x}}{2}\right)}^2=\frac{3\mathrm{\pi }{\left(\mathrm{x}\right)}^2}{4}$
Area of region C = ${\mathrm{\pi r}}_C^2-{\mathrm{\pi r}}_{\mathrm{B}}^2-{\mathrm{\pi r}}_A^2=\mathrm{\pi }{\left(\frac{3\mathrm{x}}{2}\right)}^2-\mathrm{\pi }{\left(\mathrm{x}\right)}^2-\mathrm{\pi }{\left(\frac{\mathrm{x}}{2}\right)}^2=\mathrm{\pi }{\left(\frac{3\mathrm{x}}{2}\right)}^2-\frac{3{\mathrm{\pi x}}^2}{4}=\frac{5{\mathrm{\pi x}}^2}{4}$
Thus, ratio of the areas of regions A, B and C will be
$\frac{{\mathrm{\pi x}}^2}{4}$:$\frac{3\mathrm{\pi }{\left(\mathrm{x}\right)}^2}{4}$:$\frac{5{\mathrm{\pi x}}^2}{4}$
⇒1:3:5