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Area of a Circle

Three circles...

Question

Three circles of radii $r_{1}, r_{2}$ and $r_{3}$ are drawn concentric to each other. The radii $r_{1}$ and $r_{2}$ are such that the area of the circle with radius $r_{1}$ is equal to the area between the circles of radius $r_{2}$ and $r_{1}$ . The area between the circles of radii $r_{3}$ and $r_{2}$ is equal to area between the circles of radii $r_{2}$ and $r_{r}$ . What is the value of $r_{1} : r_{2} : r_{3} ?$

1 : \sqrt{2} : \sqrt{3}

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1 : 2 : 3

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1 : \sqrt{2} : 3

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1 : 2 : \sqrt{3}

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Solution

The correct option is B $1 : \sqrt{2} : \sqrt{3}$
Given $π r_{1}^{2} = π ({r_{2}}^{2} - r {_{1}}^{2})$
$\Rightarrow 2 r {_{1}}^{2} = r {_{2}}^{2} \Rightarrow \frac{r {_{1}}^{2}}{r {_{2}}^{2}} = \frac{1}{2}$
$\frac{r_{1}}{r_{2}} = \frac{1}{\sqrt{2}} \Rightarrow r_{1} : r_{2} = 1 : \sqrt{2}$
$a n d π (r {_{3}}^{2} - r {_{2}}^{2}) = π (r {_{2}}^{2} - r {_{1}}^{2}) \Rightarrow r {_{3}}^{2} = 2 r {_{2}}^{2} - r {_{1}}^{2} = 2 r {_{2}}^{2} - \frac{r {_{2}}^{2}}{2} = \frac{3 r {_{2}}^{2}}{2} (∴ 2 r {_{1}}^{2} = r {_{2}}^{2})$
$\Rightarrow \frac{r {_{2}}^{2}}{r {_{3}}^{2}} = \frac{2}{3} \Rightarrow \frac{r_{2}}{r_{3}} = \frac{\sqrt{2}}{\sqrt{3}} \Rightarrow r_{2} : r_{3} = \sqrt{2} : \sqrt{3} . . . (i i)$
$∴$ From (i) and (ii) $r_{1} : r_{2} : r_{3} = 1 : \sqrt{2} : \sqrt{3}$

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