If the sum of circumference of three circles with radii $r_{1}, r_{2}$ and $r_{3}$ is equal to the circumference of a circle of radius $R$ , then which of the following is true?

π r_{1}^{2} + π r_{2}^{2} + π r_{3}^{2} = π R^{2}

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r_{1} + r_{2} > R

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r_{2} + r_{3} > R

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r_{1} + r_{2} + r_{3} = R

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Solution

The correct option is D $r_{1} + r_{2} + r_{3} = R$
Given that radii of 3 circles be $r_{1}, r_{2},$ and $r_{3}$ .
Thus, the circumferences of circles will be $2 π r_{1}, 2 π r_{2}, 2 π r_{3}$ respectively.
Given: $2 π r_{1} + 2 π r_{2} + 2 π r_{3} = 2 π R$
$\Rightarrow 2 π (r_{1} + r_{2} + r_{3}) = 2 π R$
$\Rightarrow r_{1} + r_{2} + r_{3} = R$
Hence, the correct answer is option d.

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