If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then
(a) $R_{1} + R_{2} = R$
(b) $R_{1} + R_{2} < R$
(c) $R_{1}^{2} + R_{2}^{2} < R^{2}$
(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

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Solution

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$
Because the sum of the areas of two circles with radii $R_{1} and R_{2}$ is equal to the area of a circle with radius R, we have:
$π {R_{1}}^{2} + π {R_{2}}^{2} = π R^{2} \Rightarrow π ({R_{1}}^{2} + {R_{2}}^{2}) = π R^{2} \Rightarrow {R_{1}}^{2} + {R_{2}}^{2} = R^{2}$

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Similar questions

If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle of radius R then

(a) $R_{1} + R_{2} = R$

(b) $R_{1} + R_{2} > R$

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

R_{1}

R_{2}

R_{1} + R_{2} = R

R_{1}^{2} + R_{2}^{2} = R^{2}

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R_{2}

r_{1}

r_{2}

r

r = r_{1} + r_{2}

{r_{1}}^{2} + {r_{2}}^{2} = r^{2}

r_{1} + r_{2} < r

{r_{1}}^{2} + {r_{2}}^{2} < r^{2}

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