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(a + b)^2 Expansion and Visualisation

If the sum of...

Question

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

R_{1} + R_{2} = R

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

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R_{1} + R_{2} < R

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

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Solution

The correct option is C $R_{1}^{2} + R_{2}^{2} = R^{2}$
$a r . c i r c l e w i t h r a d i u s R_{1} = π {R_{1}}^{2} a n d a r . c i r c l e w i t h r a d i u s R_{2} = π {R_{2}}^{2} . ∴ s u m o f t h e a r e a s = π ({R_{1}}^{2} + {R_{2}}^{2}) . A g a i n a r . c i r c l e w i t h r a d i u s R = π R^{2} . ∴ B y t h e g i v e n c o n d i t i o n π ({R_{1}}^{2} + {R_{2}}^{2}) = π R^{2} ⟹ ({R_{1}}^{2} + {R_{2}}^{2}) = R^{2} . A n s - O p t i o n B .$

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Similar questions

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

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

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}

R_{1} + R_{2} = R

R_{1} + R_{2} < R

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

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

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}

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