Question 1
If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the areas of a circle of radius R, then:
(A) $R_{1} + R_{2} = R$
(B) $R_{1}^{2} + R_{2}^{2} = R^{2}$
(C) $R_{1} + R_{2} < R$
(D) $R_{1}^{2} + R_{2}^{2} < R_{2}$

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Solution

According to the given condition

Area of circle = Area of first circle + Area of second circle

$∴ π R^{2} = π R_{1}^{2} + π R_{2}^{2}$

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

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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}

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

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