If the sum of the circumferences of two circles with radii R₁ and R₂ is equal to the circumference of a circle of radius R, then
(a) $R_{1} + R_{2} = R$
(b) $R_{1} + R_{2} > R$
(c) $R_{1} + R_{2} < R$
(d) none of these

Open in App

Solution

(a) $R_{1} + R_{2} = R$
Because the sum of the circumferences of two circles with radii $R_{1} and R_{2}$ is equal to the circumference of a circle with radius R, we have:

$2 π R_{1} + 2 π R_{2} = 2 π R \Rightarrow 2 π (R_{1} + R_{2}) = 2 π R \Rightarrow R_{1} + R_{2} = R$

Suggest Corrections

Similar questions

R_{1}

R_{2}

R_{1} + R_{2} = R

R_{1} + R_{2} > R

R_{1} + R_{2} < R

R_{1}, R_{2} a n d R

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

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 $R_{1}$ 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

Join BYJU'S Learning Program