The ratio of radii of two circles is 2:3 and | vedclass

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(A) The area of a sector of a circle is given by the formula $A = \frac{	heta}{360^\circ} 	imes \pi r^2$,where $r$ is the radius and $	heta$ is the

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$10:9$

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(A) The area of a sector of a circle is given by the formula $A = \frac{	heta}{360^\circ} 	imes \pi r^2$,where $r$ is the radius and $	heta$ is the central angle.Let the radii of the two circles be $r_1$ and $r_2$,and the central angles be $	heta_1$ and $	heta_2$.Given: $\frac{r_1}{r_2} = \frac{2}{3}$ and $\frac{	heta_1}{	heta_2} = \frac{5}{2}$.The ratio of the areas of the two sectors is $\frac{A_1}{A_2} = \frac{\frac{	heta_1}{360^\circ} 	imes \pi r_1^2}{\frac{	heta_2}{360^\circ} 	imes \pi r_2^2} = \left( \frac{	heta_1}{	heta_2} ight) 	imes \left( \frac{r_1}{r_2} ight)^2$.Substituting the given values: $\frac{A_1}{A_2} = \left( \frac{5}{2} ight) 	imes \left( \frac{2}{3} ight)^2 = \left( \frac{5}{2} ight) 	imes \left( \frac{4}{9} ight) = \frac{20}{18} = \frac{10}{9}$.Thus,the ratio of the areas of the sectors is $10:9$.

The ratio of radii of two circles is $2:3$ and the ratio of the angles at the centre of two minor sectors of those circles is $5:2$. Then,the ratio of the areas of those sectors is:

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