Two minor sectors of two distinct circles hav | vedclass

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Question

Ratio of the areas of minor sectors $=\frac{\pi r_{1}^{2} 	heta_{1} / 360}{\pi r_{2}^{2} 	heta_{2} / 360}$

$	herefore \frac{4}{9}=\frac{r_{1}^{2

Vedclass · Accepted Answer

$2:3$

Vedclass · Answer

Ratio of the areas of minor sectors $=\frac{\pi r_{1}^{2} 	heta_{1} / 360}{\pi r_{2}^{2} 	heta_{2} / 360}$

$	herefore \frac{4}{9}=\frac{r_{1}^{2}}{r_{2}^{2}} $   $\left(\because 	heta_{1}=	heta_{2}ight)$

$ 	herefore \frac{r_{1}}{r_{2}}=\frac{2}{3}$

$	herefore r_{1}: r_{2}=2: 3$

Two minor sectors of two distinct circles have the measure of the angle at the centre equal. If the ratio of their areas is $4: 9,$ then ratio of the radii of the circles is ........

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