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(A) The length of an arc of a circle is given by the formula $L = \frac{	heta}{360^{\circ}} 	imes 2\pi r$,where $	heta$ is the angle subtended by t

Vedclass · Accepted Answer

$3:1$

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(A) The length of an arc of a circle is given by the formula $L = \frac{	heta}{360^{\circ}} 	imes 2\pi r$,where $	heta$ is the angle subtended by the arc at the center and $r$ is the radius of the circle.Since the two circles are congruent,they have the same radius,let it be $r$.Let $L_1$ be the length of arc $AXB$ and $L_2$ be the length of arc $A'YB'$.Given that the angle subtended by arc $AXB$ at center $O$ is $	heta_1 = 75^{\circ}$.So,$L_1 = \frac{75^{\circ}}{360^{\circ}} 	imes 2\pi r$.Given that the angle subtended by arc $A'YB'$ at center $O'$ is $	heta_2 = 25^{\circ}$.So,$L_2 = \frac{25^{\circ}}{360^{\circ}} 	imes 2\pi r$.The ratio of the lengths of the arcs is $\frac{L_1}{L_2} = \frac{\frac{75^{\circ}}{360^{\circ}} 	imes 2\pi r}{\frac{25^{\circ}}{360^{\circ}} 	imes 2\pi r} = \frac{75^{\circ}}{25^{\circ}} = \frac{3}{1}$.Therefore,the ratio is $3:1$.

In the figure,two congruent circles have centers $O$ and $O'$. Arc $AXB$ subtends an angle of $75^{\circ}$ at the center $O$ and arc $A'YB'$ subtends an angle of $25^{\circ}$ at the center $O'$. Then the ratio of the lengths of arcs $AXB$ and $A'YB'$ is:

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