Circle Q and circle R have a central angle measuring 75°. The ratio of circle Q's radius to circle R's radius is 2:5. Which ratio represents the area of the sector for circle R to the area of the sector for circle Q?

In this question, it is given that Circle Q and circle R have a central angle measuring 75°. The ratio of circle Q's radius to circle R's radius is 2:5. And the formula of area of sector is[tex] Area = ( \frac{ \theta}{360 }) \pi r^2 [/tex]And the radius are in the ratio 2:5.Let the radius are 2x and 5x. SO area of sectors are[tex] A_{1} = ( \frac{75}{360} ) \pi (2x)^2, A_{2}= ( \frac{75}{360} ) \pi (5x)^2 [/tex]And the ratio is[tex] \frac{A_{1} }{A_{2}} = \frac { ( \frac{75}{360} pi *4x^2}{ ( \frac{75}{360}) pi*25x^2} [/tex][tex] \frac{A_{1}}{A_{2}} = \frac{4}{25} [/tex]So for the ratio of the area of sectors of circle R to Q, it is[tex] \frac{A_{2}}{A_{1}} = \frac{25}{4} [/tex]