Question

Question #933e0

Answer 1

Let the radius of the large circle be `R` inch and that of small circle be `r` inch.
By the problem the ratio of their circumferences

`(2pir)/(2piR)=2/3`

`r=2/3R`

Now Two radii making an angle of `60^@` intercept a portion of the circular ring which has area of 150 square inches. So

`60^@/350^@(piR^2-pir^2)=150"sqinch"`

`=>pi/6(R^2-r^2)=150`

`=>(R^2-4/9R^2)=900/pi`

`=>5/9R^2=900/pi`

`=>R^2=900/pixx9/5`

`=>R^2=(9^2xx20)/pi`

`=>R=18sqrt(5/pi)" inch"`