Question #933e0

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Answer 1 · 2017-03-11T07:34:22

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

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

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

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

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

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

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

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