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Two circles touch each other internally at point A as shown in the figure: (http://imgur.com/hmzgMCT) O is the centre of bigger circle. If CB = 9 cm and DE = 5 cm. Find the area of the crescent shaped part of the figure. Take the value of pi as 22/7

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  • $\begingroup$ Do you mind placing the image directly in your question? $\endgroup$ – K. Jiang Feb 22 '16 at 15:41
  • $\begingroup$ K. Jiang how do I do so? $\endgroup$ – Uday Kulkarni Feb 22 '16 at 15:44
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Assume the radius is R and r, so $2R-2r=CB=\frac{9}{2}$, and $$CD^2=OC^2+OD^2=(R-5)^2+(R-9)^2$$ $$AD^2=OA^2+OD^2=R^2+(R-5)^2$$ $$CD^2+AD^2=4r^2$$ which gives $$R^2+2(R-5)^2+(R-9)^2=4(R-\frac{9}{2})^2$$ and R=25 is found. Then use $R=25$, $r=\frac{41}{2}$ so that $$A=\frac{22}{7}\times \frac{819}{4}=643.5$$ hope it can help

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