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Question Number 9298 by tawakalitu last updated on 29/Nov/16

Answered by mrW last updated on 29/Nov/16

$$\mathrm{radius}\mathrm{r}=(7+3)/2=5\mathrm{cm}$$$$\mathrm{height}\mathrm{of}\mathrm{segment}\mathrm{h}=3\mathrm{cm}$$$$\mathrm{area}\mathrm{of}\mathrm{segment}={\mathrm{r}}^2\cos {}^{-1}\left(\frac{\mathrm{r}-\mathrm{h}}{\mathrm{r}}\right)-(\mathrm{r}-\mathrm{h})\sqrt{2\mathrm{rh}-{\mathrm{h}}^2}$$$$=5^2\cos {}^{-1}\left(\frac{5-3}{5}\right)-(5-3)\sqrt{2\times 5\times 3-3^2}$$$$=25\cos {}^{-1}\left(\frac{2}{5}\right)-2\sqrt{21}$$$$\mathrm{the}\mathrm{shaded}\mathrm{area}\mathrm{is}\mathrm{half}\mathrm{of}\mathrm{the}\mathrm{segment}$$$$=\frac{25}{2}\cos {}^{-1}\left(\frac{2}{5}\right)-\sqrt{21}\approx 9.91{\mathrm{cm}}^2$$

Commented by tawakalitu last updated on 29/Nov/16

$$\mathrm{i}\mathrm{really}\mathrm{appreciate}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Commented by Nayon last updated on 29/Mar/17

$$Howdidyourealisethatshaded$$$$areaishalfofthesegment?$$

Commented by mrW1 last updated on 29/Mar/17

$$thepictureshowsonlythehalfofthe$$$$circle.$$

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