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Question Number 12292 by sin (x) last updated on 18/Apr/17

Answered by mrW1 last updated on 18/Apr/17

$$S=\frac{1}{6}\times \pi r^2+\frac{r^2}{2}(\frac{\pi }{3}-\sin \frac{\pi }{3})$$$$=\frac{\pi r^2}{3}-\frac{r^2}{2}\times \frac{\sqrt{3}}{2}$$$$=\frac{\pi r^2}{3}-\frac{\sqrt{3}{r}^2}{4}$$$$=\frac{\pi 6^2}{3}-\frac{\sqrt{3}\times 6^2}{4}$$$$=12\pi -9\sqrt{3}$$

Answered by ajfour last updated on 18/Apr/17

$$=2\left(\frac{\pi r^2}{6}\right)-\frac{\sqrt{3}}{4}{r}^2$$$$=12\pi -9\sqrt{3}.$$

Commented by ajfour last updated on 18/Apr/17

$$sectorareaABDwitbcentreA$$$$+sectorareaBDAwithcentreB$$$$-areaof△ABD.$$

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