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Question Number 68503 by TawaTawa last updated on 12/Sep/19

Commented by Prithwish sen last updated on 12/Sep/19

$$2\mathrm{x}\frac{\pi 6^2}{4}-2\left[36(\frac{\pi }{3}-\frac{\sqrt{3}}{4})\right]=18[\sqrt{3}-\frac{\pi }{3}]\backsim 12.33$$$$\mathbf{please}\mathbf{check}.$$

Commented by TawaTawa last updated on 12/Sep/19

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{answer},\mathrm{but}\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{explain}\mathrm{more}.\mathrm{God}\mathrm{bless}$$$$\mathrm{you}\mathrm{sir}$$

Commented by mr W last updated on 12/Sep/19

$$2[\frac{1}{2}\times \frac{\pi }{6}\times 6^2-\frac{6^2}{2}(\frac{\pi }{3}-\sin \frac{\pi }{3})]$$$$=6(3\sqrt{3}-\pi )$$

Commented by TawaTawa last updated on 13/Sep/19

$$\mathrm{Sir},\mathrm{is}\mathrm{it}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{minus}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{quarter}\mathrm{circle}?$$

Commented by TawaTawa last updated on 13/Sep/19

$$\mathrm{Or}\mathrm{please}\mathrm{sir}\mathrm{help}\mathrm{me}\mathrm{label}\mathrm{the}\mathrm{diagram}$$

Commented by mr W last updated on 13/Sep/19

Commented by mr W last updated on 13/Sep/19

$$areaof\frac{\pi }{6}-sector=\frac{6^2}{2}\left(\frac{\pi }{6}\right)$$$$areaofgreensegment=\frac{6^2}{2}(\frac{\pi }{3}-\sin \frac{\pi }{3})$$

Commented by TawaTawa last updated on 13/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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