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Question Number 87671 by TawaTawa1 last updated on 05/Apr/20

Commented by TawaTawa1 last updated on 05/Apr/20

$$\mathrm{I}\mathrm{got}10.5{\mathrm{m}}^2\mathrm{but}\mathrm{am}\mathrm{not}\mathrm{sure}.$$$$\mathrm{Please}\mathrm{help}\mathrm{me}\mathrm{check}.$$

Commented by Tony Lin last updated on 05/Apr/20

$$7^2\times \frac{2\pi }{3}\times \frac{1}{2}-7^2\times \frac{\sqrt{3}}{4}$$$$=49(\frac{\pi }{3}-\frac{\sqrt{3}}{4})$$$$\approx 30\left({cm}^2\right)$$$$=0.003\left(m^2\right)$$$$0.0049-0.003=0.0019\left(m^2\right)$$

Commented by TawaTawa1 last updated on 05/Apr/20

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

Answered by mr W last updated on 05/Apr/20

$$A_{shade}=7\times 7-2\times \frac{\pi \times 7^2}{6}+\frac{7^2\times \sqrt{3}}{4}$$$$=49-\frac{49\pi }{3}+\frac{49\sqrt{3}}{4}$$$$\approx 18.9{cm}^2=0.00189m^2$$

Commented by TawaTawa1 last updated on 05/Apr/20

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

Commented by TawaTawa1 last updated on 05/Apr/20

$$\mathrm{Sir},\mathrm{have}\mathrm{been}\mathrm{trying}\mathrm{to}\mathrm{understand}-2\times \frac{\pi \times 7^2}{6}+\frac{7^2\times \sqrt{3}}{4}$$$$\mathrm{But}\mathrm{since}\mathrm{i}\mathrm{did}\mathrm{not}\mathrm{understand}.$$$$\mathrm{Please}\mathrm{help}\mathrm{me}\mathrm{explain}\mathrm{when}\mathrm{you}\mathrm{are}\mathrm{chanced}\mathrm{sir}.$$$$\mathrm{Please}.$$

Commented by mr W last updated on 05/Apr/20

Commented by mr W last updated on 05/Apr/20

Commented by mr W last updated on 05/Apr/20

$$A_1=square=7\times 7$$$$A_2=sector=\frac{circle}{6}=\frac{\pi \times 7^2}{6}$$$$A_3=equilateraltriangle=\frac{1}{2}\times 7\times \frac{\sqrt{3}\times 7}{2}=\frac{7^2\sqrt{3}}{4}$$$$A_{shade}=A_1-2A_2+A_3$$$$=7\times 7-2\times \frac{7^2\pi }{6}+\frac{7^2\sqrt{3}}{4}$$$$=49-\frac{49\pi }{3}+\frac{49\sqrt{3}}{4}$$

Commented by TawaTawa1 last updated on 05/Apr/20

$$\mathrm{Hahaha}.\mathrm{Am}\mathrm{happy}.$$$$\mathrm{I}\mathrm{understand}\mathrm{now},\mathrm{God}\mathrm{will}\mathrm{be}\mathrm{helping}\mathrm{you}\mathrm{too}\mathrm{sir}.$$

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