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Question Number 86528 by TawaTawa1 last updated on 29/Mar/20

Commented by TawaTawa1 last updated on 29/Mar/20

$$\mathrm{Student}\mathrm{sir}.$$

Commented by jagoll last updated on 29/Mar/20

$$\sin {60}^{\mathrm{o}}=\frac{\mathrm{h}}{12}\Rightarrow \mathrm{h}=6\sqrt{3}$$$$\cos {60}^{\mathrm{o}}=\frac{\mathrm{TS}}{12}\Rightarrow \mathrm{TS}=6$$$$\mathrm{TO}=6+14+6=26$$$$\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{portion}$$$$=\left(\frac{14+26}{2}\right)\times 6\sqrt{3}-\frac{1}{2}\times \frac{22}{7}\times 49$$$$=120\sqrt{3}-77$$$$=120\times 1.3-77=207.85-77$$$$=130.85$$

Commented by TawaTawa1 last updated on 29/Mar/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}$$

Commented by jagoll last updated on 29/Mar/20

$$\mathrm{thank}\mathrm{you}$$

Commented by john santu last updated on 29/Mar/20

$$\mathrm{miss}\mathrm{tawa}\mathrm{are}\mathrm{you}\mathrm{a}\mathrm{student}\mathrm{or}$$$$\mathrm{a}\mathrm{teacher}?$$

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