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Question Number 165514 by Tawa11 last updated on 02/Feb/22

Answered by som(math1967) last updated on 03/Feb/22

$$\mathit{a})\mathit{l}\mathit{e}\mathit{t}\mathit{r}\mathit{a}\mathit{d}\mathit{i}\mathit{u}\mathit{s}\mathit{O}\mathit{C}=\mathit{x}$$$$\therefore \mathit{O}\mathit{D}=(\mathit{x}-2)$$$$\mathit{f}\mathit{r}\mathit{o}\mathit{m}△\mathit{O}\mathit{E}\mathit{D}$$$${\mathit{x}}^2=6^2+{(\mathit{x}-2)}^2$$$${\mathit{x}}^2=36+{\mathit{x}}^2-4\mathit{x}+4\Rightarrow \mathit{x}=10$$$$\mathit{b})$$$$\mathrm{\angle }\mathit{E}\mathit{O}\mathit{C}={\sin }^{-1}\frac{6}{10}={37}^{°}°\left(\mathit{a}\mathit{p}\mathit{r}\mathit{o}\mathit{x}\right)$$$$\mathit{a}\mathit{r}\mathit{e}\mathit{a}\mathit{o}\mathit{f}\mathit{s}\mathit{e}\mathit{c}\mathit{t}\mathit{o}\mathit{r}$$$$=\frac{37}{360}\times \mathit{\pi }\times {10}^2\mathit{s}\mathit{q}\mathit{m}$$

Commented by som(math1967) last updated on 03/Feb/22

Commented by Tawa11 last updated on 03/Feb/22

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

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