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Question Number 160654 by Avijit007 last updated on 04/Dec/21

Answered by som(math1967) last updated on 04/Dec/21

$$letradiusofsemicirclercm$$$$\therefore {\mathit{r}}^2={(\mathit{r}-1)}^2+3^2$$$$2\mathit{r}-1=9$$$$\therefore \mathit{r}=5\mathit{c}\mathit{m}$$$$\mathit{O}R=5-1=4cm$$$$\mathit{A}r\mathit{e}\mathit{a}\mathit{o}\mathit{f}\mathit{s}\mathit{h}\mathit{a}\mathit{d}\mathit{e}\mathit{d}\mathit{r}\mathit{e}\mathit{g}\mathit{i}\mathit{o}\mathit{n}$$$$=\frac{1}{4}\times \mathit{\pi }\times 5^2-4\times 3$$$$=(6.25\mathit{\pi }-12){\mathit{c}\mathit{m}}^2$$$$\mathit{B}\mathit{a}\mathit{n}\mathit{s}$$

Commented by Tawa11 last updated on 05/Dec/21

$$\mathrm{Great}\mathrm{sir}$$

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