Question Number 227730 by mr W last updated on 15/Feb/26
Answered by ebm last updated on 16/Feb/26
Answered by fantastic2 last updated on 16/Feb/26
Commented by fantastic2 last updated on 16/Feb/26
$$\left({r}−\sqrt{\mathrm{9}−{r}^{\mathrm{2}} }\right)\left({r}+\sqrt{\mathrm{9}−{r}^{\mathrm{2}} }\right)=\mathrm{3}×\mathrm{2} \\ $$$${r}^{\mathrm{2}} −\mathrm{9}+{r}^{\mathrm{2}} =\mathrm{6} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} =\mathrm{15} \\ $$$${r}^{\mathrm{2}} /\mathrm{2}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$$$\therefore\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}} =\frac{\mathrm{15}\pi}{\mathrm{4}} \\ $$
Commented by mr W last updated on 16/Feb/26
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Answered by mr W last updated on 16/Feb/26
Commented by mr W last updated on 16/Feb/26
$$\frac{{R}}{\mathrm{3}}=\frac{\mathrm{2}+\mathrm{3}}{\mathrm{2}{R}} \\ $$$$\Rightarrow{R}^{\mathrm{2}} =\frac{\mathrm{15}}{\mathrm{2}} \\ $$$${area}\:{of}\:{semicircle}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{15}\pi}{\mathrm{4}} \\ $$