Question Number 208690 by Tawa11 last updated on 21/Jun/24
Answered by mr W last updated on 21/Jun/24
$$\left(\mathrm{2}{r}−\mathrm{3}\right)×\mathrm{3}=\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$${shaded}\:{area}=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{13}}{\mathrm{6}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{13}}{\mathrm{6}}\right)×\mathrm{2}\approx\mathrm{3}.\mathrm{04} \\ $$
Commented by Tawa11 last updated on 21/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{sir}, \\ $$$$\mathrm{You}\:\mathrm{did}\:\mathrm{this}\:\mathrm{same}\:\mathrm{approach}\:\mathrm{in}\:\mathrm{my} \\ $$$$\mathrm{previous}\:\mathrm{question}. \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{procedure}. \\ $$$$ \\ $$$$\mathrm{How}:\:\:\:\left(\mathrm{2r}\:\:−\:\:\mathrm{3}\right)\:×\:\mathrm{3}\:\:=\:\:\mathrm{2}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{You}\:\mathrm{did}:\:\:\mathrm{a}^{\mathrm{2}} \:\:=\:\:\left(\mathrm{r}\:\:−\:\:\mathrm{a}\right)\left(\mathrm{r}\:\:+\:\:\mathrm{a}\right)\:\:\mathrm{in}\:\mathrm{previous}\:\mathrm{question}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{did}\:\mathrm{the}\:\mathrm{same}\:\mathrm{thing}\:\mathrm{here}\:\mathrm{too}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 21/Jun/24
$$\mathrm{Great},\:\mathrm{I}\:\mathrm{now}\:\mathrm{understand}\:\mathrm{sir}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 21/Jun/24
Commented by mr W last updated on 21/Jun/24
Commented by Tawa11 last updated on 21/Jun/24
$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{see}\:\mathrm{question}\:\mathrm{208467}\:\mathrm{if}\:\mathrm{you}\:\mathrm{have} \\ $$$$\mathrm{alternative}\:\mathrm{approach}\:\mathrm{sir}. \\ $$