Question Number 86256 by TawaTawa1 last updated on 27/Mar/20
Answered by ajfour last updated on 27/Mar/20
$$\pi{r}^{\mathrm{2}} =\pi{a}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$
Answered by ajfour last updated on 27/Mar/20
Commented by ajfour last updated on 27/Mar/20
$${area}\:{of}\:\bigtriangleup=\:\frac{{s}^{\mathrm{2}} }{\mathrm{2}}=\:{a}^{\mathrm{2}} \:\:\left({given}\right) \\ $$$${AB}+{BC}\:=\:\frac{{s}}{\:\sqrt{\mathrm{2}}}+{r}\:=\:{r}\sqrt{\mathrm{2}} \\ $$$${r}=\frac{{s}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:=\:\frac{{s}}{\mathrm{2}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right) \\ $$$${Area}\:{of}\:{circle}\:=\:\pi{r}^{\mathrm{2}} \\ $$$$\:\:\:\:=\:\frac{\pi{s}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)\:=\:\pi{a}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:. \\ $$
Commented by TawaTawa1 last updated on 28/Mar/20
$$\mathrm{Wow},\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$