Question Number 72997 by yannickmendes_33 last updated on 05/Nov/19
$${The}\:{area}\:{of}\:{the}\:{equilateral}\:{triangle}\:{is}\:{equal}\:{to}\:\frac{\sqrt{\mathrm{16}}\sqrt{\mathrm{8}}}{\mathrm{3}\sqrt{\pi}} \\ $$$${Calculate}\:{the}\:{area}\:{of}\:{the}\:{circle}\:{inscribed}\:{in}\:{the}\:{triangle}. \\ $$$$\: \\ $$
Answered by Kunal12588 last updated on 05/Nov/19
$${area}\:{of}\:{equilateral}\:\bigtriangleup\:=\:\frac{\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\sqrt{\mathrm{16}}\sqrt{\mathrm{8}}}{\mathrm{3}\sqrt{\pi}}=\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{\pi}} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{\mathrm{3}\pi}} \\ $$$${area}\:{of}\:{circle}\:=\:\pi\:{r}^{\mathrm{2}} \\ $$$$\frac{{h}_{\mathrm{1}} }{{h}_{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{1}} \\ $$$$\Rightarrow{h}_{\mathrm{1}} =\mathrm{2}{h}_{\mathrm{2}} \\ $$$${h}_{\mathrm{1}} +{h}_{\mathrm{2}} ={h} \\ $$$$\Rightarrow\mathrm{3}{h}_{\mathrm{2}} ={h} \\ $$$$\Rightarrow{h}_{\mathrm{2}} =\frac{{h}}{\mathrm{3}} \\ $$$${h}_{\mathrm{2}} ={r}=\frac{{h}}{\mathrm{3}} \\ $$$${ar}\:{circle}=\pi{r}^{\mathrm{2}} \\ $$$$=\pi\left(\frac{{h}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$=\pi\left(\frac{\mathrm{1}}{\mathrm{3}}×\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\pi×\frac{\mathrm{1}}{\mathrm{3}}×\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\pi×\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{\mathrm{3}\pi}} \\ $$$$=\frac{\mathrm{8}\sqrt{\mathrm{6}\pi}}{\mathrm{27}} \\ $$