Question Number 190191 by normans last updated on 29/Mar/23
Answered by a.lgnaoui last updated on 30/Mar/23
$${the}\:{Area}\:{divided}\:{by}\:\mathrm{4}\:{portions} \\ $$$$\mathrm{1}\bullet{head}\:{of}\:{bird} \\ $$$$\:\:\:{A}_{\mathrm{1}} =\frac{\mathrm{1}×\pi}{\mathrm{2}} \\ $$$$\mathrm{2}\bullet{tronc} \\ $$$${A}_{\mathrm{2}} =\frac{\mathrm{2}^{\mathrm{2}} \pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}=\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\mathrm{3}\bullet{Right}\:{Wing}\left({portion}\right) \\ $$$${A}_{\mathrm{3}} =\mathrm{2}×\left[\frac{\pi}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right]=\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\:{A}_{\mathrm{4}} =\mathrm{1}−\frac{\pi}{\mathrm{4}} \\ $$$$\:\:{total}\:{Area}=\frac{\pi}{\mathrm{2}}+\frac{\mathrm{3}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{4}−\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\:\:\:\:=\frac{\mathrm{25}\pi+\mathrm{12}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{Area}}=\mathrm{7},\mathrm{112}{cm}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 30/Mar/23