Question Number 193256 by cherokeesay last updated on 08/Jun/23
Answered by som(math1967) last updated on 09/Jun/23
Commented by som(math1967) last updated on 09/Jun/23
$${DE}=\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }=\mathrm{4}{cm} \\ $$$${AD}=\mathrm{5}{cm}\:,{AB}={AF}+{FB}=\mathrm{5}{cm}+\mathrm{5}{cm}=\mathrm{10}{cm} \\ $$$${let}\:\angle{ABC}=\theta\:\therefore\angle{DAB}=\mathrm{180}−\theta \\ $$$${purple}\:{region}\:{area}\: \\ $$$$\:={Ar}\:{of}\:{semicircle}+{Ar}.\:{of}\:{ABCD} \\ $$$$−{Ar}.\:{of}\:{sectorADF}−{Ar}\:{of}\:{sectorBCF} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\pi×\mathrm{5}^{\mathrm{2}} +\mathrm{10}×\mathrm{4}−\frac{\mathrm{180}−\theta}{\mathrm{360}}×\pi×\mathrm{5}^{\mathrm{2}} \\ $$$$\:−\frac{\theta}{\mathrm{360}}×\pi×\mathrm{5}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{25}\pi}{\mathrm{2}}\:+\mathrm{40}−\mathrm{25}\pi\left(\frac{\mathrm{180}−\theta+\theta}{\mathrm{360}}\right) \\ $$$$=\mathrm{40}{cm}^{\mathrm{2}} \\ $$
Commented by cherokeesay last updated on 09/Jun/23
$${thank}\:{you}\:! \\ $$