Question Number 219068 by Spillover last updated on 19/Apr/25
Answered by mr W last updated on 20/Apr/25
Commented by mr W last updated on 20/Apr/25
$${R}=\mathrm{2}{r} \\ $$$${h}=\sqrt{{a}\left(\mathrm{2}{r}−{a}\right)} \\ $$$${a}^{\mathrm{2}} +\left({a}+{h}\right)^{\mathrm{2}} =\left(\mathrm{2}{r}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{a}\sqrt{{a}\left(\mathrm{2}{r}−{a}\right)}+\mathrm{2}{ar}=\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\mathrm{2}{a}\sqrt{{a}\left(\mathrm{2}{r}−{a}\right)}=\mathrm{4}{r}^{\mathrm{2}} −\mathrm{2}{ar}−{a}^{\mathrm{2}} \\ $$$$\mathrm{5}{a}^{\mathrm{4}} −\mathrm{4}{ra}^{\mathrm{3}} −\mathrm{4}{a}^{\mathrm{2}} {r}^{\mathrm{2}} −\mathrm{16}{r}^{\mathrm{3}} {a}+\mathrm{16}{r}^{\mathrm{4}} =\mathrm{0} \\ $$$${let}\:\lambda=\frac{{a}}{{r}} \\ $$$$\mathrm{5}\lambda^{\mathrm{4}} −\mathrm{4}\lambda^{\mathrm{3}} −\mathrm{4}{r}^{\mathrm{2}} −\mathrm{16}\lambda+\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow\lambda\approx\mathrm{0}.\mathrm{8326} \\ $$$${A}_{{green}} =\pi\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\pi{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${A}_{{total}} =\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}=\pi{r}^{\mathrm{2}} \\ $$$${fraction}\:{of}\:{shaded}\:=\frac{\lambda^{\mathrm{2}} }{\mathrm{4}}\approx\mathrm{0}.\mathrm{173} \\ $$
Commented by Spillover last updated on 20/Apr/25
$${thank}\:{you} \\ $$
Answered by Spillover last updated on 20/Apr/25
