Question Number 225658 by Jubr last updated on 05/Nov/25
Answered by ajfour last updated on 06/Nov/25
$${p}={R}\mathrm{sin}\:\alpha \\ $$$$\pi−\alpha=\frac{\mathrm{5}}{\mathrm{2}}\:\:\Rightarrow\:\:\mathrm{2}\alpha=\mathrm{2}\pi−\mathrm{5} \\ $$$$\angle{AOC}=\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\alpha\right)=\pi−\mathrm{2}\alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\pi−\left(\mathrm{2}\pi−\mathrm{5}\right)=\mathrm{5}−\pi \\ $$$${A}_{{shade}} =\frac{\mathrm{1}}{\mathrm{2}}{R}^{\mathrm{2}} \left(\mathrm{5}−\pi\right)−{bp} \\ $$$${R}=\mathrm{12}{cm},\: \\ $$$${bp}=\mathrm{144sin}\:\left(\pi−\frac{\mathrm{5}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\pi−\frac{\mathrm{5}}{\mathrm{2}}\right){cm} \\ $$$$\:\:\:\:=\mathrm{72sin}\:\left(\mathrm{2}\pi−\mathrm{5}\right){cm}^{\mathrm{2}} \\ $$$${A}_{{shade}} =\mathrm{72}\left\{\left(\mathrm{5}−\pi\right)−\mathrm{sin}\:\left(\mathrm{2}\pi−\mathrm{5}\right)\right\}{cm}^{\mathrm{2}} \\ $$$$\:\:=\mathrm{72}\left(\mathrm{5}−\pi+\mathrm{sin}\:\mathrm{5}\right){cm}^{\mathrm{2}} \\ $$$$\:\approx\mathrm{202}.\mathrm{85}\:{cm}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by ajfour last updated on 06/Nov/25
https://youtu.be/Kf_-c7AGuXs?si=jruJ379rEHYdg7qh
Commented by mr W last updated on 09/Nov/25
Commented by mr W last updated on 09/Nov/25
$${AF}=\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{r}−{r}^{\mathrm{2}} } \\ $$$${FE}=\sqrt{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{2}{r}−{r}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{r}−\mathrm{2}{r}^{\mathrm{2}} } \\ $$$${CD}=\sqrt{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\sqrt{\mathrm{1}−\mathrm{2}{r}} \\ $$$${AE}={AD} \\ $$$$\mathrm{2}\sqrt{{r}−{r}^{\mathrm{2}} }+\mathrm{2}\sqrt{{r}−\mathrm{2}{r}^{\mathrm{2}} }=\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}{r}} \\ $$$$\Rightarrow{r}\approx\mathrm{0}.\mathrm{31290632} \\ $$