Question Number 226014 by fantastic2 last updated on 18/Nov/25
Commented by fantastic2 last updated on 18/Nov/25
$${i}\:{get}\:\mathrm{3}.{What}\:{do}\:{you}\:{get} \\ $$$${mrW}\:{sir}? \\ $$
Answered by mr W last updated on 18/Nov/25
$${R}={radius}\:{big}\:{semi}−{circle} \\ $$$${a}={radius}\:{of}\:{small}\:{black}\:{circle} \\ $$$${b}={radius}\:{of}\:{small}\:{red}\:{circle} \\ $$$${b}=\frac{{R}}{\mathrm{4}} \\ $$$${a}=\frac{{R}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\frac{{red}\:{circle}\:{area}}{{black}\:{circle}\:{area}}=\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{red}\:{circle}\:{area}\:=\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{4}=\mathrm{3} \\ $$
Answered by fantastic2 last updated on 18/Nov/25
Commented by fantastic2 last updated on 18/Nov/25
$${r}\:{of}\:{small}\:{black}\:{C}={r} \\ $$$${r}\:{of}\:{half}\:{black}\:{C}={R} \\ $$$${r}\:{of}\:{small}\:{red}\:{C}={R} \\ $$$$\pi{r}^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}}{\:\sqrt{\pi}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{3}\pi{R}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{R}^{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\:\sqrt{\pi}}\:\Rightarrow{R}^{\mathrm{2}} =\frac{\mathrm{48}}{\pi} \\ $$$${l}\:{of}\:{C}\:{tangent}\:=\sqrt{\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \mathrm{cos}\:\mathrm{120}^{\mathrm{0}} } \\ $$$$={R}\sqrt{\mathrm{3}} \\ $$$${h}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{R}\sqrt{\mathrm{3}}{h}=\frac{\mathrm{1}}{\mathrm{2}}{R}^{\mathrm{2}} \mathrm{sin}\:\mathrm{120}^{\mathrm{0}} \\ $$$${h}=\frac{{R}}{\mathrm{2}} \\ $$$${h}+\mathrm{2}{R}={R} \\ $$$$\Rightarrow{R}=\frac{{R}}{\mathrm{4}} \\ $$$${A}=\pi{R}^{\mathrm{2}} =\frac{\pi×\frac{\mathrm{48}}{\pi}}{\mathrm{16}}=\mathrm{3} \\ $$