Question Number 156940 by amin96 last updated on 17/Oct/21
Answered by mr W last updated on 17/Oct/21
Commented by mr W last updated on 17/Oct/21
![r=radius of semicircle A_(green) =((π(2r)^2 )/4)−(((2r)^2 )/2)−[((πr^2 )/4)−(r^2 /2)] A_(green) =((3(π−2)r^2 )/4) θ=60° OA=(r/(tan (θ/2))) OB=2r AB=b=10 OB^2 =(2r)^2 =((r/(tan (θ/2))))^2 +b^2 −2b×(r/(tan (θ/2)))cos θ (4−(1/(tan^2 30)))r^2 +20×(r/(tan 30))cos 60−100=0 r^2 +10(√3)r−100=0 r=5((√7)−(√3)) A_(green) =((75(π−2)(5−(√(21))))/2)≈17.8698](https://www.tinkutara.com/question/Q156945.png)
$${r}={radius}\:{of}\:{semicircle} \\ $$$${A}_{{green}} =\frac{\pi\left(\mathrm{2}{r}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{\left(\mathrm{2}{r}\right)^{\mathrm{2}} }{\mathrm{2}}−\left[\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\right] \\ $$$${A}_{{green}} =\frac{\mathrm{3}\left(\pi−\mathrm{2}\right){r}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$$$\theta=\mathrm{60}° \\ $$$${OA}=\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${OB}=\mathrm{2}{r} \\ $$$${AB}={b}=\mathrm{10} \\ $$$${OB}^{\mathrm{2}} =\left(\mathrm{2}{r}\right)^{\mathrm{2}} =\left(\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{b}×\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\mathrm{cos}\:\theta \\ $$$$\left(\mathrm{4}−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\mathrm{30}}\right){r}^{\mathrm{2}} +\mathrm{20}×\frac{{r}}{\mathrm{tan}\:\mathrm{30}}\mathrm{cos}\:\mathrm{60}−\mathrm{100}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} +\mathrm{10}\sqrt{\mathrm{3}}{r}−\mathrm{100}=\mathrm{0} \\ $$$${r}=\mathrm{5}\left(\sqrt{\mathrm{7}}−\sqrt{\mathrm{3}}\right) \\ $$$${A}_{{green}} =\frac{\mathrm{75}\left(\pi−\mathrm{2}\right)\left(\mathrm{5}−\sqrt{\mathrm{21}}\right)}{\mathrm{2}}\approx\mathrm{17}.\mathrm{8698} \\ $$
Commented by Tawa11 last updated on 17/Oct/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by amin96 last updated on 17/Oct/21
$${cool}\:{sir}.\:{bravo} \\ $$