Question-212827

Question Number 212827 by Spillover last updated on 25/Oct/24

Answered by A5T last updated on 25/Oct/24

Commented by A5T last updated on 25/Oct/24

(y/((2r)/( (√3))))=(r/((4r)/( (√3))))⇒y=((2r(√3))/3)×((r(√3))/(4r))=(r/2) ⇒(2r)^2 =(r^2 /4)+(8−2r(√3)+(√(r^2 −(r^2 /4))))^2 ⇒r=4(√3)−((4(√(15)))/3) ⇒[green]=((8×8sin60)/2)−((3πr^2 )/2)−((3×2r×((2r)/( (√3))))/2) =64(√(15))−((400(√3))/3)+(48(√5)−112)π

$$\frac{{y}}{\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}}=\frac{{r}}{\frac{\mathrm{4}{r}}{\:\sqrt{\mathrm{3}}}}\Rightarrow{y}=\frac{\mathrm{2}{r}\sqrt{\mathrm{3}}}{\mathrm{3}}×\frac{{r}\sqrt{\mathrm{3}}}{\mathrm{4}{r}}=\frac{{r}}{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}{r}\right)^{\mathrm{2}} =\frac{{r}^{\mathrm{2}} }{\mathrm{4}}+\left(\mathrm{8}−\mathrm{2}{r}\sqrt{\mathrm{3}}+\sqrt{{r}^{\mathrm{2}} −\frac{{r}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{4}\sqrt{\mathrm{3}}−\frac{\mathrm{4}\sqrt{\mathrm{15}}}{\mathrm{3}} \\ $$$$\Rightarrow\left[{green}\right]=\frac{\mathrm{8}×\mathrm{8}{sin}\mathrm{60}}{\mathrm{2}}−\frac{\mathrm{3}\pi{r}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{3}×\mathrm{2}{r}×\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$$=\mathrm{64}\sqrt{\mathrm{15}}−\frac{\mathrm{400}\sqrt{\mathrm{3}}}{\mathrm{3}}+\left(\mathrm{48}\sqrt{\mathrm{5}}−\mathrm{112}\right)\pi \\ $$

Commented by Spillover last updated on 25/Oct/24

$${great}.{thanks} \\ $$

Commented by ajfour last updated on 25/Oct/24

((r−y)/(2r))=((r−(r/2))/(2r))=sin θ=(1/4) 8=((2r)/( (√3)))+2rcos θ+rcos (π/6)+(r/( (√3))) ⇒ r(((3(√3))/2)+((√(15))/2))=8 r=((4(3−(√5)))/( (√3))) G=16(√3)−(3/2)×(((2r)^2 )/( (√3)))−((3πr^2 )/2) =16(√3)−(2(√3)+((3π)/2))×((32)/3)(7−3(√5))

$$\frac{{r}−{y}}{\mathrm{2}{r}}=\frac{{r}−\frac{{r}}{\mathrm{2}}}{\mathrm{2}{r}}=\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{8}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}+\mathrm{2}{r}\mathrm{cos}\:\theta+{r}\mathrm{cos}\:\frac{\pi}{\mathrm{6}}+\frac{{r}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:{r}\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{15}}}{\mathrm{2}}\right)=\mathrm{8} \\ $$$${r}=\frac{\mathrm{4}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}{\:\sqrt{\mathrm{3}}} \\ $$$${G}=\mathrm{16}\sqrt{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{2}}×\frac{\left(\mathrm{2}{r}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{3}\pi{r}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:=\mathrm{16}\sqrt{\mathrm{3}}−\left(\mathrm{2}\sqrt{\mathrm{3}}+\frac{\mathrm{3}\pi}{\mathrm{2}}\right)×\frac{\mathrm{32}}{\mathrm{3}}\left(\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}\right) \\ $$$$\: \\ $$

Commented by ajfour last updated on 25/Oct/24

right, its okay now, thank you.

Commented by A5T last updated on 25/Oct/24

$${G}\approx\mathrm{2}.\mathrm{2636} \\ $$

Answered by mr W last updated on 25/Oct/24

Commented by mr W last updated on 25/Oct/24

(2r)^2 =(r+((2x)/( (√3))))^2 +(r+(x/( (√3))))^2 −(r+((2x)/( (√3))))(r+(x/( (√3)))) x^2 +(√3)rx−3r^2 =0 ⇒x=(((√3)((√5)−1)r)/2) ((4r)/( (√3)))+(((√3)((√5)−1)r)/2)+((2r)/( (√3)))=8 ⇒r=((4(√3)(3−(√5)))/3)≈1.764 green area =(((√3)×8^2 )/4)−((3×4r^2 )/(2×(√3)))−((3πr^2 )/2) =(((√3)×8^2 )/4)−(((4(√3)+3π)r^2 )/2) =16(√3)−((16(4(√3)+3π)(7−3(√5)))/3)

$$\left(\mathrm{2}{r}\right)^{\mathrm{2}} =\left({r}+\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left({r}+\frac{{x}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} −\left({r}+\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}}\right)\left({r}+\frac{{x}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{rx}−\mathrm{3}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){r}}{\mathrm{2}} \\ $$$$\frac{\mathrm{4}{r}}{\:\sqrt{\mathrm{3}}}+\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){r}}{\mathrm{2}}+\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}=\mathrm{8} \\ $$$$\Rightarrow{r}=\frac{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}{\mathrm{3}}\approx\mathrm{1}.\mathrm{764} \\ $$$${green}\:{area}\: \\ $$$$=\frac{\sqrt{\mathrm{3}}×\mathrm{8}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{3}×\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{2}×\sqrt{\mathrm{3}}}−\frac{\mathrm{3}\pi{r}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{\sqrt{\mathrm{3}}×\mathrm{8}^{\mathrm{2}} }{\mathrm{4}}−\frac{\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\pi\right){r}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\mathrm{16}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\pi\right)\left(\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}\right)}{\mathrm{3}} \\ $$

Commented by A5T last updated on 25/Oct/24