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Question-218384

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Question Number 218384 by Spillover last updated on 08/Apr/25
Answered by Nicholas666 last updated on 09/Apr/25
percentage of red area: ((15π(√3))/(32.3)).100%=((5π(√3))/(32))=100% =((5.π.1,732)/(32)).100%=((27,206)/(32)).100%=85,02%
$${percentage}\:{of}\:{red}\:{area}: \\ $$$$\frac{\mathrm{15}\pi\sqrt{\mathrm{3}}}{\mathrm{32}.\mathrm{3}}.\mathrm{100\%}=\frac{\mathrm{5}\pi\sqrt{\mathrm{3}}}{\mathrm{32}}=\mathrm{100\%} \\ $$$$=\frac{\mathrm{5}.\pi.\mathrm{1},\mathrm{732}}{\mathrm{32}}.\mathrm{100\%}=\frac{\mathrm{27},\mathrm{206}}{\mathrm{32}}.\mathrm{100\%}=\mathrm{85},\mathrm{02\%} \\ $$$$ \\ $$
Answered by Spillover last updated on 09/Apr/25
Commented by mr W last updated on 11/Apr/25
but this is not the same figure as in the question!
$${but}\:{this}\:{is}\:{not}\:{the}\:{same}\:{figure}\:{as}\: \\ $$$${in}\:{the}\:{question}! \\ $$
Answered by mr W last updated on 10/Apr/25
Commented by mr W last updated on 10/Apr/25
R, r=radii of big and small circle a=side length of equilateral triangle a=6r+2(√3)r=2(3+(√3))r a=(√3)r+2r+(√((R+r)^2 −(R−r)^2 ))+(√3)R =((√3)+2)r+2(√(Rr))+(√3)R ((√3)+2)r+2(√(Rr))+(√3)R=2(3+(√3))r (√3)R+2(√(Rr))−(4+(√3))r=0 ((√R)/( (√r)))=((−1+(√(1+(√3)(4+(√3)))))/( (√3)))=((−1+2(√(1+(√3))))/( (√3))) (R/r)=(((−1+2(√(1+(√3))))/( (√3))))^2 =((5+4(√3)−4(√(1+(√3))))/3) area of triangle: A_Δ =(((√3)a^2 )/4)=(((√3)×4(3+(√3))^2 r^2 )/4)=6(2(√3)+3)r^2 red area: A_(red) =6πr^2 +2πR^2 =6πr^2 +π(((5+4(√3)−4(√(1+(√3))))/3))^2 r^2 ((red)/Δ)=((6π+π(((5+4(√3)−4(√(1+(√3))))/3))^2 )/(6(2(√3)+3)))≈74%
$${R},\:{r}={radii}\:{of}\:{big}\:{and}\:{small}\:{circle} \\ $$$${a}={side}\:{length}\:{of}\:{equilateral}\:{triangle} \\ $$$${a}=\mathrm{6}{r}+\mathrm{2}\sqrt{\mathrm{3}}{r}=\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{3}}\right){r} \\ $$$${a}=\sqrt{\mathrm{3}}{r}+\mathrm{2}{r}+\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }+\sqrt{\mathrm{3}}{R} \\ $$$$\:\:=\left(\sqrt{\mathrm{3}}+\mathrm{2}\right){r}+\mathrm{2}\sqrt{{Rr}}+\sqrt{\mathrm{3}}{R} \\ $$$$\:\left(\sqrt{\mathrm{3}}+\mathrm{2}\right){r}+\mathrm{2}\sqrt{{Rr}}+\sqrt{\mathrm{3}}{R}=\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{3}}\right){r} \\ $$$$\:\sqrt{\mathrm{3}}{R}+\mathrm{2}\sqrt{{Rr}}−\left(\mathrm{4}+\sqrt{\mathrm{3}}\right){r}=\mathrm{0} \\ $$$$\frac{\sqrt{{R}}}{\:\sqrt{{r}}}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)}}{\:\sqrt{\mathrm{3}}}=\frac{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{R}}{{r}}=\left(\frac{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\frac{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{4}\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}}{\mathrm{3}} \\ $$$${area}\:{of}\:{triangle}: \\ $$$${A}_{\Delta} =\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\sqrt{\mathrm{3}}×\mathrm{4}\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {r}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{6}\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\right){r}^{\mathrm{2}} \\ $$$${red}\:{area}: \\ $$$${A}_{{red}} =\mathrm{6}\pi{r}^{\mathrm{2}} +\mathrm{2}\pi{R}^{\mathrm{2}} =\mathrm{6}\pi{r}^{\mathrm{2}} +\pi\left(\frac{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{4}\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}}{\mathrm{3}}\right)^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$$\frac{{red}}{\Delta}=\frac{\mathrm{6}\pi+\pi\left(\frac{\mathrm{5}+\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{4}\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}}{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{6}\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\right)}\approx\mathrm{74\%} \\ $$

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