Question-188451

Question Number 188451 by normans last updated on 01/Mar/23

Answered by mr W last updated on 01/Mar/23

Commented by mr W last updated on 01/Mar/23

r=radius s=side length of blue (equilateral) triangle (r/(2.5))=((2.5(√3)−r)/(2.5(√3))) ⇒r=((2.5(√3))/( (√3)+1))=((2.5(3−(√3)))/2) (1/3)×((s(√3))/2)=2.5(√3)−2r=5(√3)−7.5 ⇒s=30−15(√3) area of blue triangle: A=(((√3)s^2 )/4)=((1575(√3))/4)−674≈6.995

$${r}={radius} \\ $$$${s}={side}\:{length}\:{of}\:{blue}\:\left({equilateral}\right)\:{triangle} \\ $$$$\frac{{r}}{\mathrm{2}.\mathrm{5}}=\frac{\mathrm{2}.\mathrm{5}\sqrt{\mathrm{3}}−{r}}{\mathrm{2}.\mathrm{5}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}.\mathrm{5}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}=\frac{\mathrm{2}.\mathrm{5}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{2}.\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{2}{r}=\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{7}.\mathrm{5} \\ $$$$\Rightarrow{s}=\mathrm{30}−\mathrm{15}\sqrt{\mathrm{3}} \\ $$$${area}\:{of}\:{blue}\:{triangle}: \\ $$$${A}=\frac{\sqrt{\mathrm{3}}{s}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{1575}\sqrt{\mathrm{3}}}{\mathrm{4}}−\mathrm{674}\approx\mathrm{6}.\mathrm{995} \\ $$

Answered by mr W last updated on 01/Mar/23

Commented by mr W last updated on 02/Mar/23

2r+2×(r/( (√3)))=5 ⇒r=((5(√3))/(2((√3)+1))) side of big equilateral: a=3×5=15 side of small equilateral=s (((√3)a)/6)=(((√3)s)/6)+2r ⇒s=a−((12r)/( (√3)))=15−((30)/( (√3)+1))=30−15(√3) area =(((√(3 ))s^2 )/4)=((1575(√3))/4)−675≈6.995

$$\mathrm{2}{r}+\mathrm{2}×\frac{{r}}{\:\sqrt{\mathrm{3}}}=\mathrm{5} \\ $$$$\Rightarrow{r}=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)} \\ $$$${side}\:{of}\:{big}\:{equilateral}:\:{a}=\mathrm{3}×\mathrm{5}=\mathrm{15} \\ $$$${side}\:{of}\:{small}\:{equilateral}={s} \\ $$$$\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}}=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{6}}+\mathrm{2}{r} \\ $$$$\Rightarrow{s}={a}−\frac{\mathrm{12}{r}}{\:\sqrt{\mathrm{3}}}=\mathrm{15}−\frac{\mathrm{30}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}=\mathrm{30}−\mathrm{15}\sqrt{\mathrm{3}} \\ $$$${area}\:=\frac{\sqrt{\mathrm{3}\:}{s}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{1575}\sqrt{\mathrm{3}}}{\mathrm{4}}−\mathrm{675}\approx\mathrm{6}.\mathrm{995} \\ $$

Answered by HeferH last updated on 02/Mar/23

Commented by HeferH last updated on 02/Mar/23

(r/( (√3))) + r = (5/2) 2r + 2r(√3) =5 r = ((5(√3))/(2(√3)+2)) = ((5(√3)(2(√3)−2))/(12−4)) = ((30−10(√3))/8) = ((15−5(√3))/4) ((5(√3))/2) − ((15−5(√3))/2) = ((10(√3)−15)/2) side of blue triangle = 2(√3) ∙ (((10(√3)−15)/2)) =30 − 15(√3) A = (((30−15(√3))^2 (√3))/4) ≈ 6.9u^2

$$\frac{{r}}{\:\sqrt{\mathrm{3}}}\:+\:{r}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\:\mathrm{2}{r}\:+\:\mathrm{2}{r}\sqrt{\mathrm{3}}\:=\mathrm{5} \\ $$$$\:{r}\:=\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}}\:=\:\frac{\mathrm{5}\sqrt{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}\right)}{\mathrm{12}−\mathrm{4}}\:=\:\frac{\mathrm{30}−\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{8}}\:=\:\frac{\mathrm{15}−\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\: \\ $$$$\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\:\frac{\mathrm{15}−\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\frac{\mathrm{10}\sqrt{\mathrm{3}}−\mathrm{15}}{\mathrm{2}}\: \\ $$$$\:{side}\:{of}\:{blue}\:{triangle}\:=\:\mathrm{2}\sqrt{\mathrm{3}}\:\centerdot\:\left(\frac{\mathrm{10}\sqrt{\mathrm{3}}−\mathrm{15}}{\mathrm{2}}\right)\:=\mathrm{30}\:−\:\mathrm{15}\sqrt{\mathrm{3}} \\ $$$$\:{A}\:=\:\frac{\left(\mathrm{30}−\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}}\:\approx\:\mathrm{6}.\mathrm{9}{u}^{\mathrm{2}} \\ $$

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