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Question-191275

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Question Number 191275 by Mingma last updated on 22/Apr/23
Answered by mr W last updated on 22/Apr/23
Commented by mr W last updated on 22/Apr/23
r=1, R=4 AD=(√((2R)^2 +h^2 ))=(√(4R^2 +h^2 )) DC=h−(√((R+r)^2 −(R−r)^2 ))=h−2(√(Rr)) ((OB)/h)=((AB)/(2R))=(R/( (√(4R^2 +h^2 )))) ⇒OB=((Rh)/( (√(4R^2 +h^2 )))) ⇒AB=((2R^2 )/( (√(4R^2 +h^2 )))) BC=(√(4R^2 +h^2 ))−h+2(√(Rr))−((2R^2 )/( (√(4R^2 +h^2 )))) BC=(√((R+r)^2 −(((Rh)/( (√(4R^2 +h^2 ))))−r)^2 )) (√(4R^2 +h^2 ))−h+2(√(Rr))−((2R^2 )/( (√(4R^2 +h^2 ))))=(√((R+r)^2 −(((Rh)/( (√(4R^2 +h^2 ))))−r)^2 )) let μ=(r/R), λ=(h/R) (√(4+λ^2 ))−λ+2(√μ)−(2/( (√(4+λ^2 ))))=(√((1+μ)^2 −((λ/( (√(4+λ^2 ))))−μ)^2 )) area of blue triangle Δ=(((2R)h)/2)=λR^2 with μ=(r/R)=(1/4), we get λ=1.5 ⇒Δ=1.5×4^2 =24
$${r}=\mathrm{1},\:{R}=\mathrm{4} \\ $$$${AD}=\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} }=\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} } \\ $$$${DC}={h}−\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }={h}−\mathrm{2}\sqrt{{Rr}} \\ $$$$\frac{{OB}}{{h}}=\frac{{AB}}{\mathrm{2}{R}}=\frac{{R}}{\:\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }} \\ $$$$\Rightarrow{OB}=\frac{{Rh}}{\:\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{2}{R}^{\mathrm{2}} }{\:\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }} \\ $$$${BC}=\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }−{h}+\mathrm{2}\sqrt{{Rr}}−\frac{\mathrm{2}{R}^{\mathrm{2}} }{\:\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }} \\ $$$${BC}=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left(\frac{{Rh}}{\:\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }}−{r}\right)^{\mathrm{2}} } \\ $$$$\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }−{h}+\mathrm{2}\sqrt{{Rr}}−\frac{\mathrm{2}{R}^{\mathrm{2}} }{\:\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }}=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left(\frac{{Rh}}{\:\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }}−{r}\right)^{\mathrm{2}} } \\ $$$${let}\:\mu=\frac{{r}}{{R}},\:\lambda=\frac{{h}}{{R}} \\ $$$$\sqrt{\mathrm{4}+\lambda^{\mathrm{2}} }−\lambda+\mathrm{2}\sqrt{\mu}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}+\lambda^{\mathrm{2}} }}=\sqrt{\left(\mathrm{1}+\mu\right)^{\mathrm{2}} −\left(\frac{\lambda}{\:\sqrt{\mathrm{4}+\lambda^{\mathrm{2}} }}−\mu\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${area}\:{of}\:{blue}\:{triangle} \\ $$$$\Delta=\frac{\left(\mathrm{2}{R}\right){h}}{\mathrm{2}}=\lambda{R}^{\mathrm{2}} \\ $$$$ \\ $$$${with}\:\mu=\frac{{r}}{{R}}=\frac{\mathrm{1}}{\mathrm{4}},\:{we}\:{get}\:\lambda=\mathrm{1}.\mathrm{5} \\ $$$$\Rightarrow\Delta=\mathrm{1}.\mathrm{5}×\mathrm{4}^{\mathrm{2}} =\mathrm{24} \\ $$
Commented by Mingma last updated on 22/Apr/23
Excellent, sir!

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