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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

$$r=1,R=4$$$$AD=\sqrt{{\left(2R\right)}^2+h^2}=\sqrt{4R^2+h^2}$$$$DC=h-\sqrt{{(R+r)}^2-{(R-r)}^2}=h-2\sqrt{Rr}$$$$\frac{OB}{h}=\frac{AB}{2R}=\frac{R}{\sqrt{4R^2+h^2}}$$$$\Rightarrow OB=\frac{Rh}{\sqrt{4R^2+h^2}}$$$$\Rightarrow AB=\frac{2R^2}{\sqrt{4R^2+h^2}}$$$$BC=\sqrt{4R^2+h^2}-h+2\sqrt{Rr}-\frac{2R^2}{\sqrt{4R^2+h^2}}$$$$BC=\sqrt{{(R+r)}^2-{(\frac{Rh}{\sqrt{4R^2+h^2}}-r)}^2}$$$$\sqrt{4R^2+h^2}-h+2\sqrt{Rr}-\frac{2R^2}{\sqrt{4R^2+h^2}}=\sqrt{{(R+r)}^2-{(\frac{Rh}{\sqrt{4R^2+h^2}}-r)}^2}$$$$let\mu =\frac{r}{R},\lambda =\frac{h}{R}$$$$\sqrt{4+{\lambda }^2}-\lambda +2\sqrt{\mu }-\frac{2}{\sqrt{4+{\lambda }^2}}=\sqrt{{(1+\mu )}^2-{(\frac{\lambda }{\sqrt{4+{\lambda }^2}}-\mu )}^2}$$$$$$$$areaofbluetriangle$$$$\mathrm{\Delta }=\frac{\left(2R\right)h}{2}=\lambda R^2$$$$$$$$with\mu =\frac{r}{R}=\frac{1}{4},weget\lambda =1.5$$$$\Rightarrow \mathrm{\Delta }=1.5\times 4^2=24$$

Commented by Mingma last updated on 22/Apr/23

Excellent, sir!

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