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Question Number 203370 by mr W last updated on 17/Jan/24

Commented by mr W last updated on 17/Jan/24

$$Q203319$$

Answered by mr W last updated on 17/Jan/24

$$\cos \theta =\frac{1}{3}$$$$smalltetrahedronPQRS:$$$$edgelengtha=2r$$$$heighth=\frac{\sqrt{6}a}{3}=\frac{2\sqrt{6}r}{3}$$$$$$$$bigtetrahedronABCD:$$$$heightH=r+h+h_1$$$$with{h}_1=\frac{r}{\cos \theta }=3r$$$$edgelengthb$$$$H=\frac{\sqrt{6}b}{3}=r+\frac{2\sqrt{6}r}{3}+3r$$$$\Rightarrow b=2(\sqrt{6}+1)r$$

Commented by mr W last updated on 18/Jan/24

$$ormethod2:$$$$\frac{H}{4}-\frac{h}{4}=r$$$$\Rightarrow H=h+4r$$$$\frac{b}{a}=\frac{H}{h}=1+\frac{4r}{h}=1+\sqrt{6}$$$$\Rightarrow b=2(1+\sqrt{6})r$$

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