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Question-203370




Question Number 203370 by mr W last updated on 17/Jan/24
Commented by mr W last updated on 17/Jan/24
Q203319
$${Q}\mathrm{203319} \\ $$
Answered by mr W last updated on 17/Jan/24
cos θ=(1/3)  small tetrahedron PQRS:  edge length a=2r  height h=(((√6)a)/3)=((2(√6)r)/3)    big tetrahedron ABCD:  height H=r+h+h_1   with h_1 =(r/(cos θ))=3r  edge length b  H=(((√6)b)/3)=r+((2(√6)r)/3)+3r  ⇒b=2((√6)+1)r
$$\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${small}\:{tetrahedron}\:{PQRS}: \\ $$$${edge}\:{length}\:{a}=\mathrm{2}{r} \\ $$$${height}\:{h}=\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{6}}{r}}{\mathrm{3}} \\ $$$$ \\ $$$${big}\:{tetrahedron}\:{ABCD}: \\ $$$${height}\:{H}={r}+{h}+{h}_{\mathrm{1}} \\ $$$${with}\:{h}_{\mathrm{1}} =\frac{{r}}{\mathrm{cos}\:\theta}=\mathrm{3}{r} \\ $$$${edge}\:{length}\:{b} \\ $$$${H}=\frac{\sqrt{\mathrm{6}}{b}}{\mathrm{3}}={r}+\frac{\mathrm{2}\sqrt{\mathrm{6}}{r}}{\mathrm{3}}+\mathrm{3}{r} \\ $$$$\Rightarrow{b}=\mathrm{2}\left(\sqrt{\mathrm{6}}+\mathrm{1}\right){r} \\ $$
Commented by mr W last updated on 18/Jan/24
or method 2:  (H/4)−(h/4)=r  ⇒H=h+4r  (b/a)=(H/h)=1+((4r)/h)=1+(√6)  ⇒b=2(1+(√6))r
$${or}\:{method}\:\mathrm{2}: \\ $$$$\frac{{H}}{\mathrm{4}}−\frac{{h}}{\mathrm{4}}={r} \\ $$$$\Rightarrow{H}={h}+\mathrm{4}{r} \\ $$$$\frac{{b}}{{a}}=\frac{{H}}{{h}}=\mathrm{1}+\frac{\mathrm{4}{r}}{{h}}=\mathrm{1}+\sqrt{\mathrm{6}} \\ $$$$\Rightarrow{b}=\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{6}}\right){r} \\ $$

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