Question Number 62456 by ajfour last updated on 21/Jun/19
Commented by ajfour last updated on 21/Jun/19
$${If}\:{the}\:{regular}\:{tetrahedron}\:{with} \\ $$$${edge}\:{length}\:{unity}\:{inscribes}\:{a} \\ $$$${hemisphere}\:{such}\:{that}\:{three}\:{faces} \\ $$$${are}\:{tangent}\:{to}\:{the}\:{spherical} \\ $$$${surface}\:{while}\:{fourth}\:{serves}\:{to} \\ $$$${be}\:{the}\:{diameter}\:{plane},\:{then}\:{find} \\ $$$${radius}\:{of}\:{hemisphere}. \\ $$
Answered by mr W last updated on 21/Jun/19
Commented by mr W last updated on 21/Jun/19
$${edge}\:{length}\:{of}\:{tetrahedron}={a} \\ $$$${volume}\:{of}\:{tetrahedron}\:{V}=\frac{\sqrt{\mathrm{2}}{a}^{\mathrm{3}} }{\mathrm{12}} \\ $$$${area}\:{of}\:{each}\:{face}\:{A}_{{S}} =\frac{\mathrm{1}}{\mathrm{2}}×{a}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${volume}\:{of}\:{small}\:{pyramid}\:{O}−{ABC}=\:{V}_{\mathrm{1}} \\ $$$${V}_{\mathrm{1}} =\frac{{A}_{{S}} {R}}{\mathrm{3}}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} {R}}{\mathrm{12}} \\ $$$$\mathrm{3}{V}_{\mathrm{1}} ={V} \\ $$$$\mathrm{3}×\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} {R}}{\mathrm{12}}=\frac{\sqrt{\mathrm{2}}{a}^{\mathrm{3}} }{\mathrm{12}} \\ $$$$\Rightarrow{R}=\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{9}} \\ $$