Question Number 22525 by ajfour last updated on 19/Oct/17
Commented by ajfour last updated on 19/Oct/17
$${A}\:{hemisphere}\:{is}\:{inscribed}\:{in}\:{a} \\ $$$${regular}\:{tetrahedron}\:,{of}\:{edge}\:\boldsymbol{{a}}\:, \\ $$$${such}\:{that}\:{three}\:{faces}\:{of}\:{tetrahedron} \\ $$$${are}\:{tangent}\:{to}\:{its}\:{spherical} \\ $$$${surface},\:{and}\:{the}\:{fourth}\:{serves} \\ $$$${as}\:{a}\:{plane}\:{of}\:{the}\:{diameter}. \\ $$$$\boldsymbol{{D}}{etermine}\:{the}\:{total}\:{surface} \\ $$$${area}\:{of}\:{the}\:{hemisphere}. \\ $$
Commented by mrW1 last updated on 19/Oct/17
Commented by mrW1 last updated on 19/Oct/17
$$\mathrm{AB}=\mathrm{AC}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{a} \\ $$$$\mathrm{AO}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{AB}=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\mathrm{a} \\ $$$$\mathrm{OC}=\sqrt{\mathrm{AC}^{\mathrm{2}} −\mathrm{AO}^{\mathrm{2}} }=\sqrt{\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{12}}}\:\mathrm{a}=\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\:\mathrm{a} \\ $$$$\frac{\mathrm{R}}{\mathrm{AO}}=\frac{\mathrm{OC}}{\mathrm{AC}} \\ $$$$\Rightarrow\mathrm{R}=\mathrm{AO}×\frac{\mathrm{OC}}{\mathrm{AC}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\:\mathrm{a}×\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}×\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{9}}\:\mathrm{a} \\ $$$$\mathrm{Area}=\mathrm{2}\pi\mathrm{R}^{\mathrm{2}} =\frac{\mathrm{12}\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{81}} \\ $$
Commented by ajfour last updated on 20/Oct/17
$$\:{Thanks}\:{for}\:{the}\:{better}\:{view}\:{sir}. \\ $$