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Question Number 22525 by ajfour last updated on 19/Oct/17

Commented by ajfour last updated on 19/Oct/17

$$Ahemisphereisinscribedina$$$$regulartetrahedron,ofedge\mathit{a},$$$$suchthatthreefacesoftetrahedron$$$$aretangenttoitsspherical$$$$surface,andthefourthserves$$$$asaplaneofthediameter.$$$$\mathit{D}eterminethetotalsurface$$$$areaofthehemisphere.$$

Commented by mrW1 last updated on 19/Oct/17

Commented by mrW1 last updated on 19/Oct/17

$$\mathrm{AB}=\mathrm{AC}=\frac{\sqrt{3}}{2}\mathrm{a}$$$$\mathrm{AO}=\frac{1}{3}\mathrm{AB}=\frac{\sqrt{3}}{6}\mathrm{a}$$$$\mathrm{OC}=\sqrt{{\mathrm{AC}}^2-{\mathrm{AO}}^2}=\sqrt{\frac{3}{4}-\frac{1}{12}}\mathrm{a}=\frac{\sqrt{6}}{3}\mathrm{a}$$$$\frac{\mathrm{R}}{\mathrm{AO}}=\frac{\mathrm{OC}}{\mathrm{AC}}$$$$\Rightarrow \mathrm{R}=\mathrm{AO}\times \frac{\mathrm{OC}}{\mathrm{AC}}=\frac{\sqrt{3}}{6}\mathrm{a}\times \frac{\sqrt{6}}{3}\times \frac{2}{\sqrt{3}}=\frac{\sqrt{6}}{9}\mathrm{a}$$$$\mathrm{Area}=2\pi {\mathrm{R}}^2=\frac{12\pi {\mathrm{a}}^2}{81}$$

Commented by ajfour last updated on 20/Oct/17

$$Thanksforthebetterviewsir.$$

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