Tính V/πa³

Gọi O là tâm đáy \(\Rightarrow AO=\dfrac{a\sqrt{3}}{3}\)

\(SA=\dfrac{AO}{cos60^0}=\dfrac{2a\sqrt{3}}{3}\)

\(SO=\sqrt{SA^2-AO^2}=a\)

\(\Rightarrow R=\dfrac{SA^2}{2SO}=\dfrac{2a}{3}\)

\(V=\dfrac{4}{3}\pi R^3=\dfrac{32\pi a^3}{81}\)

\(\Rightarrow\dfrac{V}{\pi a^3}=\dfrac{32}{81}\)

Gọi O là tâm đáy, M là trung điểm AB

\(OA=\dfrac{a\sqrt{3}}{3}\)  ; \(OM=\dfrac{1}{2}OA=\dfrac{a\sqrt{3}}{6}\)

\(\widehat{SMO}=45^0\Rightarrow SO=OM=\dfrac{a\sqrt{3}}{6}\)

\(SA=\sqrt{SO^2+OA^2}=\dfrac{a\sqrt{15}}{6}\)

\(\Rightarrow R=\dfrac{SA^2}{2SO}=\dfrac{5a\sqrt{3}}{12}\)

\(V=\dfrac{4}{3}\pi R^3=\dfrac{125\pi a^3\sqrt{3}}{432}\)

\(AC=2a\sqrt{2}.\sqrt{2}=4a\Rightarrow OA=\dfrac{1}{2}AC=2a\)

\(\Rightarrow SO=\sqrt{SA^2-OA^2}=2a\sqrt{3}\)

\(\Rightarrow R=\dfrac{SA^2}{2SO}=\dfrac{4a\sqrt{3}}{3}\)

\(\Rightarrow V=\dfrac{4}{3}\pi R^3=...\)

\(AC=2a\sqrt{2}.\sqrt{2}=4a\) \(\Rightarrow OA=\dfrac{1}{2}AC=2a\)

\(\widehat{SAO}=30^0\Rightarrow\left\{{}\begin{matrix}SO=AO.tan30^0=\dfrac{2a\sqrt{3}}{3}\\SA=\dfrac{AO}{cos30^0}=\dfrac{4a\sqrt{3}}{3}\end{matrix}\right.\)

\(\Rightarrow R=\dfrac{SA^2}{2SO}=\dfrac{4a\sqrt{3}}{3}\)

\(V=\dfrac{4}{3}\pi R^3=\dfrac{256\pi a^3\sqrt{3}}{27}\)

Gọi M là trung điểm AB \(\Rightarrow\widehat{SMO}=45^0\)

\(OM=\dfrac{1}{2}AB=a\sqrt{2}\)

\(SO=OM.tan45^0=a\sqrt{2}\)

\(OA=\dfrac{1}{2}AC=2a\)

\(\Rightarrow SA=\sqrt{SO^2+OA^2}=a\sqrt{6}\)

\(\Rightarrow R=\dfrac{SA^2}{2SO}=\dfrac{3a\sqrt{2}}{2}\)

\(V=\dfrac{4}{3}\pi R^3=9\sqrt{2}\pi a^3\)

\(BC=\sqrt{AB^2+AC^2}=2a\)

Gọi M là trung điểm BC \(\Rightarrow AM=\dfrac{1}{2}BC=a\)

GỌi N là trung điểm SA \(\Rightarrow AN=\dfrac{1}{2}SA=a\)

Dựng hình chữ nhật AMIN \(\Rightarrow\) I là tâm mặt cầu ngoại tiếp

\(R=IA=\sqrt{AM^2+AN^2}=a\sqrt{2}\)

\(\Rightarrow V=\dfrac{4}{3}\pi R^3=...\)