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

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

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

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

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

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

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

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

Gọi O là tâm đáy, M là trung điểm AB

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

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

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

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

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

Tính V/πa³

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

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

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

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

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

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

Đáp án đúng : A

Kẻ \(AH\perp BC\)

Áp dụng hệ thức lượng: \(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{4}{3a^2}\Rightarrow AH=\dfrac{a\sqrt{3}}{2}\)

\(tan\widehat{SHA}=\dfrac{2}{\sqrt{3}}=\dfrac{SA}{AH}\Rightarrow SA=\dfrac{AH.2}{\sqrt{3}}=a\)

Gọi M là trung điểm BC và N là trung điểm SA, dựng hình chữ nhật AMIN \(\Rightarrow\) I là tâm mặt cầu ngoại tiếp

\(AN=\dfrac{1}{2}SA=\dfrac{a}{2}\) ; \(AM=\dfrac{1}{2}BC=\dfrac{1}{2}\sqrt{AB^2+AC^2}=a\)

\(\Rightarrow R=IA=\sqrt{AM^2+AN^2}=\dfrac{a\sqrt{5}}{2}\)

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

Gọi O là tâm đáy \(\Rightarrow OA=\dfrac{2}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{3}\)

\(\Rightarrow SO=\sqrt{SA^2-OA^2}=\dfrac{a\sqrt[]{33}}{3}\)

\(\Rightarrow R=\dfrac{SA^2}{2SO}=\dfrac{2a\sqrt{33}}{11}\)

\(V=\dfrac{4}{3}\pi R^3=\dfrac{32a^3\sqrt{3}\pi}{11\sqrt{11}}\)