Question

Answer 1

Kẻ \(SH\perp AC\) \(\Rightarrow SH\perp\left(ABCD\right)\)

\(\Rightarrow\widehat{SAH}\) là góc giữa SA và đáy

\(\Rightarrow\widehat{SAH}=60^0\)

\(\Rightarrow\widehat{SCH}=90^0-60^0=30^0\) (do SAC vuông tại S)

Ta có: \(\left\{{}\begin{matrix}cot\widehat{SCH}=\dfrac{CH}{SH}\\cot\widehat{SAH}=\dfrac{AH}{SH}\end{matrix}\right.\) 

\(\Rightarrow cot30^0+cot60^0=\dfrac{CH+AH}{SH}=\dfrac{AC}{SH}\)

\(\Rightarrow SH=\dfrac{AC}{cot30^0+cot60^0}=\dfrac{AB\sqrt{2}}{cot30^0+cot60^0}=\dfrac{a\sqrt{3}}{2}\)

\(\Rightarrow V=\dfrac{1}{3}SH.AB^2=\dfrac{a^3\sqrt{3}}{3}\)

Answer 2