Cho hình chóp $S.ABC$ có đáy là tam giác $ABC$ vuông tại $A$...

Trong kẻ . Ta có: $\left\{ \begin{aligned}
& AH\bot BC \\
& SB\bot AH \\
\end{aligned} \right.$$\Rightarrow AH\bot \left( SBC \right) \Rightarrow AH\bot SC\left( 1 \right)\Delta SACAK\bot SC\left( 2 \right)\left( 1 \right),\left( 2 \right)\Rightarrow SC\bot \left( AKH \right)\Rightarrow SC\bot HK\left( SAC \right)\left( SBC \right)\widehat{AKH}\widehat{AKH}=60{}^\circ AC=\sqrt{B{{C}^{2}}-A{{B}^{2}}}=a\sqrt{3}A{{C}^{2}}=CH.BC\Rightarrow CH=\dfrac{A{{C}^{2}}}{BC}=\dfrac{3{{a}^{2}}}{2a}=\dfrac{3a}{2}\Rightarrow AH=\sqrt{A{{C}^{2}}-C{{H}^{2}}}=\dfrac{a\sqrt{3}}{2}\Delta AKHHHK=AH.\cot 60{}^\circ =\dfrac{a}{2}CK=\sqrt{C{{H}^{2}}-H{{K}^{2}}}=a\sqrt{2}\Delta SBC\backsim \Delta HKC\left( g.g \right)\dfrac{SB}{HK}=\dfrac{BC}{KC}=\dfrac{2a}{a\sqrt{2}}=\sqrt{2}\Rightarrow SB=HK.\sqrt{2}=\dfrac{a}{\sqrt{2}}S.ABCV=\dfrac{1}{3}SB.{{S}_{\Delta ABC}}=\dfrac{1}{3}\dfrac{a}{\sqrt{2}}.\dfrac{1}{2}.a.\sqrt{3}.a=\dfrac{{{a}^{3}}\sqrt{6}}{12}$.