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

Trong kẻ .
Do $\left\{ \begin{aligned}
& CH\bot AB \\
& CH\bot SA \\
\end{aligned} \right.$$\Rightarrow CH\bot \left( SAB \right) \Rightarrow CH\bot SB\left( 1 \right)BC=\sqrt{A{{B}^{2}}-A{{C}^{2}}}=a\sqrt{3}CH=\dfrac{CA.CB}{AB}=\dfrac{a\sqrt{3}}{2}BH=\sqrt{B{{C}^{2}}-C{{H}^{2}}}=\dfrac{3}{2}a\Delta SABHK\bot SB\Rightarrow CK\bot SB\left( 2 \right)\left( 1 \right),\left( 2 \right)\left( SAB \right)\left( SBC \right)\widehat{CKH}=60{}^\circ \Delta CKHHK=CH.\cot 60{}^\circ =\dfrac{a}{2}BK=\sqrt{B{{H}^{2}}-H{{K}^{2}}}=a\sqrt{2}\Delta SAB\Delta HKB\dfrac{SA}{HK}=\dfrac{AB}{BK}=\dfrac{2a}{a\sqrt{2}}\Rightarrow SA=\dfrac{a}{\sqrt{2}}S.ABCV=\dfrac{1}{3}SA.{{S}_{\Delta ABC}}=\dfrac{1}{3}\dfrac{a}{\sqrt{2}}.\dfrac{1}{2}.a.\sqrt{3}.a=\dfrac{{{a}^{3}}\sqrt{6}}{12}$.