Đáp án:
$V_{SABC}=\dfrac{a^3}6$
Giải thích các bước giải:
Ta có: $\Delta SAB=\Delta SAC$ (cạnh huyền - cạnh góc vuông)
$\Delta SAB$ dựng $BH\bot SA\Rightarrow CH\bot SA$ và $BH=CH$
$\Rightarrow\widehat{((SAB),(SAC))}=(BH,CH)$
Th1: $\widehat{BHC}=90^o\Rightarrow BHC\bot$ cân đỉnh H suy ra $HB^2+HC^2=BC^2=AB^2+AC^2=2a^2\Rightarrow HB=a\sqrt2>a=BA,\Delta SAB$ có đường cao > cạnh góc vuông vô lý (loại)
Vậy $\widehat{BHC}=120^o\Rightarrow\Delta BHC$ cân đỉnh H
$BC^2=BH^2+CH^2-2BH.CH.\cos\widehat{BHC}$
$\Rightarrow 2a^2=2BH^2(1-\cos120^o)\Rightarrow BH^2=\dfrac{2a^2}3$
$\Rightarrow BH=\dfrac{a\sqrt2}{\sqrt3}$
$\Delta BHA\bot H: AH^2=AB^2-BH^2=a^2-\dfrac{2a^2}3=\dfrac{a^2}3\Rightarrow AH=\dfrac a{\sqrt3}$
$\Delta SAB\bot B, BH\bot SA:AB^2=AH.SA\Leftrightarrow a^2=\dfrac a{\sqrt3}.SA$
$\Rightarrow SA=a\sqrt3$
Ta có: $SH\bot BH, CH\Rightarrow SH\bot(BHC)$
$AH\bot BH, AH\bot CH\Rightarrow AH\bot (BHC)$
$\Rightarrow V_{SABC}=V_{SBHC}+V_{ABHC}=\dfrac13.SH.S_{BHC}+\dfrac13.AH.S_{BHC}$
$=\dfrac13.S_{BHC}.(SH+AH)=\dfrac13.S_{BHC}.SA$
$=\dfrac13.\dfrac12.\dfrac{a\sqrt2}{\sqrt3}.\dfrac{a\sqrt2}{\sqrt3}.\sin{120^o}.a\sqrt3$
$=\dfrac{a^3}6$
