Đáp án:
\({V_{S.ABC}} = \frac{{{a^3}\sqrt 3 }}{8};\,\,{S_{tp}} = \frac{{3{a^2}}}{2}\left( {1 + \frac{{\sqrt 3 }}{2}} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
Goi\,\,M\,\,la\,TD\,\,cua\,\,BC \Rightarrow AM \bot BC\\
Lai\,\,co\,\,SA \bot BC \Rightarrow BC \bot \left( {SAM} \right) \Rightarrow BC \bot SM\\
\Rightarrow \widehat {\left( {\left( {SBC} \right);\left( {ABC} \right)} \right)} = \widehat {\left( {SM;AM} \right)} = \widehat {SMA} = {60^0}\\
\Delta ABC\,\,deu\, \Rightarrow AM = \frac{{a\sqrt 3 }}{2}\\
SA \bot \left( {ABC} \right) \Rightarrow SA \bot AM \Rightarrow \Delta SAM\,\,vuong\,\,tai\,\,A.\\
\Rightarrow SA = AM.\tan {60^0} = \frac{{a\sqrt 3 }}{2}.\sqrt 3 = \frac{{3a}}{2}\\
{S_{\Delta ABC}} = \frac{{{a^2}\sqrt 3 }}{4} \Rightarrow {V_{S.ABC}} = \frac{1}{3}.\frac{{3a}}{2}.\frac{{{a^2}\sqrt 3 }}{4} = \frac{{{a^3}\sqrt 3 }}{8}\\
{S_{\Delta SAB}} = \frac{1}{2}SA.AB = \frac{1}{2}.\frac{{3a}}{2}.a = \frac{{3{a^2}}}{4}\\
{S_{\Delta SAC}} = \frac{1}{2}.SA.AC = \frac{1}{2}.\frac{{3a}}{2}.a = \frac{{3{a^2}}}{4}\\
{S_{\Delta SBC}} = \frac{1}{2}SM.BC = \frac{1}{2}\frac{{AM}}{{\cos {{60}^0}}}.BC = \frac{1}{2}.\frac{{\frac{{a\sqrt 3 }}{2}}}{{\frac{1}{2}}}.a = \frac{{{a^2}\sqrt 3 }}{2}\\
\Rightarrow {S_{tp}} = {S_{\Delta SAB}} + {S_{\Delta SAC}} + {S_{\Delta SBC}} + {S_{\Delta ABC}}\\
= \frac{{3{a^2}}}{4} + \frac{{3{a^2}}}{4} + \frac{{{a^2}\sqrt 3 }}{2} + \frac{{{a^2}\sqrt 3 }}{4}\\
= \frac{{3{a^2}}}{2} + \frac{{3{a^2}\sqrt 3 }}{4} = \frac{{3{a^2}}}{2}\left( {1 + \frac{{\sqrt 3 }}{2}} \right)
\end{array}\)
