Lời giải:
Kẻ \(AK\perp SB(1)\)
Ta có: \(\left\{\begin{matrix} SA\perp (ABCD)\rightarrow SA\perp BC\\ AB\perp BC(\text{do ABCD là hình vuông)}\end{matrix}\right.\Rightarrow (SAB)\perp BC\)
Mà \(AK\subset (SAB)\Rightarrow AK\perp BC(2)\)
Từ \((1);(2)\Rightarrow AK\perp (SBC)\)
Do đó \(d(A,(SBC))=AK\)
Theo hệ thức lượng trong tam giác vuông:
\(\frac{1}{AK^2}=\frac{1}{SA^2}+\frac{1}{AB^2}=\frac{1}{3a^2}+\frac{1}{a^2}\)\(\rightarrow AK=\frac{\sqrt{3}}{2}a\Rightarrow d(A,(SBC))=\frac{\sqrt{3}a}{2}\)
\(V_{chopS.ABCD}=\dfrac{1}{3}.S_{ABCD}.h_s=\dfrac{a^3\sqrt{3}}{3}\)
\(V_{chopS.ABC}=\dfrac{1}{2}V_{chopS.ABCD}=\dfrac{a^3\sqrt{3}}{6}\)
\(V_{chopS.ABC}=V_{chopA.SBC}=\dfrac{1}{3}.S_{SBC}.h_a\) {ha đường cao hạ từ A đến đáy SBC}
\(S_{SBC}=\dfrac{1}{2}BC.SB=\dfrac{1}{2}a.\sqrt{3a^2+a^2}=a^2\)
\(h_a=\dfrac{3.V_{chopS.ABC}}{S_{SBC}}=\dfrac{3.\dfrac{a^3\sqrt{3}}{6}}{a^2}=\dfrac{a\sqrt{3}}{2}\)
