Đáp án: $V_{SAGD}=\dfrac{4a^3\sqrt2}{9\sqrt3}$
Giải thích các bước giải:
Gọi $H$ là trung điểm cạnh $BC$,
Dựng $G$ sao cho $SG=\dfrac{2}{3}SH$
$\Rightarrow G$ là trọng tâm $\Delta SBC$
Gọi $J$ là trung điểm cạnh $AD\Rightarrow HJ\bot AD$
$\Rightarrow\widehat{(SBC),(ABCD)}=\widehat{(SH,HJ)}=\widehat{SHJ}=60^o$
$SJ=\sqrt{SA^2+AJ^2}=\sqrt{a^2+a^2}=a\sqrt2$
$\tan\widehat{SHJ}=\dfrac{SJ}{HJ}\Rightarrow HJ=\dfrac{a\sqrt2}{\tan60^o}=\dfrac{a\sqrt2}{\sqrt3}$
$HJ=AB=\dfrac{a\sqrt2}{\sqrt3}$
$\Rightarrow V_{SABCD}=\dfrac{1}{3}.SA.S_{ABCD}=\dfrac{1}{3}a.2a.\dfrac{a\sqrt2}{\sqrt3}=\dfrac{2a^3\sqrt2}{3\sqrt3}$
Ta có: $\dfrac{d_{(G,(SAD))}}{d_{H,(SAD)}}=\dfrac{GI}{HJ}=\dfrac{SG}{SH}=\dfrac{2}{3}$ ($GI\parallel HJ$)
$\Rightarrow \dfrac{V_{SAGD}}{V_{SABCD}}=\dfrac{\dfrac{1}{3}.d_{(G,(SAD))}.S_{SAD}}{\dfrac{1}{3}.d_{(H,(SAD))}.S_{SAD}}=\dfrac{2}{3}$
$\Rightarrow V_{SAGD}=\dfrac{2}{3}V_{SABCD}=\dfrac{2}{3}\dfrac{2a^3\sqrt2}{3\sqrt3}=\dfrac{4a^3\sqrt2}{9\sqrt3}$
