xiu231104

Xét $(ABCD)$

Từ $A$ vẽ $AE \perp CD,E \in CD$

Kẻ $CF \perp AD; F \in AD$

$\Rightarrow F$ trung điểm $AD$ ($AF=FD=a$)

$\Rightarrow \Delta ACD$ cân tại $C$

Ta có:

$\left\{\begin{matrix} (SCD)\cap (ABCD)=CD\\ SA \perp CD\\ SE \perp CD \end{matrix}\right. \Rightarrow \widehat{[(SCD);(ABCD)]}=\widehat{SEA}=60^{o}$

$\sin \widehat{ADC}=\frac{CF}{CD}=\frac{\sqrt{2}}{2}$

Mặt khác: $\sin \widehat{ADC}=\frac{AE}{AD}\Leftrightarrow AE=AD.\sin \widehat{ADC}=a\sqrt{2}$

$\Rightarrow SA=AE.\tan 60=a\sqrt{6}$

$S_{ABCD}=\frac{1}{2}.(AD+BC).AD=\frac{3a^{2}\sqrt{2}}{2}$

$\Rightarrow V_{S.ABCD}=\frac{1}{3}.SA.S_{ABCD}=a^{3}\sqrt{3}$

diện tích abcd bằng 3a*a/2 chứ