Đáp án:

$V_{SABCD}=\dfrac{a^3\sqrt{2}}{8}$

Giải thích các bước giải:

Xét hình thoi $ABCD$ có $AB=BC=CD=DA=a$

Vì $\widehat{ABC}=120^o\to \widehat{ABD}=\widehat{DBC}=60^o$

Mà $BA=AD\to \Delta ABD$ đều, $\Delta CBD$ đều

$\to BD=BA=a$

Gọi $AC\perp BD=O$

$\to OA=OC=\dfrac{a\sqrt{3}}{2}\to AC=a\sqrt3$

Vẽ $DH\perp SC$

Vì $SA\perp ABCD\to SD=SB(=\sqrt{SA^2+a^2})$
Mà $CD=CB\to \Delta SDC=\Delta SBC(c.c.c)$

Do $DH\perp SC=H\to BH\perp SC=H, DH=BH, SC\perp (HBD)\to SC\bot OH$

$\to \widehat{DHB}=\widehat{(SBC,SCD)}=60^o$

Hoặc $\widehat{DHB}=180^o-\widehat{(SBC,SCD)}=120^o$

Nếu $\widehat{DHB}=60^o$

$\to \Delta HBD$ đều cạnh $DB=a$

$\to \Delta HBD=\Delta CBD\to OH=OC$

Lại có $SC\perp HO\to\Delta OHC\bot H\to OH=OC$(vô lý)

$\to \widehat{DHB}=60^o$ (loại)

$\to\widehat{DHB}=120^o$

$\to DH^2+BH^2-BD^2=2\cos\widehat{DHB}\cdot HD\cdot HB$

$\to 2DH^2-a^2=2\cos(120^o)\cdot HD^2$

$\to DH=\dfrac{a\sqrt{3}}{3}$ vì $DH>0$

$\to HO=\sqrt{DH^2-DO^2}=\dfrac{\sqrt{3}a}{6}$

$\to HC=\sqrt{OC^2-OH^2}=\dfrac{\sqrt{6}a}{3}$

Mà $\widehat{OHC}=\widehat{SAC}=90^o$

$\to \Delta CHO\sim\Delta CAS(g.g)$

$\to \dfrac{CH}{CA}=\dfrac{HO}{AS}$

$\to AS=\dfrac{CA}{CH}\cdot OH$

$\to AS=\dfrac{a\sqrt{3}}{\dfrac{\sqrt{6}a}{3}}\cdot \dfrac{\sqrt{3}a}{6}$

$\to AS=\dfrac{\sqrt{6}a}{4}$

$\to V_{SABCD}=\dfrac13\cdot AS\cdot S_{ABCD}$

$\to V_{SABCD}=\dfrac13\cdot AS\cdot\dfrac12 BD\cdot AC$

$\to V_{SABCD}=\dfrac{a^3\sqrt{2}}{8}$