Giải thích các bước giải:
a.Gọi $AC\cap BD=O\to SO\perp ABCD, E$ là trung điểm BC$\to OE=\dfrac12DC=a$
$\to \widehat{SEO}=30^o$
$\to \tan30^o=\dfrac{SO}{OE}\to SO=\dfrac{a\sqrt{3}}{3}$
$\to V=\dfrac13SO.S_{ABCD}=\dfrac{a\sqrt{3}}{9}.4a^2=\dfrac{4a^3\sqrt{3}}{9}$
b.Gọi $BG\cap SD=H, AG\cap SC=F$
Vì $G$ là trọng tâm $\Delta SAC\to \dfrac{SG}{SO}=\dfrac23$
$\to G$ là trọng tâm $\Delta SBD\to H,F$ là trung điểm SD,SC
Ta có :
$\dfrac{V_{SAFB}}{S_{SABC}}=\dfrac{SA}{SA}.\dfrac{SF}{SC}.\dfrac{SB}{SB}=\dfrac12$
$\to V_{SAFB}=\dfrac12V_{ABC}=\dfrac14S_{ABCD}$
$\dfrac{V_{SAHF}}{V_{SADC}}=\dfrac{SA}{SA}.\dfrac{SF}{SC}.\dfrac{SH}{SD}=\dfrac14$
$\to V_{SAHF}=\dfrac14V_{SADC}=\dfrac18V_{SABCD}$
$\to V_{SAHB}=\dfrac38V_{SABCD}$
$\to V_1=\dfrac38V_{SABCD}\to V_2=\dfrac58V_{SABCD}$
$\to \dfrac{V_1}{V_2}=\dfrac35$
c.Gọi $AI\cap SC=L, BI\cap SD=K$
Vì $A,I,L$ thẳng hàng
$\to\dfrac{AC}{AO}.\dfrac{IO}{IS}.\dfrac{LS}{LC}=1$
$\to2.\dfrac{1}{3}.\dfrac{LS}{LC}=1$
$\to\dfrac{SL}{CL}=\dfrac{3}{2}$
$\to\dfrac{SL}{CL+SL}=\dfrac{3}{2+3}$
$\to\dfrac{SL}{SC}=\dfrac{3}{5}$
Tương tự $\to \dfrac{SK}{SD}=\dfrac35$
Tương tự câu b
$\to \dfrac{V_{SABLK}}{V_{ABCD}}=\dfrac12(\dfrac{SL}{SC}+\dfrac{SK}{SD}.\dfrac{SL}{SC})$
$\to \dfrac{V_{SABLK}}{V_{ABCD}}=\dfrac{12}{25}$
$\to V_1=\dfrac{12}{25}V, V_2=\dfrac{13}{25}\to \dfrac{V_1}{V_2}=\dfrac{12}{13}$
