Ta có
$$V_{S.ABCD} = V_{S.AOB} + V_{S.BOC} + V_{S.COD} + V_{S.DOA}$$
$$= \dfrac{1}{3} d(S, (AOB)) . S_{AOB} + \dfrac{1}{3} d(S, (BOC)) . S_{BOC} + \dfrac{1}{3} d(S, (COD)). S_{COD} + \dfrac{1}{3} d(S, (DOA)) . S_{DOA}$$
$$= \dfrac{1}{3} d(S, (ABCD)) . S_{AOB} + \dfrac{1}{3} d(S, (ABCD)) . S_{BOC} + \dfrac{1}{3} d(S, (ABCD)). S_{COD} + \dfrac{1}{3} d(S, (ABCD)) . S_{DOA}$$
$$= \dfrac{1}{3} d(S, (ABCD)) .(S_{AOB} + S_{BOC} + S_{COD} + S_{DOA})$$
Do O là tâm của hình vuông ABCD nên ta có
$$S_{AOB} = S_{BOC} = S_{COD} = S_{DOA}$$
Vậy ta có
$$V_{S.ABCD} = \dfrac{1}{3} d(S, (ABCD)) . 4 S_{AOB}$$
$$= 4. \dfrac{1}{3} d(S, (ABCD)) . S_{AOB}$$
$$ = 4 V_{S.AOB}$$
Vậy $V_{S.ABCD} = 4.2a^3 = 8a^3$.
