Cách 1:

Ta có:

$\begin{align}  & \frac{\sqrt{2}}{\sqrt{5}}=\tan \alpha =\frac{SA}{AC}=\frac{SA}{\sqrt{5}}=>SA=\sqrt{2} \\  & \frac{{{V}_{C.SMN}}}{{{V}_{C.SAB}}}=\frac{CN}{CA}.\frac{CM}{CB}.\frac{SC}{SC}=\frac{1}{3}.\frac{1}{2}=\frac{1}{6} \\ \end{align}$

$=>{{V}_{SABMN}}=\frac{5}{6}{{V}_{S.ABC}}=\frac{5}{12}{{V}_{S.ABCD}}=\frac{5}{12}.\frac{1}{3}.a.2a.SA=\frac{5}{6}.\frac{1}{3}{{a}^{2}}.\sqrt{2}a=\frac{5\sqrt{2}}{18}{{a}^{3}}$

Cách 2:

Đặt hệ trục tọa độ Oxyz

A(0;0;0),B(0;a;0),C(2a;a;0);D(2a;0;0);S(0;0;$\sqrt{2}$a)

=>M(a;a;0)

N($\frac{4}{3}a;\frac{2}{3}a;0$)

=>V=$\frac{1}{6}|\text{ }\!\![\!\!\text{ }\overrightarrow{SA},\overrightarrow{SB}\text{ }\!\!]\!\!\text{ }.\overrightarrow{SC}|=\frac{5\sqrt{2}}{18}{{a}^{3}}$