$V_{SABC}$= 1/3.SA. $S_{ABC}$= $\frac{a^{3}.\sqrt{3}}{6}$ (1)Có AM là đường cao trong tam giác SAB vuông tại A => $\frac{1}{AM^{2}}=\frac{1}{AB^2}+\frac{1}{AS^2}$ => $AM^2= \frac{4a^2}{5}$ => $SM^2= SA^2- AM^2= \frac{16a^2}{5} => SM= \frac{4a}{\sqrt{5}} => SM/SB= 4/5$
Tương tự SN/ SC= 4/5
=> $\frac{V_{SAMN}}{V_{SABC}}= \frac{SA}{SA}.\frac{SM}{SB}.\frac{SN}{SC}=\frac{16}{25}$ (2)
(1)(2)=> $V_{SAMN}$= $\frac{8\sqrt{3}.a^{3}}{75}$
=> $V_{AMNBC}= V_{SABC}- V_{SAMN}= \frac{3\sqrt{3}.a^{3}}{50}$