$V_{SABC}$= 1/3. SA.$S_{ABC}$= 1/6.2a. $\frac{a\sqrt{3}}{2}$.a= $\frac{a^{3}\sqrt{3}}{6}$Có AM là đường cao trong $\Delta $SAB vuông tại A => $\frac{1}{AM^{2}}$= $\frac{1}{AS^{2}}$+$\frac{1}{AB^{2}}$=> $AM^{2}$= $\frac{4a^{2}}{5}$=> $SM^{2}$= $\frac{16a^{2}}{5}$ => SM= $\frac{4a}{\sqrt{5}}$Tương t ự: SN= $\frac{4a}{\sqrt{5}}$Mà: SB=SC= a$\sqrt{5}$=> $\frac{SM}{SB} $=$\frac{SN}{SC}$=$\frac{4}{5}$=> $\frac{V_{SAMN}}{V_{SABC}}$= $\frac{SM}{SB} $.$\frac{SN}{SC}$=$\frac{16}{25}$$V_{SAMN}$ = $\frac{8\sqrt{3}.a^{3}}{75}$=> $V_{ABCNM}$= $V_{SABC}$- $V_{SAMN}$ = $\frac{3\sqrt{3}.a^{3}}{50}$
$V_{SABC}$= 1/3. SA.$S_{ABC}$= 1/6.2a. $\frac{a\sqrt{3}}{2}$.a= $\frac{a^{3}\sqrt{3}}{6}$Có AM là đường cao trong $\Delta $SAB vuông tại A => $\frac{1}{AM^{2}}$= $\frac{1}{AS^{2}}$+$\frac{1}{AB^{2}}$=> $AM^{2}$= $\frac{4a^{2}}{5}$=> $SM^{2}$= $\frac{16a^{2}}{5}$ => SM= $\frac{4a}{\sqrt{5}}$Tương t ự: SN= $\frac{4a}{\sqrt{5}}$Mà: SB=SC= a$\sqrt{5}$=> $\frac{SM}{SB} $=$\frac{SN}{SC}$=$\frac{4}{5}$=> $\frac{V_{SAMN}}{V_{SABC}}$= $\frac{SM}{SB} $.$\frac{SN}{SC}$=$\frac{16}{25}$$V_{SAMN}$ = $\frac{8\sqrt{3}.a^{3}}{75}$=> $V_{ABCNM}$= $V_{SABC}$- $V_{SAMN}$ = $\frac{3\sqrt{3}.a^{3}}{50}$
$V_{SABC}$= 1/3. SA.$S_{ABC}$= 1/6.2a. $\frac{a\sqrt{3}}{2}$.a= $\frac{a^{3}\sqrt{3}}{6}$Có AM là đường cao trong $\Delta $SAB vuông tại A => $\frac{1}{AM^{2}}$= $\frac{1}{AS^{2}}$+$\frac{1}{AB^{2}}$=> $AM^{2}$= $\frac{4a^{2}}{5}$=> $SM^{2}$= $\frac{16a^{2}}{5}$ => SM= $\frac{4a}{\sqrt{5}}$Tương t ự: SN= $\frac{4a}{\sqrt{5}}$Mà: SB=SC= a$\sqrt{5}$=> $\frac{SM}{SB} $=$\frac{SN}{SC}$=$\frac{4}{5}$=> $\frac{V_{SAMN}}{V_{SABC}}$= $\frac{SM}{SB} $.$\frac{SN}{SC}$=$\frac{16}{25}$$V_{SAMN}$ = $\frac{8\sqrt{3}.a^{3}}{75}$=> $V_{ABCNM}$= $V_{SABC}$- $V_{SAMN}$ = $\frac{3\sqrt{3}.a^{3}}{50}$