Cho hình chóp $S.ABCD$ có $ABCD$ là hình chữ nhật cạnh $AB=1,AD=2.\text{ }SA$ vuông góc với mặt phẳng $\left( ABCD \right)$ và $SA=2.$ Gọi...

Ta có: ${{V}_{S.ABD}}=\dfrac{1}{6}AS.AB.AD=\dfrac{1}{6}\times 2\times 2\times 1=\dfrac{2}{3}.$
+) $\dfrac{BP}{BD}=\dfrac{A{{B}^{2}}}{B{{D}^{2}}}=\dfrac{A{{B}^{2}}}{A{{B}^{2}}+A{{D}^{2}}}=\dfrac{1}{5}\Rightarrow BP=\dfrac{1}{5}BD,$ suy ra:
${{S}_{\Delta ABP}}=\dfrac{1}{5}{{S}_{\Delta ABD}}=\dfrac{1}{5}\times \dfrac{1}{2}.AB.AD=\dfrac{1}{5};{{S}_{\Delta APD}}=\dfrac{4}{5}{{S}_{\Delta ABD}}=\dfrac{4}{5}\times \dfrac{1}{2}.AB.AD=\dfrac{4}{5}.$
Tam giác $SAD$ vuông cân tại $A$ nên $\dfrac{SN}{SD}=\dfrac{1}{2}\Rightarrow d\left( N;\left( ABCD \right) \right)=\dfrac{1}{2}SA=1.$
+) $\dfrac{BM}{BS}=\dfrac{B{{A}^{2}}}{B{{S}^{2}}}=\dfrac{B{{A}^{2}}}{S{{A}^{2}}+A{{B}^{2}}}=\dfrac{1}{5}\Rightarrow d\left( M;\left( ABCD \right) \right)=\dfrac{1}{5}SA=\dfrac{2}{5}.$
Suy ra: ${{V}_{M.ABP}}=\dfrac{1}{3}d\left( M;\left( ABCD \right) \right).{{S}_{\Delta ABP}}=\dfrac{1}{3}.\dfrac{2}{5}.\dfrac{1}{5}=\dfrac{2}{75}.$
${{V}_{N.APD}}=\dfrac{1}{3}d\left( N;\left( ABCD \right) \right).{{S}_{\Delta ADP}}=\dfrac{1}{3}.1.\dfrac{4}{5}=\dfrac{4}{15}.$
${{V}_{S.AMN}}=\dfrac{SM}{SB}.\dfrac{SN}{SC}.{{V}_{S.ABD}}=\dfrac{4}{5}.\dfrac{1}{2}.\dfrac{2}{3}=\dfrac{4}{15}.$
Vậy ${{V}_{A.MNP}}={{V}_{S.ABD}}-{{V}_{M.ABP}}-{{V}_{N.APD}}-{{V}_{S.AMN}}=\dfrac{2}{3}-\dfrac{2}{75}-\dfrac{4}{15}-\dfrac{4}{15}=\dfrac{8}{75}.$