Google
Câu hỏi: Cho khối lăng trụ $ABC.{A}'{B}'{C}'$ có thể tích $V$, trên các cạnh $A{A}',B{B}',C{C}'$ lần lượt lấy các điểm $M,N,P$ sao cho $AM=\dfrac{1}{2}A{A}',BN=\dfrac{2}{3}B{B}',CP=\dfrac{1}{6}C{C}'$. Tính thể tích khối đa diện $AB{C}'MNP$ ?
A. $\dfrac{4V}{9}$.
B. $\dfrac{V}{2}$.
C. $\dfrac{5V}{9}$.
D. $\dfrac{2V}{5}$.
A. $\dfrac{4V}{9}$.
B. $\dfrac{V}{2}$.
C. $\dfrac{5V}{9}$.
D. $\dfrac{2V}{5}$.
Ta có: ${{V}_{ABCMNP}}={{V}_{N.ACB}}+{{V}_{N.ACPM}}$.
${{V}_{N.ACB}}=\dfrac{BN}{B{B}'}.{{V}_{{B}'ACB}}=\dfrac{BN}{B{B}'}.\dfrac{1}{3}{{V}_{ABC{A}'{B}'{C}'}}$.
$\dfrac{{{V}_{NACPM}}}{{{V}_{{B}'AC{C}'{A}'}}}=\dfrac{{{S}_{ACPM}}}{{{S}_{AC{C}'{A}'}}}=\dfrac{\dfrac{1}{2}\left( CP+AM \right)}{A{A}'}=\dfrac{1}{2}\left( \dfrac{CP}{C{C}'}+\dfrac{AM}{A{A}'} \right)$
$\Rightarrow {{V}_{NACPM}}=\dfrac{1}{2}\left( \dfrac{CP}{C{C}'}+\dfrac{AM}{A{A}'} \right).\dfrac{2}{3}{{V}_{ABC{A}'{B}'{C}'}}$
Suy ra: ${{V}_{ABCMNP}}=\dfrac{1}{3}\left( \dfrac{AM}{A{A}'}+\dfrac{CP}{C{C}'}+\dfrac{BN}{B{B}'} \right).{{V}_{ABC{A}'{B}'{C}'}}$
Vậy ${{V}_{ABCMNP}}=\dfrac{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{1}{6}}{3}.V=\dfrac{4V}{9}$.
${{V}_{N.ACB}}=\dfrac{BN}{B{B}'}.{{V}_{{B}'ACB}}=\dfrac{BN}{B{B}'}.\dfrac{1}{3}{{V}_{ABC{A}'{B}'{C}'}}$.
$\dfrac{{{V}_{NACPM}}}{{{V}_{{B}'AC{C}'{A}'}}}=\dfrac{{{S}_{ACPM}}}{{{S}_{AC{C}'{A}'}}}=\dfrac{\dfrac{1}{2}\left( CP+AM \right)}{A{A}'}=\dfrac{1}{2}\left( \dfrac{CP}{C{C}'}+\dfrac{AM}{A{A}'} \right)$
$\Rightarrow {{V}_{NACPM}}=\dfrac{1}{2}\left( \dfrac{CP}{C{C}'}+\dfrac{AM}{A{A}'} \right).\dfrac{2}{3}{{V}_{ABC{A}'{B}'{C}'}}$
Suy ra: ${{V}_{ABCMNP}}=\dfrac{1}{3}\left( \dfrac{AM}{A{A}'}+\dfrac{CP}{C{C}'}+\dfrac{BN}{B{B}'} \right).{{V}_{ABC{A}'{B}'{C}'}}$
Vậy ${{V}_{ABCMNP}}=\dfrac{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{1}{6}}{3}.V=\dfrac{4V}{9}$.
Đáp án A.