Cho hình lăng trụ ABC. A'B'C' và M, N là hai điểm lần lượt bên cạnh CA, CB sao cho MN song song với AB và CM/CA=k . Mặt phẳng (MNB'A') chia khối lăng trụ ABC. A'B'C' thành hai phần...

Đáp án A

Ta có:  $\{\begin{array}{l}\left(MN{B}^{\text{'}}{A}^{\text{'}}\right)\cap \left(AC{C}^{\text{'}}{A}^{\text{'}}\right)=A^{\text{'}}M\\ \left(MN{B}^{\text{'}}{A}^{\text{'}}\right)\cap \left(BC{C}^{\text{'}}{B}^{\text{'}}\right)=B^{\text{'}}N\\ \left(AC{C}^{\text{'}}{A}^{\text{'}}\right)\cap \left(BC{C}^{\text{'}}{B}^{\text{'}}\right)=C{C}^{\text{'}}\end{array}\Rightarrow A^{\text{'}}M,B^{\text{'}}N,C{C}^{\text{'}}$đồng quy tại S

Áp dụng định lí Ta-lét ta có: $\frac{SM}{S{A}^{\text{'}}}=\frac{MN}{A^{\text{'}}{B}^{\text{'}}}=\frac{MN}{AB}=\frac{CM}{CA}=k=\frac{SN}{S{B}^{\text{'}}}=\frac{SC}{S{C}^{\text{'}}}$

$\Rightarrow \frac{V_{S.MNC}}{V_{S.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}}=\frac{SM}{S{A}^{\text{'}}}.\frac{SN}{S{B}^{\text{'}}}.\frac{SC}{S{C}^{\text{'}}}=k^3\Rightarrow \frac{V_1}{V_{S.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}}=1-k^3\Rightarrow V_1=(1-k^3)V_{S.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}$

Ta có: $\frac{SC}{S{C}^{\text{'}}}=k\Rightarrow \frac{S{C}^{\text{'}}-C{C}^{\text{'}}}{S{C}^{\text{'}}}=k\iff \frac{C{C}^{\text{'}}}{S{C}^{\text{'}}}=1-k$

$\frac{V_{S.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}}{V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}}=\frac{\frac{1}{3}S{C}^{\text{'}}}{C{C}^{\text{'}}}=\frac{1}{3(1-k)}\Rightarrow V_{S.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=\frac{V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}}{3(1-k)}$

$\Rightarrow V_1=(1-k^3)V_{S.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=(1-k^3)\frac{V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}}{3(1-k)}=\frac{1+k+k^2}{3}{V}_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}$