Cho lăng trụ tam giác đều ABC.A'B'C' cạnh đáy bằng a, chiều cao bằng 3a. Mặt phẳng

Đáp án B

Gọi H là trung điểm của $A^{\text{'}}{C}^{\text{'}}$ , tam giác $\Delta A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$  đều nên $B^{\text{'}}H\perp A^{\text{'}}{C}^{\text{'}}.$

Ta có: $\left\{\begin{array}{l}{B}^{\text{'}}H\perp A^{\text{'}}{C}^{\text{'}}\\ HI\perp A^{\text{'}}{C}^{\text{'}}\end{array}\right.\Rightarrow A^{\text{'}}{C}^{\text{'}}\perp \left(B^{\text{'}}HI\right)\Rightarrow \left(P\right)\equiv \left(B^{\text{'}}HI\right).$

$\Delta A^{\text{'}}EH~\Delta A^{\text{'}}{C}^{\text{'}}C\Rightarrow \frac{A^{\text{'}}E}{A^{\text{'}}H}=\frac{A^{\text{'}}{C}^{\text{'}}}{A^{\text{'}}C}\Rightarrow A^{\text{'}}E=\frac{A^{\text{'}}{C}^{\text{'}}.A^{\text{'}}H}{A^{\text{'}}C}=\frac{a\sqrt{10}}{20}.$

$\Delta A^{\text{'}}IH~\Delta A^{\text{'}}{C}^{\text{'}}C\Rightarrow \frac{IH}{A^{\text{'}}H}=\frac{A^{\text{'}}C}{C^{\text{'}}C}\Rightarrow IH=\frac{A^{\text{'}}C.A^{\text{'}}H}{C^{\text{'}}C}=\frac{a\sqrt{10}}{6}.$

$S_{B^{\text{'}}HI}=\frac{1}{2}{B}^{\text{'}}H.HI=\frac{a^2\sqrt{30}}{24}\Rightarrow V_1=\frac{1}{3}.S_{B^{\text{'}}HI}.A^{\text{'}}E=\frac{1}{3}.\frac{a^2\sqrt{30}}{24}.\frac{a\sqrt{10}}{20}=\frac{a^3\sqrt{3}}{144}.$

$V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=S_{ABC}.A{A}^{\text{'}}=\frac{a^2\sqrt{3}}{4}.3a=\frac{3a^3\sqrt{3}}{4},V_2=\frac{107}{144}.a^3\sqrt{3}$ do đó $\frac{V_1}{V_2}=\frac{1}{107}.$