Cho hình lăng trụ đứng ABC.A'B'C' có các cạnh  $AB=AA\text{'}=2a$, đáy ABC  là tam giác vuông cân tại A Trên cạnh AA' lấy điểm I sao cho  $AI=\frac{1}{4}AA\text{'}.$  Gọi M,N lần lượt là các điểm đối xứng với B và C qua I. Thể tích khối đa diện $AMNA\text{'}B\text{'}C\text{'}$ bằng

 

Ta có $MN//BC\Rightarrow MN//B\text{'}C\text{'}$ và $MN=BC=B\text{'}C\text{'}$, suy ra tứ giác MNB'C' là hình bình hành.

Gọi $J=MB\text{'}\cap NC\text{'}$ suy ra J là tâm hình bình hành và $J\in AA\text{'}.$

                    $V_{AMNA\text{'}B\text{'}C\text{'}}=V_{A.MNB\text{'}C\text{'}}+V_{A\text{'}.MNB\text{'}C\text{'}}$

Do IJ là đường trung bình của tam giác MBB' nên $IJ=\frac{1}{2}BB\text{'}=a\Rightarrow A\text{'}J=\frac{1}{2}a.$

+) $B\text{'}J=C\text{'}J=\sqrt{A\text{'}B{\text{'}}^2+A\text{'}{J}^2}=\sqrt{4a^2+\frac{a^2}{4}}=\frac{a\sqrt{17}}{2},B\text{'}C\text{'}=2a\sqrt{2},$ đặt $p=\frac{JB\text{'}+JC\text{'}+B\text{'}C\text{'}}{2}$

+) $S_{\Delta JB\text{'}C\text{'}}=\sqrt{p\left(p-IB\text{'}\right)\left(p-JC\text{'}\right)\left(p-B\text{'}C\text{'}\right)}=\frac{3a^2\sqrt{2}}{2}\Rightarrow S_{MNB\text{'}C\text{'}}=4S_{\Delta JB\text{'}C\text{'}}=6a^2\sqrt{2}.$

+) $$$d\left(A\text{'};\left(JB\text{'}C\text{'}\right)\right).S_{\Delta JB\text{'}C\text{'}}=3V_{A\text{'}.JB\text{'}C\text{'}}=A\text{'}J.S_{\Delta A\text{'}B\text{'}C\text{'}}\Rightarrow d\left(A\text{'};\left(JB\text{'}C\text{'}\right)\right)=\frac{A\text{'}J.S_{\Delta A\text{'}B\text{'}C\text{'}}}{S_{\Delta JB\text{'}C\text{'}}}=\frac{a\sqrt{2}}{3}.$

+) $V_{A.JB\text{'}C\text{'}}=V_{A.A\text{'}B\text{'}C\text{'}}-V_{J.A\text{'}B\text{'}C\text{'}}=\frac{1}{3}AJ.S_{\Delta A\text{'}B\text{'}C\text{'}}=a^3.$

+) $d\left(A;\left(JB\text{'}C\text{'}\right)\right).S_{\Delta JB\text{'}C\text{'}}=3V_{A.JB\text{'}C\text{'}}=3a^3\Rightarrow d\left(A;\left(JB\text{'}C\text{'}\right)\right)=\frac{3a^3}{S_{\Delta JB\text{'}C\text{'}}}=a\sqrt{2}.$

Vậy $V_{AMNA\text{'}B\text{'}C\text{'}}=V_{A,MNB\text{'}C\text{'}}+V_{A\text{'}.MNB\text{'}C\text{'}}=\frac{1}{3}\left(d\left(A\text{'};\left(JB\text{'}C\text{'}\right)\right)\right)+d\left(A;\left(JB\text{'}C\text{'}\right)\right).S_{MNB\text{'}C\text{'}}$

                     $=\frac{1}{3}\left(\frac{a\sqrt{2}}{3}+a\sqrt{2}\right).6a^2\sqrt{2}=\frac{16a^3}{3}.$

Chọn A.