Cho khối lăng trụ tam giác ABC.A'B'C' mà

* Hướng dẫn giải

Đáp án D

Phương pháp giải:

- Sử dụng định lí: $a//\left(P\right)\supset b\Rightarrow d\left(a;b\right)=\left(a;\left(P\right)\right)$ ( a,b là 2 đường thẳng chéo nhau).

- Tính $V_{C^{\text{'}}.AB{B}^{\text{'}}{A}^{\text{'}}}=\frac{1}{3}d\left(C^{\text{'}};\left(AB{B}^{\text{'}}{A}^{\text{'}}\right)\right).S_{AB{B}^{\text{'}}{A}^{\text{'}}}$.

- Tính $V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=\frac{3}{2}{V}_{C^{\text{'}}.AB{B}^{\text{'}}{A}^{\text{'}}}$.

Giải chi tiết:

Vì CC'//AA' nên $C{C}^{\text{'}}//\left(AB{B}^{\text{'}}{A}^{\text{'}}\right)\supset A{B}^{\text{'}}$.

Do đó $d\left(C{C}^{\text{'}};A{B}^{\text{'}}\right)=d\left(C{C}^{\text{'}};\left(AB{B}^{\text{'}}{A}^{\text{'}}\right)\right)=d\left(C^{\text{'}};\left(AB{B}^{\text{'}}{A}^{\text{'}}\right)\right)=7$.

Khi đó ta có: $V_{C^{\text{'}}.AB{B}^{\text{'}}{A}^{\text{'}}}=\frac{1}{3}d\left(C^{\text{'}};\left(AB{B}^{\text{'}}{A}^{\text{'}}\right)\right).S_{AB{B}^{\text{'}}{A}^{\text{'}}}=\frac{1}{3}\mathrm{.7.4}=\frac{28}{3}$.

Mà $V_{C^{\text{'}}.AB{B}^{\text{'}}{A}^{\text{'}}}=\frac{2}{3}{V}_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}\Rightarrow V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=\frac{3}{2}{V}_{C^{\text{'}}.AB{B}^{\text{'}}{A}^{\text{'}}}=\frac{3}{2}.\frac{28}{3}=14$.