Cho khối lăng trụ đứng ABC.A'B'C' có đáy ABC là tam giác vuông tại B với

Phương pháp giải:

+) So sánh thể tích của khối tứ diện IABCIABC với thể tích của khối lăng trụ.

+) Tính thể tích khối lăng trụ.

Giải chi tiết:

Ta có: ${\text{A}}^{\text{'}}\text{M}\Vert \text{AC}\Rightarrow \frac{{\text{A}}^{\text{'}}\text{M}}{\text{AC}}=\frac{{\text{A}}^{\text{'}}\text{I}}{\text{IC}}=\frac{1}{2}\Rightarrow \frac{\text{IC}}{{\text{A}}^{\text{'}}\text{C}}=\frac{2}{3}$.

Vì $I{A}^{\text{'}}\cap \left(ABC\right)=C\Rightarrow \frac{d(I;(ABC\left)\right)}{d\left(A^{\text{'}};\left(ABC\right)\right)}=\frac{IC}{A^{\text{'}}C}=\frac{2}{3}$.

$$\begin{gathered} \Rightarrow \frac{{\text{V}}_{\text{I.ABC}}}{{\text{V}}_{{\text{ABC.A}}^{\text{'}}{\text{B}}^{\text{'}}{\text{C}}^{\text{'}}}}=\frac{\frac{1}{3} \text{d}(\text{I};(\text{ABC}\left)\right).{\text{S}}_{\text{ABC}}}{\text{d}\left({\text{A}}^{\text{'}};\left(\text{ABC}\right)\right){\text{.S}}_{\text{ABC}}}=\frac{1}{3}\cdot \frac{2}{3}=\frac{2}{9} \\ \Rightarrow {\text{V}}_{\text{I}.\text{ABC}}=\frac{2}{9} {\text{V}}_{{\text{ABC.A}}^{\text{'}}{\text{B}}^{\text{'}}{\text{C}}^{\text{'}}} \end{gathered}$$

Ta có: ${\text{AA}}^{\text{'}}\perp \left(\text{ABC}\right)\Rightarrow {\text{AA}}^{\text{'}}\perp \text{AC}\Rightarrow \Delta {\text{AA}}^{\text{'}}\text{C}$ vuông tại A.

$\Rightarrow \text{AC}=\sqrt{{\text{A}}^{\text{'}}{\text{C}}^2-{\text{AA}}^{\text{'}2}}=\sqrt{9{\text{a}}^2-4{\text{a}}^2}=\text{a}\sqrt{5}$.

Xét tam giác vuông ABC có: $\text{BC}=\sqrt{{\text{AC}}^2-{\text{AB}}^2}=\sqrt{5{\text{a}}^2-{\text{a}}^2}=2\text{a}$.

$\Rightarrow {\text{S}}_{\text{ABC}}=\frac{1}{2}\text{AB.BC}=\frac{1}{2}\text{a}.2\text{a}={\text{a}}^2$.

$\Rightarrow {\text{V}}_{\text{ABC}\cdot {\text{A}}^{\text{'}}{\text{B}}^{\text{'}}{\text{C}}^{\text{'}}}={\text{AAA}}^{\text{'}}\cdot {\text{S}}_{\text{ABC}}=2\text{a}\cdot {\text{a}}^2=2{\text{a}}^3$.

Vậy ${\text{V}}_{\text{I}.\text{ABC}}=\frac{2}{9} {\text{V}}_{{\text{ABC.A}}^{\text{'}}{\text{B}}^{\text{'}}{\text{C}}^{\text{'}}}=\frac{2}{9}.2{\text{a}}^3=\frac{4{\text{a}}^3}{9}$.

Chọn C.