Cho tứ diện đều ABCD

C

- Hướng dẫn giải

Đáp án C

Gọi H là trọng tâm $\Delta BCD$ thì $AH\perp \left(BCD\right)$.

Ta có: $BH=\frac{2}{3}.\frac{3\sqrt{3}}{2}=\sqrt{3}$

$\Rightarrow AH=\sqrt{A{B}^2-B{H}^2}=\sqrt{9-3}=\sqrt{6}$ 

Do đó: $V_{ABCD}=\frac{1}{3}.AH.S_{BCD}=\frac{1}{3}.\sqrt{6}.\frac{3^2\sqrt{3}}{4}=\frac{9\sqrt{2}}{4}$.

Lại có:

$$\begin{gathered} \frac{V_{C.MNP}}{V_{C.ABD}}=\frac{\frac{1}{3}d\left(C,\left(ABD\right)\right).S_{MNP}}{\frac{1}{3}d\left(C,\left(ABD\right)\right).S_{ABD}}=\frac{S_{MNP}}{S_{ABD}} \\ =\frac{S_{ABD}-S_{SPM}-S_{DMN}-S_{BPN}}{S_{ABD}} \\ =1-\frac{1}{2}.\frac{2017}{4035}-\frac{1}{4}-\frac{1}{2}.\frac{2018}{4035}=\frac{1}{4} \end{gathered}$$

 Vậy $V_{C.MNP}=\frac{1}{4}.\frac{9\sqrt{2}}{4}=\frac{9\sqrt{2}}{16}$.