Đáp án:

$B.\ \dfrac{5}{12}$

Giải thích các bước giải:

Theo tỉ lệ thể tích khối chóp tam giác, ta có:

$\bullet\quad\dfrac{V_{S.MNC}}{V_{S.ABC}}=\dfrac{SM}{SA}\cdot \dfrac{SN}{SB}\cdot \dfrac{SC}{SC}$

$\Leftrightarrow \dfrac{V_{S.MNC}}{V_{S.ABC}}= \dfrac12\cdot \dfrac23\cdot 1$

$\Leftrightarrow \dfrac{V_{S.MNC}}{V_{S.ABC}}=\dfrac13$

mà $V_{S.ABC}=\dfrac12V_{S.ABCD}$

nên $\dfrac{V_{S.MNC}}{V_{S.ABCD}}=\dfrac16$

$\bullet\quad\dfrac{V_{S.MCD}}{V_{S.ACD}}=\dfrac{SM}{SA}\cdot \dfrac{SC}{SC}\cdot \dfrac{SD}{SD}$

$\Leftrightarrow \dfrac{V_{S.MCD}}{V_{S.ACD}}= \dfrac12\cdot 1\cdot 1$

$\Leftrightarrow \dfrac{V_{S.MCD}}{V_{S.ACD}}= \dfrac12$

mà $V_{S.ACD}=\dfrac12V_{S.ABCD}$

nên $\dfrac{V_{S.MCD}}{V_{S.ABCD}}=\dfrac14$

Ta được:

$\dfrac{V_{S.MNCD}}{V_{S.ABCD}}=\dfrac{V_{S.MNC}}{V_{S.ABCD}} +\dfrac{V_{S.MCD}}{V_{S.ABCD}}=\dfrac16+\dfrac14 =\dfrac{5}{12}$

Theo công thức về tỉ lệ thể tích ta được:

$\begin{array}{l}
{V_{S.ABCD}} = {V_{S.ABC}} + {V_{S.ADC}} = 2{V_{S.ABC}}\\
{V_{S.MNCD}} = {V_{S.MNC}} + {V_{S.MCD}}\\
\dfrac{{{V_{S.MNC}}}}{{{V_{S.ABC}}}} = \dfrac{{SM}}{{SA}}.\dfrac{{SN}}{{SB}}.\dfrac{{SC}}{{SC}} = \dfrac{1}{2}.\dfrac{2}{3}.1 = \dfrac{1}{3}\\
 \Rightarrow {V_{S.MNC}} = \dfrac{1}{3}{V_{S.ABC}}\\
\dfrac{{{V_{S.MCD}}}}{{{V_{S.ACD}}}} = \dfrac{{SM}}{{SA}}.\dfrac{{SD}}{{SD}}.\dfrac{{SC}}{{SC}} = \dfrac{1}{2}\\
 \Rightarrow {V_{S.MCD}} = \dfrac{1}{2}{V_{S.ACD}}\\
 \Rightarrow {V_{S.MNCD}} = \dfrac{1}{3}{V_{S.ABC}} + \dfrac{1}{2}{V_{S.ACD}}\\
 \Rightarrow {V_{S.MNCD}} = \dfrac{1}{6}{V_{S.ABCD}} + \dfrac{1}{4}{V_{S.ABCD}}\\
 \Rightarrow {V_{S.MNCD}} = \dfrac{5}{{12}}{V_{S.ABCD}} \to B
\end{array}$