Đáp án:

181) $V_{B'.AMC} = \dfrac{a^3}{4}$

182) $V_{A.B'C'D'}= \dfrac16$

183) $V_{ABCDMN} = \dfrac{5a^3}{2}$

Giải thích các bước giải:

181) Sửa đề: $M\in AB$ sao cho $AM=  3MB$

Ta có:

$AM = \dfrac{3}{4}AB$

$\Rightarrow S_{AMC} = \dfrac{3}{4}S_{ABC} = \dfrac{3}{4}\cdot\dfrac{1}{2}AB.BC = \dfrac{3a^2}{4}$

$\Rightarrow V_{B'.AMC} = \dfrac13S_{AMC}.AA' = \dfrac13\cdot\dfrac{3a^2}{4}\cdot a = \dfrac{a^3}{4}$

182) Ta có:

$V_{A.B'C'D'} = \dfrac{1}{3}S_{B'C'D'}.AA' = \dfrac{1}{3}\cdot\dfrac{1}{2}S_{A'B'C'D'}.AA' = \dfrac{1}{6}V_{ABCD.A'B'C'D'} = \dfrac16$

183) Ta có:

$\dfrac{SA}{SA'} = 2; \dfrac{SB}{SB} = 1;\, \dfrac{SC}{SC} = 1;\, \dfrac{SD}{SD'} = 2$

$\Rightarrow \dfrac{V_{S.MNBC}}{V_{S.ABCD}} = \dfrac{2+1+1+2}{4.2.1.1.2} =\dfrac{3}{8}$

$\Rightarrow V_{ABCDMN} = \dfrac{5}{8}V_{S.ABCD} = \dfrac{5}{8}\cdot\dfrac13\cdot (2a)^2\cdot 3a = \dfrac{5a^3}{2}$