Đáp án:
178) $V_{A.BMNC} = 10$
179) $V_{A.SB'C'} = \dfrac{a^3}{24}$
180) $V_{A'.ACD'} = 2a^3$
Giải thích các bước giải:
178) Ta có:
$\dfrac{V_{S.AMN}}{V_{S.ABC}}= \dfrac{SA}{SA}\cdot\dfrac{SM}{SB}\cdot\dfrac{SN}{SC}=1\cdot\dfrac{1}{2}\cdot\dfrac{2}{3} = \dfrac{1}{3}$
$\Rightarrow \dfrac{V_{A.BMNC}}{V_{S.ABC}} =\dfrac{2}{3}$
$\Rightarrow V_{A.BMNC} = \dfrac23V_{S.ABC}= \dfrac23\cdot\dfrac13\cdot 5.9 = 10$
179) $V_{A.SBC} = \dfrac{1}{3}S_{SBC}.AS = \dfrac{1}{3}\cdot\dfrac{1}{2}SB.SC.SA = \dfrac{a^3}{6}$
Áp dụng hệ thức lượng trong tam giác vuông ta được:
$SA^2 = AB'.AB$
$\Rightarrow \dfrac{AB'}{AB} = \dfrac{SA^2}{AB^2}$
$SA^2 = AC'.AC$
$\Rightarrow \dfrac{AC'}{AC} = \dfrac{SA^2}{AC^2}$
Ta có:
$\dfrac{V_{A.SB'C'}}{V_{S.ABC}} = \dfrac{SA}{SA}\cdot\dfrac{AB'}{AB}\cdot\dfrac{AC'}{AC} = \dfrac{SA^4}{AB^2.AC^2} = \dfrac{a^4}{(a\sqrt2)^2.(a\sqrt2)^2} = \dfrac{1}{4}$
$\Rightarrow V_{A.SB'C'} = \dfrac{1}{4}V_{A.SBC} = \dfrac{1}{4}\cdot\dfrac{a^3}{6} = \dfrac{a^3}{24}$
180) Ta có:
$V_{A'.ACD'} = \dfrac{1}{2}V_{ABCD.A'B'C'D'} - V_{C.A'D'C'} - V_{D'.ADC}$
$=\dfrac{1}{2}V - \dfrac{1}{6}V - \dfrac{1}{6}V = \dfrac{1}{6}V$
$\Rightarrow V_{A'.ACD'} = \dfrac{1}{6}.3a.2a.2a = 2a^3$
