Đáp án:
$V_{HA'MN}=\dfrac{9a^3}{32}$
Giải thích các bước giải:
$\Delta ABC\bot A: AC=\sqrt{BC^2-AB^2}=\sqrt{4a^2-a^2}=a\sqrt3$
$S_{ABC}=\dfrac12.AH.BC=\dfrac12.AB.AC$
$\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{a.a\sqrt3}{2a}=\dfrac{a\sqrt3}2$
$AC'=\sqrt{AC^2+CC'^2}=\sqrt{3a^2+3a^2}=a\sqrt6\Rightarrow MN=\dfrac{AC'}2=\dfrac{a\sqrt6}2$
Ta có:
$\begin{cases}AB\bot AC\text{ (do giả thiết cho tam giác ABC vuông tại A)}\\AB\bot AA'\text{ (do ABC.A'B'C' là lăng trụ đứng)}\end{cases}$
$\Rightarrow AB\bot(ACC'A')$
Xét $\Delta ABC\bot A$ có $AH\bot BC$
$\Rightarrow: AC^2=CH.BC\Rightarrow CH=\dfrac{AC^2}{BC^2}=\dfrac{3a^2}{2a}=\dfrac{3a}2$
$\Rightarrow\dfrac{d(H,(AMN))}{d(B,(AMN))}=\dfrac{HC}{BC}=\dfrac{d(H,(AMN))}{BA}$
$\Rightarrow d(H, (AMN))=\dfrac{BA.HC}{BC}=\dfrac{a.\dfrac{3a}2}{2a}=\dfrac{3a}4=d(H,(A'MN))$
$S_{A'MN}=S_{ACC'A}-S_{AA'M}-S_{A'C'N}-S_{CMN}$
$=(a\sqrt3)^3-\dfrac12.a\sqrt3.\dfrac{a\sqrt3}2-\dfrac12.a\sqrt3.\dfrac{a\sqrt3}2-\dfrac12.\dfrac{a\sqrt3}2.\dfrac{a\sqrt3}2$
$=3a^2-\dfrac{3a^2}4-\dfrac{3a^2}4-\dfrac{3a^2}8=\dfrac{9a^2}8$
$\Rightarrow V_{HA'MN}=\dfrac13.d(H,(A'MN)).S_{A'MN}$
$=\dfrac13.\dfrac{3a}4.\dfrac{9a^2}8$
$=\dfrac{9a^3}{32}$.
