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Câu hỏi: Cho khối lăng trụ ABC.A'B'C' có thể tích bằng V. Điểm M là thuộc cạnh A'B' sao cho ${A}'M=\dfrac{1}{3}{A}'{B}'$. Mặt phẳng (BCM) cắt đường thẳng AA' tại F, và cắt đường thẳng A'C' tại G. Thể tích khối chóp ABCMGA' bằng
A. $\dfrac{1}{27}V$
B. $\dfrac{5}{27}V$
C. $\dfrac{13}{27}V$
D. $\dfrac{1}{54}V$
Đường thẳng AA' cắt BM tại F $\Rightarrow A{A}'\cap \left( BCM \right)=\left\{ F \right\}$
Đường thẳng FC cắt A'C' tại G $\Rightarrow {A}'{C}'\cap \left( BCM \right)=\left\{ G \right\}$
${A}'M//AB\Rightarrow \dfrac{F{A}'}{FA}=\dfrac{FM}{FB}=\dfrac{{A}'M}{AB}=\dfrac{1}{3}$
${A}'G//AC\Rightarrow \dfrac{FG}{FC}=\dfrac{F{A}'}{FA}=\dfrac{1}{3}$
$\Rightarrow \dfrac{{{V}_{F{A}'MG}}}{{{V}_{FABC}}}=\dfrac{1}{3}.\dfrac{1}{3}.\dfrac{1}{3}=\dfrac{1}{27}\Rightarrow {{V}_{F{A}'MG}}=\dfrac{1}{27}{{V}_{FABC}}$
${{V}_{FABC}}=\dfrac{1}{3}{{S}_{\Delta ABC}}.d\left[ F, \left( ABC \right) \right]=\dfrac{1}{3}{{S}_{\Delta ABC}}.\dfrac{3}{2}d\left[ {A}', \left( ABC \right) \right]=\dfrac{1}{2}V\Rightarrow {{V}_{F{A}'MG}}=\dfrac{1}{27}.\dfrac{1}{2}=\dfrac{1}{54}V$
Suy ra ${{V}_{ABCMG{A}'}}={{V}_{F.ABC}}-{{V}_{F.MG{A}'}}=\dfrac{1}{2}V-\dfrac{1}{54}V=\dfrac{13}{27}V$
A. $\dfrac{1}{27}V$
B. $\dfrac{5}{27}V$
C. $\dfrac{13}{27}V$
D. $\dfrac{1}{54}V$
Đường thẳng FC cắt A'C' tại G $\Rightarrow {A}'{C}'\cap \left( BCM \right)=\left\{ G \right\}$
${A}'M//AB\Rightarrow \dfrac{F{A}'}{FA}=\dfrac{FM}{FB}=\dfrac{{A}'M}{AB}=\dfrac{1}{3}$
${A}'G//AC\Rightarrow \dfrac{FG}{FC}=\dfrac{F{A}'}{FA}=\dfrac{1}{3}$
$\Rightarrow \dfrac{{{V}_{F{A}'MG}}}{{{V}_{FABC}}}=\dfrac{1}{3}.\dfrac{1}{3}.\dfrac{1}{3}=\dfrac{1}{27}\Rightarrow {{V}_{F{A}'MG}}=\dfrac{1}{27}{{V}_{FABC}}$
${{V}_{FABC}}=\dfrac{1}{3}{{S}_{\Delta ABC}}.d\left[ F, \left( ABC \right) \right]=\dfrac{1}{3}{{S}_{\Delta ABC}}.\dfrac{3}{2}d\left[ {A}', \left( ABC \right) \right]=\dfrac{1}{2}V\Rightarrow {{V}_{F{A}'MG}}=\dfrac{1}{27}.\dfrac{1}{2}=\dfrac{1}{54}V$
Suy ra ${{V}_{ABCMG{A}'}}={{V}_{F.ABC}}-{{V}_{F.MG{A}'}}=\dfrac{1}{2}V-\dfrac{1}{54}V=\dfrac{13}{27}V$
Đáp án C.