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Câu hỏi: Cho hình lăng trụ $ABC.{A}'{B}'{C}'$. Gọi M, N, P lần lượt là các điểm thuộc các cạnh $\text{A{A}'},\text{ B{B}'},\text{ C{C}'}$ sao cho $AM=2M{A}',N{B}'=2NB,PC=P{C}'$. Gọi ${{V}_{1}},{{\text{V}}_{2}}$ lần lượt là thể tích của hai khối đa diện ABCMNP và ${A}'{B}'{C}'MNP$. Tính tỷ số $\dfrac{{{V}_{1}}}{{{V}_{2}}}$.
A. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=2$
B. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{1}{2}$
C. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=1$
D. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{2}{3}$
Đặt $V={{V}_{ABC.{A}'{B}'{C}'}}$
Ta có ${{V}_{ABCMNP}}={{V}_{P.ABNM}}+{{V}_{P.ABC}}$ mà
${{V}_{P.ABC}}=\dfrac{1}{3}d\left( P,(ABC) \right).{{S}_{\Delta ABC}}=\dfrac{1}{6}d\left( C;(ABC) \right).{{S}_{\Delta ABC}}=\dfrac{V}{6}$
$\dfrac{{{S}_{ABNM}}}{{{S}_{AB{B}'{A}'}}}=\dfrac{AM+BN}{A{A}'+B{B}'}=\dfrac{\dfrac{2}{3}A{A}'+\dfrac{1}{3}B{B}'}{A{A}'+B{B}'}=\dfrac{1}{2}\Rightarrow {{V}_{P.ABNM}}=\dfrac{1}{2}{{V}_{C.AB{B}'{A}'}}$
Mà ${{V}_{C.AB{B}'{A}'}}=\dfrac{2}{3}V$ suy ra ${{V}_{P.ABNM}}=\dfrac{1}{2}.\dfrac{2}{3}V=\dfrac{V}{3}$.
Khi đó ${{V}_{ABCMNP}}=\dfrac{V}{6}+\dfrac{V}{3}=\dfrac{V}{2}$.
Vậy $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{V}{2}:\dfrac{V}{2}=1$.
A. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=2$
B. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{1}{2}$
C. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=1$
D. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{2}{3}$
Đặt $V={{V}_{ABC.{A}'{B}'{C}'}}$
Ta có ${{V}_{ABCMNP}}={{V}_{P.ABNM}}+{{V}_{P.ABC}}$ mà
${{V}_{P.ABC}}=\dfrac{1}{3}d\left( P,(ABC) \right).{{S}_{\Delta ABC}}=\dfrac{1}{6}d\left( C;(ABC) \right).{{S}_{\Delta ABC}}=\dfrac{V}{6}$
$\dfrac{{{S}_{ABNM}}}{{{S}_{AB{B}'{A}'}}}=\dfrac{AM+BN}{A{A}'+B{B}'}=\dfrac{\dfrac{2}{3}A{A}'+\dfrac{1}{3}B{B}'}{A{A}'+B{B}'}=\dfrac{1}{2}\Rightarrow {{V}_{P.ABNM}}=\dfrac{1}{2}{{V}_{C.AB{B}'{A}'}}$
Mà ${{V}_{C.AB{B}'{A}'}}=\dfrac{2}{3}V$ suy ra ${{V}_{P.ABNM}}=\dfrac{1}{2}.\dfrac{2}{3}V=\dfrac{V}{3}$.
Khi đó ${{V}_{ABCMNP}}=\dfrac{V}{6}+\dfrac{V}{3}=\dfrac{V}{2}$.
Vậy $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{V}{2}:\dfrac{V}{2}=1$.
Đáp án C.