Google
Câu hỏi: Cho hình lăng trụ $ABC.{A}'{B}'{C}'$. Gọi M, N, P lần lượt là các điểm thuộc các cạnh $A{A}'$, $B{B}'$, $C{C}'$ sao cho $AM=2M{A}'$, $N{B}'=2NB$, $PC=P{C}'$. Gọi ${{V}_{1}}$, ${{V}_{2}}$ lần lượt là thể tích của hai khối đa diện $ABCMNP$ và ${A}'{B}'{C}'MNP$. Tính tỉ 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. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{2}{3}$
A. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=2$
B. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{1}{2}$
C. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=1$
D. $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{2}{3}$
Gọi V là thể tích khối lăng trụ $ABC.{A}'{B}'{C}'$
Ta có ${{V}_{1}}={{V}_{M.ABC}}+{{V}_{M.BCPN}}$.
${{V}_{M.ABC}}=\dfrac{1}{3}d\left( M;\left( ABC \right) \right).{{S}_{ABC}}=\dfrac{1}{3}.\dfrac{2}{3}d\left( {A}';\left( ABC \right) \right).{{S}_{ABC}}=\dfrac{2}{9}V$.
$\dfrac{{{V}_{M.BCPN}}}{{{V}_{M.BC{C}'{B}'}}}=\dfrac{{{S}_{BCPN}}}{{{S}_{BC{C}'{B}'}}}=\dfrac{\dfrac{1}{2}d\left( C;B{B}' \right).\left( BN+CP \right)}{\dfrac{1}{2}d\left( C;B{B}' \right).\left( B{B}'+C{C}' \right)}=\dfrac{BN+CP}{B{B}'+C{C}'}=\dfrac{\dfrac{1}{3}B{B}'+\dfrac{1}{2}C{C}'}{B{B}'+C{C}'}$
$\Rightarrow {{V}_{M.BCPN}}\Rightarrow \dfrac{5}{12}{{V}_{M.BC{C}'{B}'}}=\dfrac{5}{12}{{V}_{A.BC{C}'{B}'}}=\dfrac{5}{12}.2{{V}_{AB{C}'{B}'}}=\dfrac{5}{12}.2.\dfrac{1}{3}V=\dfrac{5}{18}V$
$\Rightarrow {{V}_{1}}={{V}_{M.ABC}}+{{V}_{M.BCPN}}=\dfrac{2}{9}V+\dfrac{5}{18}V=\dfrac{1}{2}V\Rightarrow {{V}_{2}}=V-\dfrac{1}{2}V=\dfrac{1}{2}V\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=1$.
Ta có ${{V}_{1}}={{V}_{M.ABC}}+{{V}_{M.BCPN}}$.
${{V}_{M.ABC}}=\dfrac{1}{3}d\left( M;\left( ABC \right) \right).{{S}_{ABC}}=\dfrac{1}{3}.\dfrac{2}{3}d\left( {A}';\left( ABC \right) \right).{{S}_{ABC}}=\dfrac{2}{9}V$.
$\dfrac{{{V}_{M.BCPN}}}{{{V}_{M.BC{C}'{B}'}}}=\dfrac{{{S}_{BCPN}}}{{{S}_{BC{C}'{B}'}}}=\dfrac{\dfrac{1}{2}d\left( C;B{B}' \right).\left( BN+CP \right)}{\dfrac{1}{2}d\left( C;B{B}' \right).\left( B{B}'+C{C}' \right)}=\dfrac{BN+CP}{B{B}'+C{C}'}=\dfrac{\dfrac{1}{3}B{B}'+\dfrac{1}{2}C{C}'}{B{B}'+C{C}'}$
$\Rightarrow {{V}_{M.BCPN}}\Rightarrow \dfrac{5}{12}{{V}_{M.BC{C}'{B}'}}=\dfrac{5}{12}{{V}_{A.BC{C}'{B}'}}=\dfrac{5}{12}.2{{V}_{AB{C}'{B}'}}=\dfrac{5}{12}.2.\dfrac{1}{3}V=\dfrac{5}{18}V$
$\Rightarrow {{V}_{1}}={{V}_{M.ABC}}+{{V}_{M.BCPN}}=\dfrac{2}{9}V+\dfrac{5}{18}V=\dfrac{1}{2}V\Rightarrow {{V}_{2}}=V-\dfrac{1}{2}V=\dfrac{1}{2}V\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=1$.
Đáp án C.