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Câu hỏi: Cho khối chóp $S.ABC$. Trên ba cạnh $SA,SB,SC$ lần lượt lấy ba điểm $A,{B}',{C}'$ sao cho $2S{A}'=SA,4S{B}'=SB,\text{ 5}S{C}'=SC$. Tính tỉ số $\dfrac{{{V}_{S.A'B'C'}}}{{{V}_{S.ABC}}}$
A. $\dfrac{1}{10}$.
B. $\dfrac{1}{40}$.
C. $\dfrac{1}{8}$.
D. $\dfrac{1}{20}$.
$2S{A}'=SA,4S{B}'=SB,\text{ 5}S{C}'=SC$ $\Rightarrow \dfrac{S{A}'}{SA}=\dfrac{1}{2},\dfrac{S{B}'}{SB}=\dfrac{1}{4},\ \dfrac{S{C}'}{SC}=\dfrac{1}{5}$.
$\dfrac{{{V}_{S.A'B'C'}}}{{{V}_{S.ABC}}}=\dfrac{S{A}'}{SA}.\dfrac{S{B}'}{SB}.\dfrac{S{C}'}{SC}=\dfrac{1}{2}.\dfrac{1}{4}.\dfrac{1}{5}=\dfrac{1}{40}$.
A. $\dfrac{1}{10}$.
B. $\dfrac{1}{40}$.
C. $\dfrac{1}{8}$.
D. $\dfrac{1}{20}$.
$\dfrac{{{V}_{S.A'B'C'}}}{{{V}_{S.ABC}}}=\dfrac{S{A}'}{SA}.\dfrac{S{B}'}{SB}.\dfrac{S{C}'}{SC}=\dfrac{1}{2}.\dfrac{1}{4}.\dfrac{1}{5}=\dfrac{1}{40}$.
Đáp án B.