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Câu hỏi: Cho khối chóp $S.ABC$. Gọi $M$ là trung điểm của $SA$ . Tỉ số thể tích $\dfrac{{{V}_{M.ABC}}}{{{V}_{S.ABC}}}$ bằng
A. $\dfrac{1}{4}$.
B. $\dfrac{1}{2}$.
C. $2$.
D. $\pi {{a}^{3}}$.
Ta có $\dfrac{{{V}_{M.ABC}}}{{{V}_{S.ABC}}}=\dfrac{\dfrac{1}{3}{{S}_{\Delta ABC}}.d\left( M;\left( ABC \right) \right)}{\dfrac{1}{3}{{S}_{\Delta ABC}}.d\left( S;\left( ABC \right) \right)}$ $=\dfrac{d\left( M;\left( ABC \right) \right)}{d\left( S;\left( ABC \right) \right)}=\dfrac{MA}{SA}$ $=\dfrac{1}{2}$.
A. $\dfrac{1}{4}$.
B. $\dfrac{1}{2}$.
C. $2$.
D. $\pi {{a}^{3}}$.
Ta có $\dfrac{{{V}_{M.ABC}}}{{{V}_{S.ABC}}}=\dfrac{\dfrac{1}{3}{{S}_{\Delta ABC}}.d\left( M;\left( ABC \right) \right)}{\dfrac{1}{3}{{S}_{\Delta ABC}}.d\left( S;\left( ABC \right) \right)}$ $=\dfrac{d\left( M;\left( ABC \right) \right)}{d\left( S;\left( ABC \right) \right)}=\dfrac{MA}{SA}$ $=\dfrac{1}{2}$.
Đáp án B.