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Câu hỏi: Cho hình chóp $S.ABCD$ có đáy là hình vuông, cạnh bên $SA$ vuông góc với đáy. Gọi $M$, $N$ là trung điểm của $SA$, $SB$. Mặt phẳng $MNCD$ chia hình chóp đã cho thành hai phần. Tỉ số thể tích hai phần $S.MNCD$ và $MNABCD$ là
A. $1$.
B. $\dfrac{4}{5}$.
C. $\dfrac{3}{4}$.
D. $\dfrac{3}{5}$.
Ta có ${{V}_{S.MNCD}}={{V}_{S.MCD}}+{{V}_{S.MNC}}$
+ $\dfrac{{{V}_{S.MCD}}}{{{V}_{S.ACD}}}=\dfrac{SM}{SA}.\dfrac{SC}{SC}.\dfrac{SD}{SD}=\dfrac{1}{2}\Rightarrow {{V}_{S.MCD}}=\dfrac{1}{2}{{V}_{S.ACD}}=\dfrac{1}{4}{{V}_{S.ABCD}}.$
+ $\dfrac{{{V}_{S.MNC}}}{{{V}_{S.ABC}}}=\dfrac{SM}{SA}.\dfrac{SN}{SB}.\dfrac{SC}{SC}=\dfrac{1}{4}\Rightarrow {{V}_{S.MNC}}=\dfrac{1}{4}{{V}_{S.ABC}}=\dfrac{1}{8}{{V}_{S.ABCD}}.$
$\Rightarrow {{V}_{S.MNCD}}={{V}_{S.MCD}}+{{V}_{S.MNC}}=\dfrac{1}{4}{{V}_{S.ABCD}}+\dfrac{1}{8}{{V}_{S.ABCD}}=\dfrac{3}{8}{{V}_{S.ABCD}}.$
$\Rightarrow {{V}_{MNABCD}}={{V}_{S.ABCD}}-{{V}_{S.MNCD}}={{V}_{S.ABCD}}-\dfrac{3}{8}{{V}_{S.ABCD}}=\dfrac{5}{8}{{V}_{S.ABCD}}.$
Do đó $\dfrac{{{V}_{S.MNCD}}}{{{V}_{MNABCD}}}=\dfrac{\dfrac{3}{8}{{V}_{S.ABCD}}}{\dfrac{5}{8}{{V}_{S.ABCD}}}=\dfrac{3}{5}.$
A. $1$.
B. $\dfrac{4}{5}$.
C. $\dfrac{3}{4}$.
D. $\dfrac{3}{5}$.
Ta có ${{V}_{S.MNCD}}={{V}_{S.MCD}}+{{V}_{S.MNC}}$
+ $\dfrac{{{V}_{S.MCD}}}{{{V}_{S.ACD}}}=\dfrac{SM}{SA}.\dfrac{SC}{SC}.\dfrac{SD}{SD}=\dfrac{1}{2}\Rightarrow {{V}_{S.MCD}}=\dfrac{1}{2}{{V}_{S.ACD}}=\dfrac{1}{4}{{V}_{S.ABCD}}.$
+ $\dfrac{{{V}_{S.MNC}}}{{{V}_{S.ABC}}}=\dfrac{SM}{SA}.\dfrac{SN}{SB}.\dfrac{SC}{SC}=\dfrac{1}{4}\Rightarrow {{V}_{S.MNC}}=\dfrac{1}{4}{{V}_{S.ABC}}=\dfrac{1}{8}{{V}_{S.ABCD}}.$
$\Rightarrow {{V}_{S.MNCD}}={{V}_{S.MCD}}+{{V}_{S.MNC}}=\dfrac{1}{4}{{V}_{S.ABCD}}+\dfrac{1}{8}{{V}_{S.ABCD}}=\dfrac{3}{8}{{V}_{S.ABCD}}.$
$\Rightarrow {{V}_{MNABCD}}={{V}_{S.ABCD}}-{{V}_{S.MNCD}}={{V}_{S.ABCD}}-\dfrac{3}{8}{{V}_{S.ABCD}}=\dfrac{5}{8}{{V}_{S.ABCD}}.$
Do đó $\dfrac{{{V}_{S.MNCD}}}{{{V}_{MNABCD}}}=\dfrac{\dfrac{3}{8}{{V}_{S.ABCD}}}{\dfrac{5}{8}{{V}_{S.ABCD}}}=\dfrac{3}{5}.$
Đáp án D.