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Câu hỏi: Cho hình chóp $S.ABCD$. Đáy $ABCD$ là hình bình hành, $M$ là trung điểm $SB,N$ thuộc cạnh $SC$ sao cho $\dfrac{SN}{SC}=\dfrac{2}{3},P$ thuộc cạnh $SD$ sao cho $\dfrac{SP}{SD}=\dfrac{3}{4}.$ $Mp\left( MNP \right)$ cắt $SA,AD,BC$ lần lượt tại $Q,E,F$. Biết thể tích khối $S.MNPQ$ bằng 1. Tính thể tích khối $ABEFQM$.
A. $\dfrac{73}{15}$
B. $\dfrac{154}{66}$
C. $\dfrac{207}{41}$
D. $\dfrac{29}{5}$
Đặt $\dfrac{SM}{SB}=x,\dfrac{SN}{SC}=y,\dfrac{SP}{SD}=z,\dfrac{SQ}{SA}=t$ thì $\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{y}+\dfrac{1}{t}\Leftrightarrow 2+\dfrac{4}{3}=\dfrac{3}{2}+\dfrac{1}{t}\Leftrightarrow t=\dfrac{6}{11}$
Mặt khác $\dfrac{{{V}_{S.MNPQ}}}{{{V}_{S.ABCD}}}=\dfrac{xyzt}{4}\left( \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t} \right)=\dfrac{5}{22}\Rightarrow {{V}_{S.ABCD}}=\dfrac{22}{5}\Rightarrow {{V}_{ABCD.MNPQ}}=\dfrac{17}{5}$
Theo định lý Menelaus trong $\Delta SAD$ ta có
$\dfrac{SQ}{QA}.\dfrac{AE}{ED}.\dfrac{DP}{PS}=1\Leftrightarrow \dfrac{6}{5}.\dfrac{AE}{ED}.\dfrac{1}{3}=1\Leftrightarrow \dfrac{AE}{ED}=\dfrac{5}{2}\Rightarrow \dfrac{AD}{DE}=\dfrac{3}{2}\Rightarrow \dfrac{{{S}_{DEF}}}{{{S}_{ABC}}}=\dfrac{2}{3}\Rightarrow \dfrac{{{S}_{DEF}}}{{{S}_{ABCD}}}=\dfrac{1}{3}$
Theo định lý Menelaus trong $\Delta SBC$ ta có
$\dfrac{SM}{MB}.\dfrac{BF}{FC}.\dfrac{CN}{NS}=1\Leftrightarrow \dfrac{BF}{FC}=2\Rightarrow \dfrac{BF}{BC}=2\Rightarrow \dfrac{{{S}_{DCF}}}{{{S}_{ABC}}}=1\Rightarrow \dfrac{{{S}_{DCF}}}{{{S}_{ABCD}}}=\dfrac{1}{2}$
Suy ra $\dfrac{{{S}_{CDEF}}}{{{S}_{ABCD}}}=\dfrac{5}{6}\Rightarrow \dfrac{{{V}_{N.CDEF}}}{{{V}_{S.ABCD}}}=\dfrac{NC}{SC}.\dfrac{{{S}_{CDEF}}}{{{S}_{ABCD}}}=\dfrac{5}{18}\Rightarrow {{V}_{N.CDEF}}=\dfrac{11}{9}$
Ta có
$\dfrac{{{V}_{N.DPE}}}{{{V}_{S.ABCD}}}=\dfrac{{{V}_{N.DPE}}}{2{{V}_{C.SAD}}}=\dfrac{1}{2}.\dfrac{SN}{SC}.\dfrac{DP}{DS}.\dfrac{DE}{AD}=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{1}{4}.\dfrac{2}{3}=\dfrac{1}{18}\Rightarrow {{V}_{N.DPE}}=\dfrac{1}{18}{{V}_{S.ABCD}}=\dfrac{11}{45}$
Vậy thể tích khối cầu cần tính là ${{V}_{ABFEQM}}={{V}_{ABCD.MNPQ}}+{{V}_{N.DPE}}+{{V}_{N.CDEF}}=\dfrac{17}{5}+\dfrac{11}{9}+\dfrac{11}{45}=\dfrac{73}{15}.$
A. $\dfrac{73}{15}$
B. $\dfrac{154}{66}$
C. $\dfrac{207}{41}$
D. $\dfrac{29}{5}$
Đặt $\dfrac{SM}{SB}=x,\dfrac{SN}{SC}=y,\dfrac{SP}{SD}=z,\dfrac{SQ}{SA}=t$ thì $\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{y}+\dfrac{1}{t}\Leftrightarrow 2+\dfrac{4}{3}=\dfrac{3}{2}+\dfrac{1}{t}\Leftrightarrow t=\dfrac{6}{11}$
Mặt khác $\dfrac{{{V}_{S.MNPQ}}}{{{V}_{S.ABCD}}}=\dfrac{xyzt}{4}\left( \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t} \right)=\dfrac{5}{22}\Rightarrow {{V}_{S.ABCD}}=\dfrac{22}{5}\Rightarrow {{V}_{ABCD.MNPQ}}=\dfrac{17}{5}$
Theo định lý Menelaus trong $\Delta SAD$ ta có
$\dfrac{SQ}{QA}.\dfrac{AE}{ED}.\dfrac{DP}{PS}=1\Leftrightarrow \dfrac{6}{5}.\dfrac{AE}{ED}.\dfrac{1}{3}=1\Leftrightarrow \dfrac{AE}{ED}=\dfrac{5}{2}\Rightarrow \dfrac{AD}{DE}=\dfrac{3}{2}\Rightarrow \dfrac{{{S}_{DEF}}}{{{S}_{ABC}}}=\dfrac{2}{3}\Rightarrow \dfrac{{{S}_{DEF}}}{{{S}_{ABCD}}}=\dfrac{1}{3}$
Theo định lý Menelaus trong $\Delta SBC$ ta có
$\dfrac{SM}{MB}.\dfrac{BF}{FC}.\dfrac{CN}{NS}=1\Leftrightarrow \dfrac{BF}{FC}=2\Rightarrow \dfrac{BF}{BC}=2\Rightarrow \dfrac{{{S}_{DCF}}}{{{S}_{ABC}}}=1\Rightarrow \dfrac{{{S}_{DCF}}}{{{S}_{ABCD}}}=\dfrac{1}{2}$
Suy ra $\dfrac{{{S}_{CDEF}}}{{{S}_{ABCD}}}=\dfrac{5}{6}\Rightarrow \dfrac{{{V}_{N.CDEF}}}{{{V}_{S.ABCD}}}=\dfrac{NC}{SC}.\dfrac{{{S}_{CDEF}}}{{{S}_{ABCD}}}=\dfrac{5}{18}\Rightarrow {{V}_{N.CDEF}}=\dfrac{11}{9}$
Ta có
$\dfrac{{{V}_{N.DPE}}}{{{V}_{S.ABCD}}}=\dfrac{{{V}_{N.DPE}}}{2{{V}_{C.SAD}}}=\dfrac{1}{2}.\dfrac{SN}{SC}.\dfrac{DP}{DS}.\dfrac{DE}{AD}=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{1}{4}.\dfrac{2}{3}=\dfrac{1}{18}\Rightarrow {{V}_{N.DPE}}=\dfrac{1}{18}{{V}_{S.ABCD}}=\dfrac{11}{45}$
Vậy thể tích khối cầu cần tính là ${{V}_{ABFEQM}}={{V}_{ABCD.MNPQ}}+{{V}_{N.DPE}}+{{V}_{N.CDEF}}=\dfrac{17}{5}+\dfrac{11}{9}+\dfrac{11}{45}=\dfrac{73}{15}.$
Đáp án A.