Cho hình chóp S.ABCD. Đáy ABCD là hình bình hành, M là trung điểm SB

Chọn A.

Đặt $\frac{SM}{SB}=x,\frac{SN}{SC}=y,\frac{SP}{SD}=z,\frac{SQ}{SA}=t$ thì $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}+\frac{1}{t}\iff 2+\frac{4}{3}=\frac{3}{2}+\frac{1}{t}\iff t=\frac{6}{11}$

Mặt khác $\frac{V_{S.MNPQ}}{V_{S.ABCD}}=\frac{xyzt}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)=\frac{5}{22}\Rightarrow V_{S.ABCD}=\frac{22}{5}\Rightarrow V_{ABCD.MNPQ}=\frac{17}{5}$

Theo định lý Menelaus trong $\Delta SAD$ ta có

$\frac{SQ}{QA}.\frac{AE}{ED}.\frac{DP}{PS}=1\iff \frac{6}{5}.\frac{AE}{ED}.\frac{1}{3}=1\iff \frac{AE}{ED}=\frac{5}{2}\Rightarrow \frac{AD}{DE}=\frac{3}{2}\Rightarrow \frac{S_{DEF}}{S_{ABC}}=\frac{2}{3}\Rightarrow \frac{S_{DEF}}{S_{ABCD}}=\frac{1}{3}$

Theo định lý Menelaus trong $\Delta SBC$ ta có

$\frac{SM}{MB}.\frac{BF}{FC}.\frac{CN}{NS}=1\iff \frac{BF}{FC}=2\Rightarrow \frac{BF}{BC}=2\Rightarrow \frac{S_{DCF}}{S_{ABC}}=1\Rightarrow \frac{S_{DCF}}{S_{ABCD}}=\frac{1}{2}$

Suy ra $\frac{S_{CDEF}}{S_{ABCD}}=\frac{5}{6}\Rightarrow \frac{V_{N.CDEF}}{V_{S.ABCD}}=\frac{NC}{SC}.\frac{S_{CDEF}}{S_{ABCD}}=\frac{5}{18}\Rightarrow V_{N.CDEF}=\frac{11}{9}$

Ta có

$\frac{V_{N.DPE}}{V_{S.ABCD}}=\frac{V_{N.DPE}}{2V_{C.SAD}}=\frac{1}{2}.\frac{SN}{SC}.\frac{DP}{DS}.\frac{DE}{AD}=\frac{1}{2}.\frac{2}{3}.\frac{1}{4}.\frac{2}{3}=\frac{1}{18}\Rightarrow V_{N.DPE}=\frac{1}{18}{V}_{S.ABCD}=\frac{11}{45}$

Vậy thể tích khối cầu cần tính là $V_{ABFEQM}=V_{ABCD.MNPQ}+V_{N.DPE}+V_{N.CDEF}=\frac{17}{5}+\frac{11}{9}+\frac{11}{45}=\frac{73}{15}.$