Lời giải chi tiết:
Ta có: (left{ begin{array}{l}left( {SAB} right) bot left( {ABCD} right) = ABSM subset left( {SAB} right),,,SM bot ABend{array} right.)( Rightarrow SM bot left( {ABCD} right)).
Trong (ABCD) kẻ (MI bot ND,,left( {I in ND} right)), trong (SMI) kẻ (MH bot SI).
Ta có: (left{ begin{array}{l}ND bot MIND bot SMend{array} right. Rightarrow ND bot left( {SMI} right) Rightarrow ND bot MH).
(left{ begin{array}{l}MH bot NDMH bot SIend{array} right. Rightarrow MH bot left( {SND} right)) ( Rightarrow dleft( {M;left( {SND} right)} right) = MH = dfrac{{3asqrt 2 }}{4}).
Ta có:
(begin{array}{l}{S_{BMN}} = dfrac{1}{2}BM.BN = dfrac{1}{2}.a.a = dfrac{{{a^2}}}{2}{S_{AMD}} = dfrac{1}{2}AM.AD = dfrac{1}{2}.a.2a = {a^2}{S_{CND}} = dfrac{1}{2}CN.CD = dfrac{1}{2}.a.2a = {a^2}{S_{ABCD}} = A{B^2} = 4{a^2} Rightarrow {S_{MND}} = {S_{ABCD}} – {S_{BMN}} – {S_{AMD}} – {S_{CND}} Rightarrow {S_{MND}} = 4{a^2} – dfrac{{{a^2}}}{2} – {a^2} – {a^2} = dfrac{{3{a^2}}}{2}end{array})
Lại có (ND = sqrt {C{N^2} + C{D^2}} = sqrt {{a^2} + 4{a^2}} = asqrt 5 ), ({S_{MND}} = dfrac{1}{2}MI.ND).
( Rightarrow MI = dfrac{{2{S_{MND}}}}{{ND}} = dfrac{{2.dfrac{{3{a^2}}}{2}}}{{asqrt 5 }} = dfrac{{3asqrt 5 }}{5}).
Áp dụng hệ thức lượng trong tam giác vuông SMI có:
(begin{array}{l}dfrac{1}{{M{H^2}}} = dfrac{1}{{S{M^2}}} + dfrac{1}{{M{N^2}}}dfrac{1}{{{{left( {dfrac{{3asqrt 2 }}{4}} right)}^2}}} = dfrac{1}{{S{M^2}}} + dfrac{1}{{{{left( {dfrac{{3asqrt 5 }}{5}} right)}^2}}} Rightarrow dfrac{1}{{S{M^2}}} = dfrac{1}{{3{a^2}}} Rightarrow SM = asqrt 3 end{array})
Gọi P là trung điểm của CD, ta có (SG cap left( {ABCD} right) = P) ( Rightarrow dfrac{{dleft( {G;left( {ABCD} right)} right)}}{{dleft( {S;left( {ABCD} right)} right)}} = dfrac{{GP}}{{SP}} = dfrac{1}{3})
( Rightarrow dleft( {G;left( {ABCD} right)} right) = dfrac{1}{3}SM = dfrac{{asqrt 3 }}{3}).
(begin{array}{l}{S_{AMND}} = {S_{ABCD}} – {S_{BMN}} – {S_{CND}}{S_{AMND}} = 4{a^2} – dfrac{1}{2}{a^2} – {a^2} = dfrac{{5{a^2}}}{2}end{array})
Vậy ({V_{G.AMND}} = dfrac{1}{3}dleft( {G;left( {ABCD} right)} right).{S_{AMND}}) ( = dfrac{1}{3}.dfrac{{asqrt 3 }}{3}.dfrac{{5{a^2}}}{2} = dfrac{{5sqrt 3 {a^3}}}{{18}}).
Chọn D.
