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Câu hỏi: . Cho khối chóp S.ABCD có đáy ABCD là tứ giác lồi, tam giác ABD đều cạnh a, tam giác BCD cân tại $C,\widehat{BCD}={{120}^{0}},SA\bot \left( ABCD \right),SA=a.$ Mặt phẳng (P) đi qua A và vuông góc với SC cắt các cạnh $SB,SC,SD$ lần lượt tại $M,N,P.$ Tính thể tích khối chóp $S.AMNP$
A. $\dfrac{{{a}^{3}}\sqrt{3}}{12}.$
B. $\dfrac{{{a}^{3}}\sqrt{3}}{42}.$
C. $\dfrac{2{{a}^{3}}\sqrt{3}}{21}.$
D. $\dfrac{{{a}^{3}}\sqrt{3}}{14}.$
Ta có: $\widehat{AB\text{D}}=\widehat{A\text{D}B}=60{}^\circ ,\widehat{CB\text{D}}=\widehat{C\text{D}B}=30{}^\circ $
Suy ra $\widehat{ABC}=\widehat{A\text{D}C}=90{}^\circ $.
Suy ra $BC\bot AB$, mà $BC\bot \text{S}A\Rightarrow CB\bot \left( SAB \right)$
Dựng $AM\bot \text{S}B$, ta có $AM\bot BC\Rightarrow AM\bot \text{S}C$.
Tương tự ta có $AP\bot \text{SD}$.
Dựng $AN\bot \text{S}C$ theo tính chất đối xứng thì
$\dfrac{{{V}_{S.AMNP}}}{{{V}_{S.ABC\text{D}}}}=\dfrac{{{V}_{S.AMN}}}{{{V}_{S.ABC}}}=\dfrac{SM}{SB}.\dfrac{SN}{SC}$
Mặt khác $SA=SM.SB\Rightarrow \dfrac{SM}{SB}=\dfrac{S{{A}^{2}}}{S{{B}^{2}}}=\dfrac{1}{2}$
Tương tự ta có $\dfrac{SN}{SC}=\dfrac{S{{A}^{2}}}{S{{C}^{2}}}=\dfrac{1}{1+A{{C}^{2}}}$
Trong đó $AI=\dfrac{a\sqrt{3}}{2},CI=IB\tan 30{}^\circ =\dfrac{a\sqrt{3}}{6}\Rightarrow AC=\dfrac{2}{3}a\sqrt{3}\Rightarrow \dfrac{SN}{SC}=\dfrac{3}{7}$
Suy ra $\dfrac{{{V}_{S.AMNP}}}{{{V}_{S.ABC\text{D}}}}=\dfrac{1}{2}.\dfrac{3}{7}=\dfrac{3}{14},{{S}_{ABC\text{D}}}=\dfrac{1}{2}AC.B\text{D}=\dfrac{{{a}^{2}}\sqrt{3}}{3}$
$\Rightarrow {{V}_{S.AMNP}}=\dfrac{3}{14}{{V}_{S.ABC\text{D}}}=\dfrac{3}{14}.\dfrac{1}{3}.SA.\dfrac{{{a}^{2}}\sqrt{3}}{3}=\dfrac{{{a}^{2}}\sqrt{3}}{42}$.
A. $\dfrac{{{a}^{3}}\sqrt{3}}{12}.$
B. $\dfrac{{{a}^{3}}\sqrt{3}}{42}.$
C. $\dfrac{2{{a}^{3}}\sqrt{3}}{21}.$
D. $\dfrac{{{a}^{3}}\sqrt{3}}{14}.$
Ta có: $\widehat{AB\text{D}}=\widehat{A\text{D}B}=60{}^\circ ,\widehat{CB\text{D}}=\widehat{C\text{D}B}=30{}^\circ $
Suy ra $\widehat{ABC}=\widehat{A\text{D}C}=90{}^\circ $.
Suy ra $BC\bot AB$, mà $BC\bot \text{S}A\Rightarrow CB\bot \left( SAB \right)$
Dựng $AM\bot \text{S}B$, ta có $AM\bot BC\Rightarrow AM\bot \text{S}C$.
Tương tự ta có $AP\bot \text{SD}$.
Dựng $AN\bot \text{S}C$ theo tính chất đối xứng thì
$\dfrac{{{V}_{S.AMNP}}}{{{V}_{S.ABC\text{D}}}}=\dfrac{{{V}_{S.AMN}}}{{{V}_{S.ABC}}}=\dfrac{SM}{SB}.\dfrac{SN}{SC}$
Mặt khác $SA=SM.SB\Rightarrow \dfrac{SM}{SB}=\dfrac{S{{A}^{2}}}{S{{B}^{2}}}=\dfrac{1}{2}$
Tương tự ta có $\dfrac{SN}{SC}=\dfrac{S{{A}^{2}}}{S{{C}^{2}}}=\dfrac{1}{1+A{{C}^{2}}}$
Trong đó $AI=\dfrac{a\sqrt{3}}{2},CI=IB\tan 30{}^\circ =\dfrac{a\sqrt{3}}{6}\Rightarrow AC=\dfrac{2}{3}a\sqrt{3}\Rightarrow \dfrac{SN}{SC}=\dfrac{3}{7}$
Suy ra $\dfrac{{{V}_{S.AMNP}}}{{{V}_{S.ABC\text{D}}}}=\dfrac{1}{2}.\dfrac{3}{7}=\dfrac{3}{14},{{S}_{ABC\text{D}}}=\dfrac{1}{2}AC.B\text{D}=\dfrac{{{a}^{2}}\sqrt{3}}{3}$
$\Rightarrow {{V}_{S.AMNP}}=\dfrac{3}{14}{{V}_{S.ABC\text{D}}}=\dfrac{3}{14}.\dfrac{1}{3}.SA.\dfrac{{{a}^{2}}\sqrt{3}}{3}=\dfrac{{{a}^{2}}\sqrt{3}}{42}$.
Đáp án B.