Google
Câu hỏi: Cho hình chóp $S.ABCD$ có đáy $ABCD$ là hình thang vuông tại $A$ và $D$ ; $AB=2a, AD=DC=a$, cạnh bên $SA=a\sqrt{2}$ và vuông góc vớI $\left( ABCD \right)$. Khoảng cách từ điểm $C$ đến mặt phẳng $\left( SBD \right)$ bằng?
A. $\dfrac{a\left( 5\sqrt{2}-2\sqrt{5} \right)}{5}$.
B. $\dfrac{a\sqrt{7}}{7}$.
C. $\dfrac{a\sqrt{5}}{5}$.
D. $\dfrac{2a\sqrt{7}}{7}$.
Ta có ${{S}_{\Delta BCD}}=\dfrac{1}{2}{{S}_{\Delta ABD}}\Rightarrow {{V}_{S.ABCD}}=3{{V}_{SBCD}}\Rightarrow d\left( C;\left( SBD \right) \right)=\dfrac{{{V}_{S.ABCD}}}{{{S}_{\Delta SBD}}}$.
Mà ${{V}_{S.ABCD}}=\dfrac{1}{3}a\sqrt{2}\dfrac{\left( a+2a \right)a}{2}=\dfrac{\sqrt{2}{{a}^{3}}}{2}$
$SD=\sqrt{{{a}^{2}}+{{\left( a\sqrt{2} \right)}^{2}}}=a\sqrt{3};SB=\sqrt{{{\left( 2a \right)}^{2}}+{{\left( a\sqrt{2} \right)}^{2}}}=a\sqrt{6};BD=\sqrt{{{a}^{2}}+{{\left( 2a \right)}^{2}}}=a\sqrt{5}$
Suy ra, $cos\widehat{BSD}=\dfrac{S{{B}^{2}}+S{{D}^{2}}-B{{D}^{2}}}{2.SB.SD}=\dfrac{2}{3\sqrt{2}}\Rightarrow \sin \widehat{BSD}=\dfrac{\sqrt{7}}{3}.$
Từ đó, ${{S}_{\Delta SBD}}=\dfrac{1}{2}SB.SD.\sin \widehat{BSD}=\dfrac{{{a}^{2}}\sqrt{14}}{2}$.
Vậy $d\left( B;\left( SBD \right) \right)=\dfrac{\sqrt{2}{{a}^{3}}}{2}.\dfrac{2}{{{a}^{2}}\sqrt{14}}=\dfrac{a\sqrt{7}}{7}.$
A. $\dfrac{a\left( 5\sqrt{2}-2\sqrt{5} \right)}{5}$.
B. $\dfrac{a\sqrt{7}}{7}$.
C. $\dfrac{a\sqrt{5}}{5}$.
D. $\dfrac{2a\sqrt{7}}{7}$.
Mà ${{V}_{S.ABCD}}=\dfrac{1}{3}a\sqrt{2}\dfrac{\left( a+2a \right)a}{2}=\dfrac{\sqrt{2}{{a}^{3}}}{2}$
$SD=\sqrt{{{a}^{2}}+{{\left( a\sqrt{2} \right)}^{2}}}=a\sqrt{3};SB=\sqrt{{{\left( 2a \right)}^{2}}+{{\left( a\sqrt{2} \right)}^{2}}}=a\sqrt{6};BD=\sqrt{{{a}^{2}}+{{\left( 2a \right)}^{2}}}=a\sqrt{5}$
Suy ra, $cos\widehat{BSD}=\dfrac{S{{B}^{2}}+S{{D}^{2}}-B{{D}^{2}}}{2.SB.SD}=\dfrac{2}{3\sqrt{2}}\Rightarrow \sin \widehat{BSD}=\dfrac{\sqrt{7}}{3}.$
Từ đó, ${{S}_{\Delta SBD}}=\dfrac{1}{2}SB.SD.\sin \widehat{BSD}=\dfrac{{{a}^{2}}\sqrt{14}}{2}$.
Vậy $d\left( B;\left( SBD \right) \right)=\dfrac{\sqrt{2}{{a}^{3}}}{2}.\dfrac{2}{{{a}^{2}}\sqrt{14}}=\dfrac{a\sqrt{7}}{7}.$
Đáp án B.