CÂU 5:Ta có:
\[\begin{array}{l}
{\rm{SH}} \bot ({\rm{ABC)}} = > {\rm{HC}} = {\rm{h}}{{\rm{c}}_{({\rm{ABC)}}}}{\rm{SC}} \\
= > \left[ {SC,(ABC)} \right] = SCH = 60 \\
\end{array}\]
Xét $\Delta$AHC có
${\rm{HC = }}\sqrt {{\rm{H}}{{\rm{A}}^2} + {\rm{A}}{{\rm{C}}^2} - 2{\rm{HA}}{\rm{.AC}}{\rm{.cos}}} = \sqrt {{{\left( {\frac{2}{3}{\rm{AB}}} \right)}^2} + {\rm{A}}{{\rm{C}}^2} - 2\frac{2}{3}{\rm{AB}}{\rm{.AC}}{\rm{.cos}}} = \frac{{a\sqrt 7 }}{3}$
$\Delta$ vuông SHC cho ta
${\rm{SH = HC}}\tan 60 = \frac{{a\sqrt 7 }}{3}\sqrt 3 = \frac{{a\sqrt {21} }}{3}$
${{\rm{V}}_{SABC}} = \frac{1}{3}{\rm{SH}}{\rm{.}}{{\rm{S}}_{ABC}} = \frac{1}{3}\frac{{a\sqrt {21} }}{3}\frac{{{a^2}\sqrt 3 }}{4} = \frac{{{a^3}\sqrt 7 }}{{12}}$
Dựng hình bình hành ACBD
Kẻ ${\rm{HI}} \bot {\rm{AD}}$
Ta có:
$\left\{ \begin{array}{l}
{\rm{SH}} \bot {\rm{AD}} \\
{\rm{HI}} \bot {\rm{AD}} \\
\end{array} \right. = > {\rm{AD}} \bot {\rm{(SHI) = > (SAD)}} \bot {\rm{(SHI)}}$
Kẻ ${\rm{HK}} \bot {\rm{SI}}$
Vậy
\[\begin{array}{l}
\left\{ \begin{array}{l}
{\rm{(SAD)}} \bot {\rm{(SHI)}} \\
{\rm{(SAD)}} \cap {\rm{(SHI) = SI}} \\
{\rm{HK}} \bot {\rm{SI}} \\
\end{array} \right. \\
= > {\rm{HK}} \bot {\rm{(SAD)}} \\
= > {\rm{d}}\left[ {{\rm{H,(SAD)}}} \right] = {\rm{HI}} \\
\end{array}\]
Gọi BJ là đường cao hạ từ B trong $\Delta$ABD,ta có:
$\frac{{{\rm{HI}}}}{{{\rm{BJ}}}} = \frac{{{\rm{HA}}}}{{{\rm{AB}}}} = > {\rm{HI = }}\frac{{a\sqrt 3 }}{3}$
$\Delta$ vuông SHI cho ta:
$\begin{array}{l}
\frac{1}{{H{K^2}}} = \frac{1}{{H{I^2}}} + \frac{1}{{H{S^2}}} \\
= > {\rm{HK = }}\frac{{a\sqrt {42} }}{{12}} \\
\end{array}$
Mà \[{\rm{d}}\left[ {{\rm{B,(SAD)}}} \right] = \frac{{{\rm{AB}}}}{{{\rm{AH}}}}{\rm{d}}\left[ {{\rm{H,(SAD)}}} \right] = \frac{{a\sqrt {42} }}{{8}}\]
Ta có:
$\begin{array}{l}
{\rm{BC//AD = > BC//(SAD)}} \\
{\rm{ = > d(SA,BC) = d}}\left[ {{\rm{C,(SAD)}}} \right] = {\rm{d}}\left[ {{\rm{B,(SAD)}}} \right] = \frac{{a\sqrt {42} }}{{8}} \\
\end{array}$
Bài viết đã được chỉnh sửa nội dung bởi longqnh: 04-07-2012 - 20:40