Question

Giải giúp mình với ạ

Answer 1

\(AC=\sqrt{AB^2+BC^2}=2a\)

\(SA\perp\left(ABCD\right)\Rightarrow\widehat{SCA}\) là góc giữa SC và (ABCD) \(\Rightarrow\widehat{SCA}=45^0\)

\(\Rightarrow SA=SC.tan45^0=2a\)

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\AB\perp BC\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)

Từ A kẻ H vuông góc SB \(\Rightarrow AH\perp\left(SBC\right)\Rightarrow AH=d\left(A;\left(SBC\right)\right)\)

\(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}\Rightarrow AH=\dfrac{SA.AB}{\sqrt{SA^2+AB^2}}=\dfrac{2a}{\sqrt{5}}\)

b. Từ A kẻ AI vuông góc BD, trong mp (SAI) kẻ AJ vuông góc SI

\(\Rightarrow AJ\perp\left(SBD\right)\Rightarrow AJ=d\left(A;\left(SBD\right)\right)\)

\(\dfrac{1}{AI^2}=\dfrac{1}{AD^2}+\dfrac{1}{AD^2}=\dfrac{1}{a^2}+\dfrac{1}{3a^2}=\dfrac{4}{3a^2}\)

\(\dfrac{1}{AJ^2}=\dfrac{1}{SA^2}+\dfrac{1}{AI^2}=\dfrac{1}{4a^2}+\dfrac{4}{3a^2}=\dfrac{19}{12a^2}\Rightarrow AJ=a\sqrt{\dfrac{12}{19}}\)

c. Gọi N là trung điểm AD 

\(DM=\sqrt{DC^2+\left(\dfrac{BC}{2}\right)^2}=\dfrac{a\sqrt{7}}{2}\Rightarrow sin\widehat{ADM}=\dfrac{MN}{DM}=\dfrac{AB}{DM}=\dfrac{2}{\sqrt{7}}\)

Kẻ \(AP\perp DM\) , kẻ \(AQ\perp SP\Rightarrow AQ\perp\left(SDM\right)\Rightarrow AQ=d\left(A;\left(SDM\right)\right)\)

\(AP=AD.sin\widehat{ADM}=\dfrac{2a\sqrt{21}}{7}\)

\(\dfrac{1}{AQ^2}=\dfrac{1}{AP^2}+\dfrac{1}{SA^2}=\dfrac{5}{6a^2}\Rightarrow AQ=a\sqrt{\dfrac{6}{5}}\)