Đáp án:
${S_{ABCD}} = \dfrac{{1050}}{{\sqrt 3 }}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {ABC} = {60^0}\\
\widehat {ACB} = {90^0}
\end{array} \right. \Rightarrow \widehat {CAB} = {30^0}\\
\Rightarrow \widehat {ACD} = \widehat {CAB} = {30^0}\\
\Rightarrow \left\{ \begin{array}{l}
AD = CD.\tan \widehat {ACD} = 30.\tan {30^0} = \dfrac{{30}}{{\sqrt 3 }}\\
AC = \dfrac{{CD}}{{\cos \widehat {ACD}}} = \dfrac{{30}}{{\cos {{30}^0}}} = \dfrac{{30}}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{{60}}{{\sqrt 3 }}
\end{array} \right.\\
\Rightarrow AB = \dfrac{{AC}}{{\cos \widehat {CAB}}} = \dfrac{{\dfrac{{60}}{{\sqrt 3 }}}}{{\cos {{30}^0}}} = \dfrac{{\dfrac{{60}}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 }}{2}}} = 40\\
\Rightarrow {S_{ABCD}} = \dfrac{1}{2}AD.\left( {AB + CD} \right) = \dfrac{1}{2}.\dfrac{{30}}{{\sqrt 3 }}.\left( {40 + 30} \right) = \dfrac{{1050}}{{\sqrt 3 }}
\end{array}$
Vậy ${S_{ABCD}} = \dfrac{{1050}}{{\sqrt 3 }}$
