Question

Answer 1

\(S_{ABD}=\dfrac{1}{2}\cdot AB\cdot AD\cdot sinBAD=\dfrac{1}{2}\cdot5\cdot5\cdot sin120=\dfrac{25}{2}\cdot sin120\)

\(S_{BCD}=\dfrac{1}{2}\cdot CB\cdot CD\cdot sinBCD\)

\(=\dfrac{1}{2}\cdot5\cdot5\cdot sin120=\dfrac{25}{2}\cdot sin120\)

\(S_{ABCD}=S_{ABD}+S_{CBD}=\dfrac{25}{2}\cdot sin120+\dfrac{25}{2}\cdot sin120\)

\(=25\cdot sin120=\dfrac{25\sqrt{3}}{2}\left(cm^2\right)\)