1: \(S_{ABCD}=\dfrac{1}{2}\cdot AH\cdot\left(AB+CD\right)\)
=>\(\left(AB+3AB\right)\cdot\dfrac{1}{2}\cdot3=30\)
=>4AB=20
=>AB=5(m)
CD=3*AB=15(m)
2:
Xét ΔEAB có AB//CD
nên \(\dfrac{EA}{ED}=\dfrac{AB}{CD}\)
=>\(\dfrac{EA}{ED}=\dfrac{1}{3}\)
Xét ΔEAB và ΔEDC có
\(\widehat{E}\) chung
\(\dfrac{EA}{ED}=\dfrac{EB}{EC}\)
Do đó: ΔEAB đồng dạng với ΔEDC
=>\(\dfrac{S_{EAB}}{S_{EDC}}=\left(\dfrac{AB}{DC}\right)^2=\dfrac{1}{9}\)
=>\(\dfrac{S_{EAB}}{S_{ABCD}}=\dfrac{1}{8}\)
=>\(S_{EAB}=\dfrac{30}{8}=3,75\left(m^2\right)\)