Ta có MC = \(\dfrac{CD}{2}=\dfrac{30}{2}=15\) (m) (do M trung điểm CD)
Lại có SBCM = \(\dfrac{1}{2}\).MC.BC
Hay 102 = \(\dfrac{1}{2}\). 15.BC
Nên BC = 13,6 (m)
⇒ AD = BC = 13,6 (m) (T/C hcn)
Vậy SABCD = \(\dfrac{1}{2}.\left(AB+DM\right).AD=\dfrac{1}{2}.\left(30+15\right).13,6=306\left(m\right)\)