\(S_{ABCD}=\left(AB+CD\right)x\dfrac{h}{2}\) (h là chiều cao, vuông góc AB và CD)
\(S_{ABCD}=\left(\dfrac{1}{3}xCD+CD\right)x\dfrac{h}{2}\)
\(S_{ABCD}=\dfrac{4}{3}xCDx\dfrac{h}{2}=4x\left(\dfrac{1}{3}xCDx\dfrac{h}{2}\right)\)
mà \(S_{ABD}=ABx\dfrac{h}{2}=\dfrac{1}{3}xCDx\dfrac{h}{2}\)
Nên \(S_{ABCD}=4xS_{ABD}\)
\(\Rightarrow S_{ABD}=S_{ABCD}:4=200:4=50\left(m^2\right)\)
Vậy diện tích tam giác ABD là \(50m^2\)