\(a,\) Ta có \(BH=HC=AE=EB=\dfrac{1}{2}AB=\dfrac{1}{2}BC=\dfrac{1}{2}\cdot10=5\left(cm\right)\)
\(S_{BHDA}=S_{ABCD}-S_{CHD}=AD^2-\dfrac{1}{2}CD\cdot CH\\ =100-\dfrac{1}{2}\cdot10\cdot5=75\left(cm^2\right)\)
\(b,S_{AHD}=S_{BHDA}-S_{AHB}=75-\dfrac{1}{2}\cdot10\cdot5=50\left(cm^2\right)\\ S_{AHE}=S_{AHB}-S_{HBE}=25-\dfrac{1}{2}\cdot5\cdot5=\dfrac{25}{2}\left(cm^2\right)\\ \Rightarrow S_{AHD}>S_{AHE}\)