area of a triangle

$A_{ABC}=\dfrac{1}{2}(2y)(5x)(\sin~B)$ $\implies$ $yx \sin~B=\dfrac{126}{5}$

$A_1=\dfrac{1}{2}(y)(2x)( \sin~B)=\dfrac{126}{5}$

$A_{ABC}=\dfrac{1}{2}(2y)(14z)(\sin~A)$ $\implies$ $yz \sin~A=9$

$A_2=\dfrac{1}{2}(y)(4z)(\sin~A)=2(9)=18$

$A_{ABC}=\dfrac{1}{2}(5x)(14z)(\sin~C)$ $\implies$ $xz \sin~C=\dfrac{18}{5}$

$A_3=\dfrac{1}{2}(3x)(10z)(\sin~C) = 15\left(\dfrac{18}{5}\right)=54$

The area of $\triangle EAD$ is $126-\dfrac{126}{5}-18-54=\dfrac{144}{5}=$ $\color{#D61F06}\boxed{28.8}$