Can you figure this out (2)

Denote the red area as a1, the others as shown in the diagram. a1+a2+72+a4+8= Half of Quadrilateral ABCD. Also, a2+79+a4+10 = Half of Quadrilateral ABCD. In other words, a1+a2+72+a4+8 =a2+79+a4+10 ► R =89-80 =9 s.u.

Let $x$ be the Area of the $\color{#D61F06}{Red}$ Region. Area of parallelogram $ABCD = Ar(ABCD).$

Notice that triangles containing $(72+b + 8)$ and $(x+ a)$ areas have a total area $A = \dfrac 12 (BE + AE) \times ($ height of $\color{#20A900}{Green}$ segment $) = \dfrac {Ar(ABCD)}{2}$

Similarly, traingle with $(a+ 79 + b+ 10)$ have area $= \dfrac 12 (AD)\times ($ height of $\color{#3D99F6}{Blue}$ segment $) = \dfrac {Ar(ABCD)}{2} = A$

Hence, $A = 72 + b + 8 + x + a = a + 79 + b + 10 \Rightarrow \boxed{x = 9}$