MHB Find the area of the red region

[anemone]

The diagram below (which is not drawn to scale) shows a parallelogram. The area of the green regions are 8 unit² , 10 unit² , 72 unit² and 79 unit² respectively. Find the area of the red region.

[TIKZ]
\coordinate (A) at (0,0);
\coordinate (B) at (8,0);
\coordinate (C) at (12,0);
\coordinate (D) at (12.75,3);
\coordinate (E) at (14,8);
\coordinate (F) at (6,8);
\coordinate (G) at (2,8);
\coordinate[label=above: \huge 79] (P) at (4.5,3);
\coordinate[label=above: \huge 72] (Q) at (9.5,6);
\coordinate[label=above: \large 8] (R) at (8.5,1.2);
\coordinate[label=above: \large 10] (S) at (11,2.9);
\draw (A) -- (C)-- (E) -- (G) -- (A);
\draw (A) -- (F);
\draw (A) -- (D);
\draw (B) -- (F);
\draw (B) -- (E);
\draw (D) -- (G);
\draw[fill=teal] (0,0) -- (7.564,1.745) -- (6.513,5.949) -- (4.983,6.644);
\draw[fill=teal] (6,8) -- (6.508,5.931) -- (10.93,3.85) -- (14,8);
\draw[fill=teal] (8,0) -- (7.564,1.76) -- (9.674,2.25);
\draw[fill=teal] (10.92,3.89) -- (12.75,3) -- (9.674,2.25);
\draw[fill=teal] (4.5,3) node[text=pink] {\huge 79};
\draw[fill=teal] (9.5,6) node[text=pink] {\huge 72};
\draw[fill=teal] (8.5,1.2) node[text=pink] {\large 8};
\draw[fill=teal] (11,2.9) node[text=pink] {\large 10};
\draw[fill=magenta] (2,8) -- (4.983,6.644) -- (6,8);
[/TIKZ]

 

[anemone]

Spoiler: Hint

Area of triangle

 

[anemone]

Spoiler: My solution

[TIKZ]
\coordinate (A) at (0,0);
\coordinate (B) at (8,0);
\coordinate (C) at (12,0);
\coordinate (D) at (12.75,3);
\coordinate (E) at (14,8);
\coordinate (F) at (6,8);
\coordinate (G) at (2,8);
\coordinate[label=above: \huge 79] (P) at (4.5,3);
\coordinate[label=above: \huge 72] (Q) at (9.5,6);
\coordinate[label=above: \large 8] (R) at (8.5,1.2);
\coordinate[label=above: \large 10] (S) at (11,2.9);
\coordinate[label=above: \huge A] (M) at (2.5,5);
\coordinate[label=above: \huge B] (N) at (6,0.48);
\coordinate[label=above: \huge C] (Z) at (5.8,6.8);
\coordinate[label=above: \huge D] (J) at (8.5,3.2);
\coordinate[label=above: \huge E] (K) at (10.5,0.9);
\coordinate[label=above: \huge F] (I) at (12.2,4.2);
\draw (A) -- (C)-- (E) -- (G) -- (A);
\draw (A) -- (F);
\draw (A) -- (D);
\draw (B) -- (F);
\draw (B) -- (E);
\draw (D) -- (G);
\draw[fill=teal] (0,0) -- (7.564,1.745) -- (6.513,5.949) -- (4.983,6.644);
\draw[fill=teal] (6,8) -- (6.508,5.931) -- (10.93,3.85) -- (14,8);
\draw[fill=teal] (8,0) -- (7.564,1.76) -- (9.674,2.25);
\draw[fill=teal] (10.92,3.89) -- (12.75,3) -- (9.674,2.25);
\draw[fill=teal] (4.5,3) node[text=pink] {\huge 79};
\draw[fill=teal] (9.5,6) node[text=pink] {\huge 72};
\draw[fill=teal] (8.5,1.2) node[text=pink] {\large 8};
\draw[fill=teal] (11,2.9) node[text=pink] {\large 10};
\draw[fill=magenta] (2,8) -- (4.983,6.644) -- (6,8);
[/TIKZ]

If we look at the parallelogram in such a way that the horizontal sides are the base, then we have

$\normalsize \text{Area of B}+\text{Area of C}+79+\text{Area of E}+\text{Area of F}+10=\text{Area of A}+\text{Area of red region}+72+8+\text{Area of D}$

If we look at the parallelogram in such a way that the slanted sides are the base, then we have

$\normalsize \text{Area of B}+\text{Area of E}+8+\text{Area of F}+\text{Area of C}+72+\text{Area of red region}=\text{Area of A}+79+10+\text{Area of D}$

Subtracting the below from the above we get

$ 9-\text{Area of red region}=-9+\text{Area of red region}\\ \\ \therefore \text{Area of red region}=9$