A geometry problem by Syed Hamza Khalid

Fast answer : $\large \color{#69047E} 79 + 10 - 72 - 8$

The key is to find a triangle or a set of triangles which equal to half of the parallelogram's area.

We will use the following principle:

We will be using the same technique in our problem.

First, let us label the diagram and then identify this:

Then we apply the principle:

Now we can form an equation by making their areas equal:

$\large \color{#3D99F6} x + a + 72 + b + 8 = a + 79 + b + 10 \\ \\ \large \color{#3D99F6} \therefore x = 79 + 10 - 72 - 8 \\ \large \color{#3D99F6}= 9$

Assign the unknown areas as $a$ , $b$ , $c$ to $g$ as shown in the figure. Let the area of parallelogram $ABCD$ be $A$ . We need to find $a$ . We note that:

$\begin{cases} a + b + 72 +c + f + 8 + g = a+b+c+f+g+80 = \frac 12 A & ...(1) \\ b + 79 + f + g + 10 + c = b+c+f+g + 89 = \frac 12A & ...(2) \end{cases}$

Note that $(1) = (2) = \frac 12 A$ :

$\begin{aligned} \implies a+b+c+f+g+80 & = b+c+f+g + 89 \\ \implies a & = \boxed{9}\end{aligned}$

The solution can be found here on Youtube channel "Mind Your Decisions"