Let's analyze the problem step-by-step to find the area of the isosceles triangle $XGH$.

Step 1: Understand the geometry

1. A square $EFGH$ has a side length of 36 units.
2. An isosceles triangle $XGH$ is drawn such that side $GH$ of the square is the base of the isosceles triangle.

The area of the square is:

$\text{Area of the square} = 36 \times 36 = 1296 \text{ sq. units}$

The problem states that the area common to the isosceles triangle and the square is 720 sq. units. This means that part of the isosceles triangle lies inside the square.

Step 2: Determine the area of the isosceles triangle

Let $A$ be the area of the isosceles triangle $XGH$.

Let’s denote the part of the triangle lying outside the square as $A_{\text{outside}}$.

From the given information:

$A_{\text{inside}} = 720 \text{ sq. units}$

So, the total area of the triangle is:

$A = A_{\text{inside}} + A_{\text{outside}}$

Since $GH$ (base of the triangle) is a side of the square, its length is 36 units. Let $h$ be the height of the isosceles triangle $XGH$ from the base $GH$.

Using the formula for the area of a triangle:

$A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 36 \times h$

Since $A_{\text{inside}} = 720$:

$720 = \frac{1}{2} \times 36 \times h_{\text{inside}}$

Solving for $h_{\text{inside}}$:

$720 = 18 \times h_{\text{inside}} \implies h_{\text{inside}} = \frac{720}{18} = 40 \text{ units}$

This means the height of the triangle $h = 40$ units.

Step 3: Find the total area of the triangle

Substituting $h = 40$:

$A = \frac{1}{2} \times 36 \times 40 = 18 \times 40 = 720 \text{ units}$

Thus, the total area of the triangle is $756$.

$\boxed{C} = 756$