Since the triangle is isosceles, the length of each leg is $\frac{4\sqrt{2}}{\sqrt{2}}=4$ units.
The area of a triangle is equal to half the product of the lengths of two sides and the sine of the angle between them.
Since the triangle is a right triangle, the angle between the legs is $90^\circ$,
so the area of the triangle is $\frac{1}{2}(4)(4)\sin(90^\circ)=\frac{1}{2}(4)(4)(1)=\boxed{8}$ square units.
The answer is: 8
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
What is the area, in square units, of an isosceles right triangle with a hypotenuse of length $4\sqrt{2}$ units?

### Answer:
