We need to find the length of side $AB$ of the given quadrilateral to the nearest mm. The quadrilateral has sides of length $16$ cm, $18$ cm, and $8$ cm given, as well as side $AB$.

1. Problem Description

We need to find the length of side $AB$ of the given quadrilateral to the nearest mm. The quadrilateral has sides of length $16$ cm, $18$ cm, and $8$ cm given, as well as side $AB$.

2. Solution Steps

First, we will construct a right triangle. Draw a vertical line from $B$ to the horizontal line from $A$, and label the intersection $C$.

Then $AC = 18 - x$, where $x$ is the length of the bottom portion. We have $16 - 8 = 8$ which is the height difference. Thus we have $BC = 16 - 8 = 8$.

Now consider the triangle $BC$ and call the bottom right corner $C$.

We are given the bottom line is $18$. And we want to subtract $AC$ from it to get $x$.

We can create a right triangle with hypotenuse $8$.

So $8^2 + x^2 = 8^2$. Thus $x^2 = 8^2 - 8^2 = 0$.

Now draw a perpendicular from $B$ to $AD$. Let $E$ be where they meet.

$BE = 18$ and $AE = 16-8=8$.

So $AB = \sqrt{8^2 + 18^2}$.

$AB = \sqrt{64 + 324} = \sqrt{388}$.

$\sqrt{388} \approx 19.6977$.

To the nearest cm, we have $19.7$ cm.

$19.6977 \text{ cm} = 196.977 \text{ mm}$. To the nearest mm, we have $197 \text{ mm}$.

3. Final Answer

197 mm