Rectangle $ABCD$ has perimeter $178$ and area $1848$. What is the length of the diagonal of the rectangle?

[i]2016 CCA Math Bonanza Individual Round #2[/i]
1. Let the length and width of the rectangle be \(a\) and \(b\) respectively. Given the perimeter of the rectangle is 178, we have:
   \[
   2a + 2b = 178 \implies a + b = 89
   \]

2. Given the area of the rectangle is 1848, we have:
   \[
   ab = 1848
   \]

3. We need to find the values of \(a\) and \(b\). These values are the roots of the quadratic equation formed by the sum and product of the roots:
   \[
   x^2 - (a+b)x + ab = 0
   \]
   Substituting \(a + b = 89\) and \(ab = 1848\), we get:
   \[
   x^2 - 89x + 1848 = 0
   \]

4. Solving the quadratic equation \(x^2 - 89x + 1848 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   x = \frac{89 \pm \sqrt{89^2 - 4 \cdot 1848}}{2}
   \]
   \[
   x = \frac{89 \pm \sqrt{7921 - 7392}}{2}
   \]
   \[
   x = \frac{89 \pm \sqrt{529}}{2}
   \]
   \[
   x = \frac{89 \pm 23}{2}
   \]
   Thus, the solutions are:
   \[
   x = \frac{89 + 23}{2} = 56 \quad \text{and} \quad x = \frac{89 - 23}{2} = 33
   \]
   Therefore, \(a = 56\) and \(b = 33\) (or vice versa).

5. The length of the diagonal \(d\) of the rectangle can be found using the Pythagorean theorem:
   \[
   d = \sqrt{a^2 + b^2}
   \]
   Substituting \(a = 56\) and \(b = 33\):
   \[
   d = \sqrt{56^2 + 33^2}
   \]
   \[
   d = \sqrt{3136 + 1089}
   \]
   \[
   d = \sqrt{4225}
   \]
   \[
   d = 65
   \]

The final answer is \(\boxed{65}\).