If you take only quadrilaterals with two common adjacent sides, you have 5x10 = 50 quadrilaterals. (5 possible quadrilaterals for each of the 10 pair of adjacent sides)
Now, taking non adjacent sides, it will be a problem similar to:
How many different pair of letters you can make from a through j, without repeat and order is not important, ie ab and ba are the same, provided two adjacent letters cannot be paired, including a and j.
You have the sides:
da (but you already have ad!, so this is rejected. I won't put other repeats)
For a total of (24 + 50) possible quadrilaterals.