Let's denote the cyclic quadrilateral as ABCD, where AB = 2, BC = 5, and angle BAC = 60 degrees.

We can start by drawing the cyclic quadrilateral like this:
```
A-----B
| |
| |
D-----C
```

Since ABCD is a cyclic quadrilateral, the opposite angles sum up to 180 degrees. Therefore, angle BCD = 180 - angle BAC = 180 - 60 = 120 degrees.

Now, we can use the Law of Cosines to find the length of CD:
CD^2 = BC^2 + BD^2 - 2 * BC * BD * cos(angle BCD)
CD^2 = 5^2 + BD^2 - 2 * 5 * BD * cos(120)
CD^2 = 25 + BD^2 + 10BD * (-1/2)
CD^2 = 25 + BD^2 - 5BD
CD^2 - BD^2 + 5BD - 25 = 0

Since ABCD is a cyclic quadrilateral, opposite sides are equal. Therefore, AD = BC = 5.

We can use the Law of Cosines again to find the length of BD:
BD^2 = AD^2 + AB^2 - 2 * AD * AB * cos(angle BAD)
BD^2 = 5^2 + 2^2 - 2 * 5 * 2 * cos(60)
BD^2 = 25 + 4 - 20 * (1/2)
BD^2 = 29 - 10
BD^2 = 19

Substituting BD^2 = 19 into the previous equation:
CD^2 - 19 + 5BD - 25 = 0
CD^2 + 5BD - 44 = 0

We can factorize this quadratic equation:
(CD + 11)(CD - 4) = 0

Since the area of the quadrilateral is 4, we know that the area of triangle ABC is half of that, which is 2. Using the formula for the area of a triangle:
Area = (1/2) * AB * AC * sin(angle BAC)
2 = (1/2) * 2 * AC * sin(60)
2 = AC * sqrt(3)/2
AC = 4/sqrt(3)

Now we can solve for the possible values of CD:
1) CD + 11 = 0
CD = -11 (not possible since lengths cannot be negative)
2) CD - 4 = 0
CD = 4

Since CD represents the length of a side, it cannot be negative. Therefore, CD = 4.

Now, we can find the length of BD:
BD^2 = 19
BD = sqrt(19)

Finally, we can find the area of the cyclic quadrilateral using the formula:
Area = (1/2) * BD * AC * sin(angle BAC)
Area = (1/2) * sqrt(19) * (4/sqrt(3)) * sin(60)
Area = (1/2) * sqrt(19) * (4/2) * (sqrt(3)/2)
Area = sqrt(19) * sqrt(3)
Area = sqrt(57)