Solving for the perimeter?

In ΔABD:
AD = x + 6
BD = 4 + [(1/2)x + 2] = (x/2) + 6
AB = (3/2)x + 6 = (3x/2) + 6

AD² + BD² = AB²   (Pythagorean theorem)
[x + 6]² + [(x/2)² + 6]² = [(3x/2)² + 6]²
[x² + 12x + 36] + [(x²/4) + 6x + 36] = (9x²/4) + 18x + 36
x² = 36
x = 6 or x = -6 (rejected)

AD = (6) + 6 = 12
DC = (1/2)(6) + 2 = 5
AB = [3(6)/2] + 6 = 15

In ΔABC:
AC² = AD² + DC²   (Pythagorean theorem)
AC² = 12² + 5²
AC = 13

Perimeter of ΔABC
= AB + BC + AC
= 15 + 4 + 13
= 32 (units)

Perimeter P will be (3/2)x + 6 + 4 + AC
AC^2 = (x + 6)^2 + (x/2 + 2)^2 = (5x^2 + 56x + 160)/4
P = (3/2)x + 10 + √[5x^2 + 56x + 160]/2

You can't use the Pythagorean Theorem on triangle ABC since it's not a right triangle.

But you can use it with triangle ACD since you are given expressions for CD and AD with AC being its hypotenuse.

a² + b² = c²
(x/2 + 2)² + (x + 6)² = c²
x²/4 + 2x + 4 + x² + 12x + 36 = c²
5x²/4 + 14x + 40 = c²

Let's factor a 1/4 from the left side:

(1/4)(5x² + 56x + 160) = c²

Square root of both sides:

(1/2)√(5x² + 56x + 160) = c

That's the length of the hypotenuse, in terms of x.  That is then the third length of triangle ABC.  Add that to the lengths of AB and BC to get the perimeter:

(3/2)x + 6 + 4 + (1/2)√(5x² + 56x + 160)
(3/2)x + 10 + (1/2)√(5x² + 56x + 160)

That's the expression for the perimeter of ABC.