A(-4,-2) B(8,-2) C(2,6)

let us calculate the length of the sides:

AB = sqrt[(8--4)^2 + (-2--2)^2]= sqrt(12^2)= 12

AC= sqrt[(2--4)^2 + (6--2)^2]= sqrt(100) = 10

BC= sqrt[(2-8)^2 + (6--2)^2]= sqrt(100)= 10

We notice that ABS is an isosceles triangle.

**The perimeter (P) = AB + AC + BC= 12+ 10 + 10 = 32**

Since it is an isosceles, where AC = BC

==> let the perpendicular line from C on AB divides AB in midpoint

Let D be the midpoint,

==> AD = BD = 12/2= 6

But :

AD^2 = AC^2 - CD^2

= 100- 36 = 64

==> **The perpendicular line from C to AB (AD) = 8**

The area of the triangle = (1/2)*AD* AB

= (1/2)*8*12= 48

**The area (a) = 48**