In an isosceles triangle ABC with AB = AC and AD is perpendicular to BC, if AD = 6 cm and the perimeter of $$\triangle$$ABC is 36 cm, then the area of $$\triangle$$ABC is:

Given, AB = AC

Let AB = AC = p

Perimeter of $$\triangle$$ABC = 36 cm

$$\Rightarrow$$  AB + AC + BC = 36

$$\Rightarrow$$  p + p + BC = 36

$$\Rightarrow$$  BC = 36 - 2p

Since AB = AC and AD is perpendicular to BC

AD will be the perpendicular bisector which bisects BC

$$\Rightarrow$$  BD = CD = $$\frac{36-2p}{2}$$ = 18 - p

In $$\triangle$$ADB,

AB$$^2$$ = BD$$^2$$ + AD$$^2$$

$$\Rightarrow$$  p$$^2$$ = (18 - p)$$^2$$ + 6$$^2$$

$$\Rightarrow$$  p$$^2$$ = 324 + p$$^2$$ - 36p + 36

$$\Rightarrow$$  36p = 360

$$\Rightarrow$$  p = 10

BC = 36 - 2p = 36 - 20 = 16

$$\therefore\ $$Area of the triangle = $$\frac{1}{2}$$ x AD x BC = $$\frac{1}{2}$$ x 6 x 16 = 48 $$cm^2$$

Hence, the correct answer is Option D