If the perimeter of an isoscelesright triangle is $$(16\sqrt{2} + 16)$$ cm, then the area of the triangle is:

The perimeter of an isosceles right triangle  is given below

from the above using Pythagoras theorem

     $$b^2 = a^2 + a^2 $$

     $$\Rightarrow b^2 = 2a^2 $$

$$\Rightarrow b = \sqrt {2} a $$

sum of all sides = $$ 2a + b $$

$$\Rightarrow 2a + \sqrt {2} a $$ (put the balue)

$$\Rightarrow a (2 + \sqrt {2}) $$

then from the  question

  $$ a (2+\sqrt {2}) = 16 \sqrt{2} + 16  $$

$$\Rightarrow a \sqrt{2} (\sqrt{2} +1) = 16 (\sqrt {2} +1) $$

$$\Rightarrow a = \frac {16}{\sqrt{2}} $$

then area of isoscaletriangle = $$ \frac{1}{2} Base \times height $$

$$\Rightarrow  \frac {1} {2} a \times a $$

$$\Rightarrow \frac {1}{2} (\frac{16} {\sqrt {2}})^2$$

$$\Rightarrow \frac{1} {2} \frac{16\times 16} {2} $$

$$\Rightarrow 64 cm^2 $$ Ans