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

64 sq.cm

We know that,

If we have isosceles right angled triangle then the sides will be in the ratio of 1 : 1 : \(\sqrt {2}\)

Let x be the side of triangle, then,

the sides will be = x : x : \(\sqrt {2}\)x

Perimeter of isosceles right triangle is = (2x + \(\sqrt {2}\)x)

(2x + \(\sqrt {2}\)x)= (16\(\sqrt {2}\) + 16)

\(\sqrt {2}\)x(\(\sqrt {2}\) + 1) = 16(\(\sqrt {2}\) + 1)

x = 8\(\sqrt {2}\)

Area of triangle = \(\frac{1}{2}\) × Base × Height = \(\frac{1}{2}\) × 8\(\sqrt {2}\) × 8\(\sqrt {2}\)= 64