If the perimeter of an isosceles right triangle is (18\(\sqrt {2}\) + 18) cm, then the area of the triangle is?

81

2a + \(\sqrt {2}\)a = 18 + 18\(\sqrt {2}\)

\(\sqrt {2}\)a(\(\sqrt {2}\) + 1) = 18 (\(\sqrt {2}\) + 1)

a = 9\(\sqrt {2}\)

area of triangle = \(\frac{1}{2}\) × b × h = \(\frac{1}{2}\) × 9\(\sqrt {2}\) × 9\(\sqrt {2}\) = 81 cm²