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