Çözüldü Isosceles Right Triangle's Circumcircle - Radical Numbers

A slightly hardened test adaptation from the Manitoba Mathematical (High School) Contest - Canada, 2003 Questions:

If the perimeter of an isosceles right triangle is 8, what is the ratio, as a multiple of π, of the area of its circumcircle to that of the triangle?
A) 5
B) 4
C) 3
D) 2
E) 1

Length of the vertical sides of the isosceles right triangle: a unit
Length of the hypotenuse of the isosceles right triangle: a·√2 unit
2a + a·√2 = 8 ⇒ a = 4·(2 - √2) units
Area of The Isosceles Right Triangle: S1 = [ 4·(2 - √2) ]^2 / 2 = 16·(3 - 2·√2) unit^2.
Length of the radius of the circumcircle of the isosceles right triangle: r unit
r = (a / 2)·√2 = [ 4·(2 - √2) / 2 ]·√2 = 4·(√2 - 1) units
Area of the circumcircle: S2 = π·[ 4·(√2 - 1) ]^2 unit^2
S2 / S1 = π·{ [ 4·(√2 - 1) ]^2 } / [ 16·(3 - 2·√2) ] = π·1.

Original Question: "If the perimeter of an isosceles right-angled triangle is 8, what is its area?"
https://server.math.umanitoba.ca/~craigen/manitobamathletics/mmc/ARCHIVE/QUESTIONS2003.pdf
(Page 1, Question 3.b)