MHB What is the Minimal Area of a Right-Angled Triangle with an Inradius of 1 Unit?

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What is the minimal area of a right-angled triangle whose inradius is 1 unit?
 
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The area of a triangle is \(A = sr\), where \(r\) is the inradius (\(r=1\)) and \(s = (a + b + c)/2\) is the semiperimeter. Apparently, the smallest area is obtained when the semiperimeter - or perimeter - is smallest. This happens, when the triangle is isosceles. The right isosceles triangle with incircle radius 1 has side length $$a = (1 + \sqrt{2})\sqrt{2} = 2 + \sqrt{2}.$$ The area of it is $$A = a^2/2 = 2 + \sqrt{2} + 1 = 3 + \sqrt{2}.$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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