Perhaps, related to "What is arccos(sqrt(Pi)/2)?"
... which is the trigonometric function that defines the vertex angle of a Pythagorean triangle that squares the circle:
arccos(.88622692545275801364908374167057..)
= 27.597112635690604451732204752339.. degrees.
For a circle having a diameter equal to 2, the triangle's long side (circle's chord; side of its square)
= sqrt(Pi) and its hypotenuse = 2 (circle's diameter), with the vertex point on the circumference.
-sqrt2
The square root of 38 = ± 6.1644146+sqrt2
That would be 3 x 3 x sqrt2 x sqrt2 = 9 x 2 The answer is 18.
4=(sqrt2)4
It is about 120.71 square units.A= 2(1 + sqrt2)S²For S = 5A = (2 + 2sqrt2) 25A = 50 + 50 (sqrt2)A = 50 + 70.71 = 120.71
one side of the square inscribed in a circle of radius r is sqrt2 * r (the square root of two times the radius) So the perimeter is 4 * sqrt2 * r
It is about 482.84 square units.A= 2(1 + sqrt2)S²For S = 10A = (2 + 2sqrt2) 100A = 200 + 200 (sqrt2)A = 200 + 282.84 = 482.84
1, 2+sqrt3, sqrt2+sqrt6
4 x sqrt2
Ramanujan got many formulas, one of them "The Algorithm for Pi" 9801*sqrt2/4412
A= 2(1 + sqrt2)S² where S is a side length
(sqrt2, 315)