JennyFluff 
💾this toot is sponsored by square-circle
fungus (penicillin producer)
Pino Carafa@hypha @JennyFluff I'm too lazy but someone should work out the area within that construction. The area of a "normal" square with sides of length 1 is, well, 1. If these lines are all length 1 as well, what is its area?
Richard@rozeboosje @hypha @JennyFluff I got
every side = 1
inner radius r
outer radius R = r+1
inner missing part m = r*Ï„-1
outer curve C = 1 = m*R/r
inner circle area a = 1/2*Ï„*r^2
inner pie area p = a * 1/(Ï„*r)
outer pie area P = 1/2*Ï„*R^2 * 1/(Ï„*R) = 1/2*R
total area A = P+a-p
Which comes to A≈0.61 (if neither me nor Wolfram Alpha messed up somewhere along the way)
@rigrig @hypha @JennyFluff This is what I'm getting:
Area of a circle is Pi * r^2
The circumference C of a circle is 2 * Pi * r
So the area within a circle is 1/2 * C * r
That then also allows us to calculate the are of a segment S, likewise, as
1/2 * S * r
All the sides are 1 so S is 1 both for circle and Wedge and "r" for the large wedge is 1+r
Circle without Wedge = 1/2 r
Wedge = 1/2 * 1 * (r + 1)
Total: r + 1/2
I reckon you're right. I'm struggling a bit figuring out what r is
@rozeboosje @hypha @JennyFluff Yeah, that’s the tricky part.
Say you have the outer segment S (=1) with radius R, and the part that’s missing from the inner circle s (m in my toot above), with radius r. They’re the same shape, only scaled by the difference in radius, so you have S = s×R/r. The inner circle would have a total circumference of 2×Pi×r, which is 1 plus the open part, so you get 2×Pi×r = 1+s, and we know S = 1.
So now we have
R = r + 1
s = 2×Pi×r - 1
S = 1 = s×R/r
and can solve for r. (But not easily, which is why I plugged it into WA and only posted the numerical result instead of the exact answer with lots of Pie-terms)