toads_tfnegative area square?
negative area makes sense if you consider it as the "area under a function" (so a defined integral of a function)
and you could just handwave it anyways.
This looks like a "find the overlapping area of three circles" question. It's extremely tedious, but pretty braindead with a symbolic calculator
Here's a very rough outline of how to find A3 (not wasting any more time on this):
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let's start with an A_s (assuming it's the outer square) of 1 and multiply by the actual A_s at the very end
https://i.imgur.com/2GSGwQ1.png
I guessed that the vague curves are sections of circle with center at a corner of the large square and radius the same as the square's side length
https://i.imgur.com/qny2VWS.png
the program gives the equations for the circles easily
https://i.imgur.com/mzP36RE.png
change the relevant half of the circle into a y=f(x) format
https://i.imgur.com/y15qo1k.png
https://i.imgur.com/53bLMWX.png
find the intersections of the relevant curves, and use them as the bounds of the defined integrals..
https://i.imgur.com/VqDoIGT.png
and just smack the integrals into a calculator, and you add them all together:
+ integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) dx
- integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) + 1/2 (-1 + 2 sqrt(x - x^2))) dx
+ integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
- integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
+ integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(2 x - x^2)) dx
- integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(1 - x^2)) dx
and multiply the result by the side length
and the final answer for A3 (unless I made mistakes....)
surprise, it's a massive clusterfuck:
(-47)*(9/64 - sqrt(3)/8 - (5 π)/12 + tan^(-1)(4 + sqrt(7)) + 1/64 (-17 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)((2 sqrt(5 - sqrt(7)))/3)) + (1/64 (sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + 8 (-3 + 4 sqrt(3) - sqrt(7) + 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (5 π)/12) - (1/64 (-9 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)(2 sqrt(3 - sqrt(7))))) + (-1/64 sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + π/3 + 1/8 (-1 + sqrt(7) - 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (1/192 (3 - 24 sqrt(3) + 24 sqrt(7) + 32 π - 96 cos^(-1)(1/8 (5 - sqrt(7))))))
= ~0.036566432691702221271670 - 0.019630731925696712432468 + 0.061918112844524846486709 - 0.0150274870976263361702352 + 0.04230790314076436373846 - 0.0173897792991522791591702
= 0.0887444503545161037349656 * (-47)
= ~-4.1709891666622568755433832..
and then just do that two more times for the final answer. It's not going to get much shorter.
Of course the question is pretty vague so this could be the solution to the wrong problem.
https://i.imgur.com/2GSGwQ1.png
I guessed that the vague curves are sections of circle with center at a corner of the large square and radius the same as the square's side length
https://i.imgur.com/qny2VWS.png
the program gives the equations for the circles easily
https://i.imgur.com/mzP36RE.png
change the relevant half of the circle into a y=f(x) format
https://i.imgur.com/y15qo1k.png
https://i.imgur.com/53bLMWX.png
find the intersections of the relevant curves, and use them as the bounds of the defined integrals..
https://i.imgur.com/VqDoIGT.png
and just smack the integrals into a calculator, and you add them all together:
+ integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) dx
- integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) + 1/2 (-1 + 2 sqrt(x - x^2))) dx
+ integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
- integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx
+ integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(2 x - x^2)) dx
- integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(1 - x^2)) dx
and multiply the result by the side length
and the final answer for A3 (unless I made mistakes....)
surprise, it's a massive clusterfuck:
(-47)*(9/64 - sqrt(3)/8 - (5 π)/12 + tan^(-1)(4 + sqrt(7)) + 1/64 (-17 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)((2 sqrt(5 - sqrt(7)))/3)) + (1/64 (sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + 8 (-3 + 4 sqrt(3) - sqrt(7) + 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (5 π)/12) - (1/64 (-9 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)(2 sqrt(3 - sqrt(7))))) + (-1/64 sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + π/3 + 1/8 (-1 + sqrt(7) - 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (1/192 (3 - 24 sqrt(3) + 24 sqrt(7) + 32 π - 96 cos^(-1)(1/8 (5 - sqrt(7))))))
= ~0.036566432691702221271670 - 0.019630731925696712432468 + 0.061918112844524846486709 - 0.0150274870976263361702352 + 0.04230790314076436373846 - 0.0173897792991522791591702
= 0.0887444503545161037349656 * (-47)
= ~-4.1709891666622568755433832..
and then just do that two more times for the final answer. It's not going to get much shorter.
Of course the question is pretty vague so this could be the solution to the wrong problem.