General solution: area of large square = a^2 area of smaller square = x^2 then (a/2 + x)^2 + (x/2)^2 = (a√2/2)^2 (expand the squares) (a^2) / 4 + a x + x^2 + (x^2) / 4 = 2 * (a^2) / 4 (multiply by 4) a^2 + 4 a x + 5 x^2 = 2 a^2 (subtract 2 a^2) 5 x^2 + 4 a x - a ^2 = 0 (abc quadratic equation) x(1,2) = (-4a +/- √((4a)^2 - 4 * 5 * (-a^2)))/ (2 * 5) = (-4a +/- √(16a^2 + 20 a^2))/ 10 = (-4a +/- √(36a^2))/10 x = (-4a + 6a)/10 or x = (-4a - 6a)/10 x = 2a/10 or x = -10a/10 x= a/5 or x = -a The second answer (x = -a) would give us the blue starting square, so that kind of makes sense. The first answer shows that there will always be a 5 times reduction in side of the square, and consequently a 25 times reduction in area of the smaller square.

Chord theorem Let side of the small square = x x(5+x)=(2.5-x/2)(2.5+x/2) x(5+x)=1/2(5-x)1/2(5+x) 4x(5+x)=(5-x)(5+x) 20x+4x^2=25-x^2 5x^2 + 20x - 25 =0 x=1 A=x^2=1 ***

Cna we say that the area of left and right white part in lower of side of blue one is equal?? If yes then answer might be different

Area is 1.

Call s the side of the small square, then Thales and Pythagoras tell us that: (5 + 2s)^2 + s^2 = (5*sqrt(2))^2 5s^2 + 20s -25 =0 5(s+5)(s-1) = 0 Hence s = 1

А=1.