SQUARES

i know that the outer square is half of inner square so....

We can split the square into 8 triangles by adding one line that runs down horizontally down the center of the big square and one line that runs down vertically down the big square also. Thus, we will find that the the big square has 8 triangles while the small square has 4 triangles. Since all the triangles are similar, this proves that the smaller square is half that of the big square

By drawing AC and BD, it becomes clear the figure is built from eight triangles; they must be congruent, because they share angles and sides. The inner square is built of four of these and must have half the surface of the outer.

The side’s ratio is (√2/2). So by similarity the area ratio is (√2/2)2 = ½. So the area of the smaller square is 30.

Consider the large square's side=2a. The area of large square=(2a)^2=60. Join BD. We know that the length of the diagonal of the small square=a side of the large square. The area of a square using diagonal length is 1/2d^2. =1/2×(2a)^2 =1/2×60 =30

I solved it applying the Pythagoras theorem to one of the triangles formed at the angles of the outer square : calling $a$ the side of the outer triangle and $b$ the side of the inner, we have $b=\sqrt{a^2+a^2}$