Easy geometry problem but simply neat.

Note that everything besides the left and the right triangle are not important. You can calculate the side lengths of the square with x = 1/2 (a²+b²) with a and b being the side lengths of the left and right triangle respectively.

Also note that there exists a square at the top iff the height of the left triangle is 1/2 of the base length of the right triangle

Due to the left rhombi sequence, the right-up side of the square is parallel and equal to the right side of the 12 cm isosceles triangle. Due to the right rhombi, the left-up side of the square is parallel and equal to the left side of the 16 cm isosceles triangle. To create rhombi, all the triangles must have the same sides (out of their basis).

So the point is to determine the initial basis angles of the 12 and 16 triangles, so that they add up to 90°.

Let x be the length of the side, and α the basis angle of the 12 triangle. The basis angle of the 16 triangle must be (90°-α).

In the 12 triangle, we have : x.cos(α) = 12/2 = 6. In the 16 triangle, we have : x.cos(90°-α) = x.sin(α) = 16/2 = 8.

So we have tan(α) = 8/6 =1.33. Then α = 53.13°.

The height of 12 triangle is 6 x 1.33 = 8, and the side is sqrt(6^2 + 8^2)= 10.

The area of the square is 100.