Middle Square Gone Missing!

Relevant wiki: Length and Area

Let's call the side of the square $A$ and the side of the big triangle $L$ .
The area of the square is $A^2$ .
Both small triangles have sides $A$ and $L-A$ ,
therefore they have the same areas.

The pink area is $\dfrac{(L-A)(A)}2 \cdot 2 = AL-A^2$ .

Now, the area of the big triangle is $\dfrac{L^2}2 = (AL-A^2)+A^2 = AL$ . And so, $2A = L$ .
Replacing we get that the pink area is $2A^2 -A^2 = A^2$ , therefore both areas are equal.

The answer is the same because if we divide the green square into 2 right triangles by the diagonals, then we get 4 congruent right triangles. So the answer is the same.

It's an easy problem. First we know that each right angle side of the triangle is equal to each side of the square= S. Then we know that area of 1triangle = 1/2bh in this case = 1/2xSxS=1/2 s² . The pink area has two triangles hence its area is 2x1/2*S²= S² Since area of square= AxA we know this is equal to SxS = S²= area of the two triangles. Hence the areas are equal

The three vertices of the square are the mid-points of the respective sides of triangle.The two pink triangles are congruent.Now imagine that the sides of the square as a mirror[the sides inside the triangle].The reflection of the two pink areas on the square will overlap and form the same square.Hence the pink area is equal to the green area.