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Problem with shaded region of square

In the picture above, I know the answer to the problem is $\frac{1}{4}a^2$. However, I'm sure why the shaded region is a square with sides with length $\frac{1}{2}a$. If I can demonstrate why, then that would be a huge help. T

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The side of the shaded square is half the side of the bigger squares. The side of the big squares is $a$ so the side of the shaded square is $a/2$ which makes its area $ a^2/4$ which is option $B$ of the given options.

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The diagonals of the large squares split them each into four congruent right triangles. Moreover, these triangles are isosceles (since the diagonals bisect each other). The diagonals meet at the centre of a large square. If you drop a perpendicular from this centre to a side of the square, you're in effect dropping a perpendicular from the vertex of an isosceles triangle to its base. Thus this perpendicular must bisect the base. Hence, the region of overlap is a square with sides $$\frac12 a.$$

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A diagonal of a square divides square into two isosceles right triangles, so the little shaded triangles have two $45$ degree angles and the remaining angle has to be a right angle. But the same is true for little white triangles because diagonals of square are perpendicular. We can also conclude that little white triangles are congruent to little shaded triangles (common side and two congruent angles). From this we can conlude that the leg of shaded triangle is half of big square side. But it's also easy to see that area of one little white triangle plus area of one shaded triangle equals one quarter of the square.

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