arranging squares

Let the side length of the smallest square be $x$ . Then the side lengths of the other squares are $x+2,x+4,$ and $x+6$ . So the base of the right triangle is

$x+x+2+x+4+x+6=4x+12$

By pythagorean theorem, we have

$(2\sqrt{221})^2=(4x+12)^2+(x+6)^2$

$884=16x^2+96x+144+x^2+12x+36$

$17x^2+108x-704=0$

Using the quadratic formula, we get

$x=4$

So the side lengths of the other squares are $6,8,$ and $10$ .

The area of the shaded region is equal to the total area of the squares minus the area of the triangle

$A=4^2+6^2+8^2+10^2-\dfrac{1}{2}(28)(10) = 216-140=\boxed{76}$