Hard geometry

I think you have to find the areas of the right triangles surrounding the grey square and use that to find the ratio. You have to use the Pythagorean theorum to find the length of the slanted line and find the part which is the length of the grey square

The hypotenuse of the large right triangle at the left =  sqrt (1^2 + (1+2)^2) = sqrt (10)

Looking  at the small right triangle at the bottom left, notice that it is congruent to  the small right triangle at the upper left

Call the longer leg of one of these triangles, x

The sin of the greater acute angle in these triangles = 3 / sqrt (10)

By the Law of Sines, we have

sin (90) /1  =  3/sqrt (10) / x

1 = [3 sqrt (10)] / x

x = 3/sqrt (10)

Call the smaller leg in either one of these triangles, y

The sin of the smaller acute angle in these triangles = 1/sqrt (10)

Again, by the Law of Sines, we have 

sin (90) / 1 =   1/sqrt (10)  / y

1 =[ 1/sqrt (10)] / y

y =  1/sqrt (10)

The length of the  side of the square =

[The hypotenuse of the large right triangle on the left]   - [  x ]  - [  y  ]  =

sqrt (10 )  - 3/sqrt (10)  - 1/sqrt (10)

sqrt (10)  - [ 3 + 1] /sqrt (10)

(10 - 4) /sqrt (10)  =    6/sqrt (10) = side of the shaded square

Shaded area =  [ 6 / sqrt (10) ] ^2 =   36 / 10  =   3.6

Area of large square  =   9

Ratio of areas =  3.6 / 9 =   36/90  = 4 / 10 =   2/ 5