In the square above, if the ten nonoverlapping regions have the areas

In the given figure, the square with area x^2 has length and width x each. So, for the rectangle with area xy on the left, the width is x and area is xy so length is y. Similarly, for the rectangle on the top with area xy, the length is x and area is xy so width is y. Therefore, the length/width of the square = x + y.
Area of the square = \((x+y)^2 = x^2 + 2xy + y^2\)
From the given figure, the area of the square is equal to the sum of the areas of the ten regions inside it.
So, \(x^2 + 2xy + y^2 = x^2 + 2xy + a^2 + 2ab + s^2 + st + st + t^2\)
On simplifying it, \(y^2 = a^2 + 2ab + (s+t)^2\)
\(y^2 = (a+b)^2 - b^2 + (s+t)^2\)--------- equation 1


Similar to how the outside square's length was found out, we can see in the inside figure has length of side a+b.
Its area = \((a+b)^2\)= sum of the areas of the constituting regions inside
So, \((a+b)^2 = a^2 + 2ab + s^2 + 2st + t^2\)
On simplifying it, \(b^2 = (s+t)^2\)----------- equation 2
Using this in equation 1,
\(y^2 = (a+b)^2\)
y = a+b (since they are non-negative)
Since b = s+t from equation 2 (since they are non-negative), y = a+b = a + s + t

Therefore, the answer is C.