Yep, it's definitely doable. For this question, the answer should be that the 22.46 length side should be shortened to 20.97, and thus the 15.41 side will be lengthened to 16.90.
If you want to know how this is achieved (or are looking for a formula to simply plug your values into spit out a result in future problems) then we'll start by substituting your numerical values for unknown constants and variables.
Let the variables be the sides that need to be changed, thus
x=15.41
y=22.46
The constants will be
a=16.12
b=4
c=14.12
d=2.83
And let's consider that the angle in the corner doesn't change, so we don't need to make that a constant.
The Area of A1 will be the South-East rectangle,
A_1=x(a+b)
The Area of A2 is the North-West rectangle which includes the triangle bit. Now, the area of the rectangle is simply y*a, but to find the area of the triangle we need to use a bit of trigonometry. The hypotenuse is d, and the angle in this right triangle is 45o, thus the other sides are both,
d\cos(45^o)=d\sin(45^o)=\frac{d}{\sqrt{2}}
And the area of a triangle is given by A=\frac{1}{2}bh where in this case b=h=\frac{d}{\sqrt{2}}
So we now have,
A_2=ya+\frac{d^2}{4}
Now we need both these areas to be equal, so we equate them,
A_1=A_2
x(a+b)=ya+\frac{d^2}{4}
But what can we do with this? Well, we have one equation with 2 variables, so we can't really do much. We need another equation relating x and y.
The only other thing we know about x and y is that their sum doesn't change, thus we can use
x+y=k in which your case, k=15.41+22.46=37.87
So now we have two equations with two unknowns, so we can find out the value of these unknowns.
x+y=k
y=k-x
Substituting this into the first equation gives
x(a+b)=(k-x)a+\frac{d^2}{4}
Now we solve for x,
x(a+b)+(x-k)a=\frac{d^2}{4}
x(a+b+a)-ka=\frac{d^2}{4}
x(2a+b)=ka+\frac{d^2}{4}
x=\frac{ka+\frac{d^2}{4}}{2a+b}
And since y=k-x
We have
y=k-\frac{ka+\frac{d^2}{4}}{2a+b}
You'll find that if you plug your values into these variables, you'll get the answers that I posted up the top.
