Your idea is correct. The area of the larger rectangle minus the area of the smaller tringle equals the area of the border.

But what are the dimensions of the large rectangle? 10m by 25m? No. Those are for the smaller rectangle.

We are looking for the uniform width of the border. We don't know it yet, so let us call it x meter wide.

Now we can get the dimensions of the larger rectangle.

width = 10 +x +x = (10 +2x) meters

length = 25 +x +x = (25 +2x) meters

So,

(10 +2x)(25 +2x) -(10)(25) = 74

10*25 +10*2x +2x*25 +2x*2x -10*25 = 74

The 10*25 cancels out,

20x +50x +4x^2 = 74

4x^2 +70x -74 = 0

Divide both sides by 2,

2x^2 +35x -37 = 0 ------------***

You can use the Quadratic Formula to solve for x, but I think you are into factoring now, so we factor that,

(2x +37)(x -1) = 0

2x +37 = 0

2x = -37

x = -37/2 = -18.5 m ----reject this because there are no negative dimensions, for one.

x -1 = 0

x = 1 meter ------------answer.

Check,

Large rectangle:

width = 10 +2(1) = 12 m

length = 25 +2(1) = 27 m

(12)(27) -(10)(25) =? 74

324 - 250 =? 74

74 =? 74

Yes, so, OK.