To determine the dimensions of the fenced yard, we need to solve for the length and width of the rectangle while satisfying the constraints:

1. The area of the yard is $400 \, \text{m}^2$: $L \cdot W = 400$

2. The perimeter of the fenced portion is $50 \, \text{m}$ (only three sides are fenced, as one side is along the house): $2W + L = 50$

Step 1: Express $L$ in terms of $W$

From the second equation: $L = 50 - 2W$

Step 2: Substitute $L$ into the area equation

Substitute $L = 50 - 2W$ into $L \cdot W = 400$: $(50 - 2W) \cdot W = 400$

Expand: $50W - 2W^2 = 400$

Rearrange into standard quadratic form: $2W^2 - 50W + 400 = 0$

Step 3: Solve the quadratic equation

Divide through by 2 to simplify: $W^2 - 25W + 200 = 0$

Use the quadratic formula: $W = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -25$, $c = 200$: $W = \frac{-(-25) \pm \sqrt{(-25)^2 - 4(1)(200)}}{2(1)}$ $W = \frac{25 \pm \sqrt{625 - 800}}{2}$ $W = \frac{25 \pm \sqrt{-175}}{2}$

Since the discriminant is negative, the problem setup might require reconsideration or simplification. Could you confirm any assumptions?