1. Determine the Grazing Radius:

The cows are tethered to the corners of a square field of side length $s = 28$ m. Since each cow can just touch the adjacent cows, the rope length ($r$) for each cow is half the side length of the square.

Radius $r = \frac{s}{2} = \frac{28 \text{ m}}{2} = 14$ m.

2. Calculate the Grazable Area Inside the Square:

Each cow grazes a quarter-circle area within the square. The total grazable area inside the square ($A_{\text{grazable}}$) is the sum of the areas of four quarter-circles of radius $r$.

Area of one quarter-circle = $\frac{1}{4} \pi r^2$.

Total grazable area $A_{\text{grazable}} = 4 \times \left( \frac{1}{4} \pi r^2 \right) = \pi r^2$.

Substituting the values ($\pi = 22/7$, $r = 14$ m):

$A_{\text{grazable}} = \frac{22}{7} \times (14 \text{ m})^2 = \frac{22}{7} \times 196 \text{ m}^2 = 22 \times 28 \text{ m}^2 = 616 \text{ m}^2$.

3. Calculate the Total Field Area:

The area of the square field ($A_{\text{square}}$) is:

$A_{\text{square}} = s^2 = (28 \text{ m})^2 = 784 \text{ m}^2$.

4. Calculate the Area Beyond the Cows' Reach:

The area inside the square that is beyond the reach of the cows ($A_{\text{beyond}}$) is the difference between the total field area and the grazable area.

$A_{\text{beyond}} = A_{\text{square}} - A_{\text{grazable}} = 784 \text{ m}^2 - 616 \text{ m}^2 = 168 \text{ m}^2$.

5. Determine the Grazing Time for the Beyond Area:

The grass in the grazable area ($A_{\text{grazable}} = 616 \text{ m}^2$) feeds the cows for $D_1 = 22$ days. We need to find how many days ($D_2$) the grass in the area beyond reach ($A_{\text{beyond}} = 168 \text{ m}^2$) would feed them if it were cut and placed within the grazable parts. Assuming the grass density and consumption rate are constant, the number of days is directly proportional to the area.

We can set up a proportion:

$\frac{A_{\text{grazable}}}{D_1} = \frac{A_{\text{beyond}}}{D_2}$

Solving for $D_2$:

$D_2 = D_1 \times \frac{A_{\text{beyond}}}{A_{\text{grazable}}}$

$D_2 = 22 \text{ days} \times \frac{168 \text{ m}^2}{616 \text{ m}^2}$

Simplify the fraction $\frac{168}{616}$. Both numbers are divisible by 7, then 8:

$\frac{168}{616} = \frac{24}{88} = \frac{3}{11}$.

$D_2 = 22 \times \frac{3}{11} = 2 \times 3 = 6$ days.

Therefore, the grass beyond the reach of the cows would feed them for 6 days.