Biff’s pet is tied to the corner of a square shed, 6 meters on each side. The rope
is 8 meters long. The area outside the shed over which the pet can wander is Nπ
square meters. Find the whole number N.

To solve the problem, we need to determine the area over which Biff's pet can wander. 

### Step 1: Visualize the problem
The shed is a square with each side measuring 6 meters, and the pet is tied to one of the corners with a rope that is 8 meters long. The area in which the pet can roam would be part of a circle with a radius equal to the length of the rope.

### Step 2: Analyze the area the pet can cover
The rope is 8 meters long, which is longer than the side of the shed (6 meters). The pet can therefore wander freely in a quarter-circle sector with radius 8 meters, but part of this quarter-circle will be obstructed by the shed.

### Step 3: Calculate the total area of the quarter-circle
The area of a full circle with radius 8 meters is:
\[
\text{Area of full circle} = \pi \times r^2 = \pi \times 8^2 = 64\pi \text{ square meters}
\]
Since the pet can only wander in a quarter-circle:
\[
\text{Area of quarter-circle} = \frac{1}{4} \times 64\pi = 16\pi \text{ square meters}
\]

### Step 4: Determine the area covered by the pet outside the shed
The pet’s roaming area outside the shed includes two segments:
1. A quarter-circle with radius 8 meters.
2. The portion of this area inside the 6-meter side of the shed, which needs to be subtracted.

The part of the quarter-circle that the pet can access outside the shed corresponds to a sector with radius 8 meters minus the sector obstructed by the shed.

The part of the sector that overlaps with the shed is a quarter-circle with radius 2 meters (since the shed's side length is 6 meters and the total radius is 8 meters). The area of this smaller quarter-circle is:
\[
\text{Area of smaller quarter-circle} = \frac{1}{4} \times \pi \times 2^2 = \frac{1}{4} \times \pi \times 4 = \pi \text{ square meters}
\]

### Step 5: Subtract the overlapping area
The area that Biff's pet can wander outside the shed is the area of the larger quarter-circle minus the area of the smaller quarter-circle:
\[
\text{Total area} = 16\pi - \pi = 15\pi \text{ square meters}
\]

### Conclusion
The value of \( N \) is 15.

So, \( N = 15 \).

Discover how to calculate the area outside a square shed where Biff’s pet can wander, given the shed's dimensions and the length of the pet’s rope. This problem involves geometry concepts such as circle sectors and area calculations.

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