What if part of the fencing is used to build a partition perpendicular to a side? Consider a rectangular region with one partition? With 2 partitions? with n partitions? (There is a surprise in this one!!) What if the partition is a diagonal of the rectangle?
To calculate the remaining length of fencing after building a partition, we need to consider the dimensions of the rectangle and the location of the partition.
1. **One Partition:**
Let's assume we have a rectangular region with length L and width W. If we build a partition perpendicular to one of the sides (say the length side), it will divide the rectangle into two smaller rectangles. The remaining length of the fence will be the perimeter of these two smaller rectangles combined:
Remaining length of fencing = Perimeter of smaller rectangle 1 + Perimeter of smaller rectangle 2
For example, if the partition divides the rectangle into two rectangles with lengths L1 and L2, and widths W1 and W2, respectively, the remaining length can be calculated as follows:
Remaining length of fencing = 2(L1 + W1) + 2(L2 + W2)
2. **Two Partitions:**
If we add another partition perpendicular to a different side (let's say the width side) of the rectangle, it will create three smaller rectangles. Now the remaining length of the fencing will be the sum of the perimeters of these three smaller rectangles:
Remaining length of fencing = Perimeter of smaller rectangle 1 + Perimeter of smaller rectangle 2 + Perimeter of smaller rectangle 3
3. **n Partitions:**
For n partitions, where n > 2, we can use a similar approach. Each partition will divide the rectangle into n+1 smaller rectangles. The remaining length of the fencing will be the sum of the perimeters of all these smaller rectangles:
Remaining length of fencing = Perimeter of smaller rectangle 1 + Perimeter of smaller rectangle 2 + ... + Perimeter of smaller rectangle n+1
The surprise in this scenario is that no matter how many partitions you add to the rectangle, the remaining length of the fence will remain constant. This is because the perimeter of each smaller rectangle is constant, and adding partitions only rearranges the length and width of each smaller rectangle.
4. **Diagonal Partition:**
If the partition is a diagonal of the rectangle, it will divide the rectangle into two right-angled triangles. The remaining length of the fencing will be the sum of the hypotenuses of these triangles:
Remaining length of fencing = Hypotenuse of triangle 1 + Hypotenuse of triangle 2
The hypotenuse of each triangle can be found using the Pythagorean theorem:
Remaining length of fencing = sqrt(L^2 + W^2) + sqrt(L^2 + W^2)
So, for a diagonal partition, the remaining length of fencing will be twice the square root of the sum of the squares of the length and width of the rectangle.