Example 6 A square piece of paper contains 100 points. Together with the vertices of the square, there are 104 points, and no three points among these 104 points are collinear. Now, the square piece of paper is entirely cut into triangles, with each vertex of these triangles being one of the 104 points, and each of the 104 points being a vertex of these triangles. If cutting along a line segment counts as one cut, how many cuts are needed to obtain these triangles, and how many triangles can be cut in total?
Analysis: If there is only one point $P_{0}$ inside the square, connecting $P_{0}$ to the four vertices results in four triangles, which clearly require the first 4 cuts. If another point $P_{1}$ is added, this point must fall within one of the triangles, let's assume it falls within $\triangle P_{0} C D$ (as shown in Figure 5), at this point, two more triangles are added, requiring 3 cuts. If another point $P_{2}, \cdots \cdots$ is added, and so on, the total number of triangles is $4 + 2 \times 99 = 202$, and the total number of cuts is $4 + 3 \times 99 = 301$.

Explanation: This problem can also examine the total degree sum of all the internal angles of the triangles, to find the number of triangles $=$ $\frac{360^{\circ} \times \text{number of points inside the square} + 90^{\circ} \times 4}{180^{\circ}}$.