C. Answer the following question accordingly. It shouldn’t take that long. Thanks in advance! **1. Explain the relationship between the number of faces and the number of edges in a triangulation.**In triangulation, each additional face typically introduces one new edge while sharing previously existing edges. For a polyhedron, Euler’s formula, which is $ V - E + F = 2 $, helps describe the relationship between vertices (V), edges (E), and faces (F). Specifically for triangulated surfaces, the number of edges relates to the number of faces by the nature of a triangular shape. Each face (triangle) contributes three edges, though these are shared with adjacent faces within the connected surface. The following is a planar graph that can be drawn in the plane without any edge crossings. We can evaluate not only the edges and vertices with the given graph but also the faces (areas that are enclosed by edges).*Graph Description:*- The graph is composed of five vertices connected by edges, forming three distinct faces.- Face 1 is a quadrilateral enclosed by four edges.- Face 2 is a triangular area enclosed by three edges.- Face 3 is another triangular area enclosed by three edges.Remember: If there are no areas enclosed by edges, as in a tree, we say there is 1 face.

note : As per our company guidelines we are supposed to answer ?️only the first question. Kindly repost other questions next time. For the question 1 we will use Euler's characteristic equation for graphs to find the relation between the number of faces and edges in a triangulation graph.Let's take an arbitrary graph having v vertices, f faces and e edges. Then as pet the Euler's characteristic equation we have : v-e+f = 2 ➞ f = 2+e-v & e = v+f-2 Now considering the triangulation of the graph, all the non-collinear vertices will be connected by edges, and all the faces will be triangular. Which means each face will contain 3 edges. And each edge will touch 2 faces. Hence we get : 2e = 3f Combining te equations we have : 2e = 3(2+e-v) & 3f = 2(v+f-2) ➞ 2e = 6+3e-3v & 3f = 2v+2f-4 ➞ e = 3v-6 & f = 2v-4 Therefore in a triangulation having v vertices we have : Total number of faces : f = 2(v-2) Total number of edges: e = 3(v-2)