it really isn't hard 5
Suppose a polygon has n vertices (and sides). From each vertex, a diagonal can be drawn to all vertices, excluding itself and the two adjacent vertices. So n-3 diagonals can be drawn from each vertex. Multiplying by the full complement of n vertices gives n(n-3). However, as things stand we have counted each diagonal twice: once at both ends. Dividing by two gives the actual number of diagonals. number of diagonals = n(n-3)/2
In a hexagon, each vertex can connect to all other non-adjacent vertices to form diagonals. Since a hexagon has 6 vertices, each vertex can form diagonals with the other 4 non-adjacent vertices. Therefore, each vertex of a hexagon can create 4 diagonals. This results in a total of 6 vertices x 4 diagonals = 24 diagonals in a hexagon.
This assumes you mean a regular octagon. There are 4 pairs where the diagonals each skip two vertices. Then there are 4 sets of three parallel diagonals where one is the main diagonal (skips three vertices) and two skip just one vertex. For each of these sets of three, you can choose three pairs of two. That makes 12 pairs all up. So there are 4 + 12 = 16 pairs.
A heptagon has seven sides, so when drawing diagonals from one vertex, it will create five triangles. This is because each diagonal drawn from a single vertex will create a triangle until it intersects the previous diagonal. Therefore, the number of triangles formed by drawing all diagonals from one vertex in a heptagon is five.
A hexagon (six-sided polygon) can be divided into 4 triangles by drawing all of the diagonals from one vertex (only three lines can be drawn in this case, since each vertex already connects to two others on the edges of the form). If you instead drew lines from the center to each vertex, you would get 6 triangles.
In an octagon, each vertex can connect to other vertices except itself and its two adjacent vertices. Since there are 8 vertices in total, a vertex can connect to (8 - 3 = 5) other vertices to form diagonals. Therefore, from each vertex of an octagon, 5 diagonals can be drawn.
16 diagonals* * * * *5. To all vertices except itself and one each on either side.
5
n-3 from each vertex.
In a 15-gon, each vertex can connect to other vertices to form diagonals. Specifically, a vertex can connect to (15 - 3 = 12) other vertices (excluding itself and its two adjacent vertices). Therefore, from one vertex of a 15-gon, 12 diagonals can be drawn.
47 sides. Take a vertex of an n-sided polygon. There are n-1 other vertices. It is already joined to its 2 neighbours, leaving n-3 other vertices not connected to it. Thus n-3 diagonals can be drawn in from each vertex. For n=50, n-3 = 50-3 = 47 diagonals can be drawn from each vertex. The total number of diagonals in an n-sided polygon would imply n-3 diagonals from each of the n vertices giving n(n-3). However, the diagonal from vertex A to C would be counted twice, once for vertex A and again for vertex C, thus there are half this number of diagonals, namely: number of diagonals in an n-sided polygon = n(n-3)/2.
Yes if drawn within an octagon from a vertex to each of the other vertices will give you 8 triangles within the octagon
The formula for the number of diagonals is: 0.5*(n^2-3n) whereas 'n' is the number of sides of the polygon
An octagon has four (4) diagonals, each connecting two opposite vertices.
There can be 14 lines in a seven side shape * * * * * That is the total number of diagonals from ALL vertices. Not what the question asked, though. From one vertex, there can be 4. One to every other vertex except for itself and one each on either side.
In a polygon with 17 sides, a diagonal can be drawn from a fixed vertex to any of the other non-adjacent vertices. From one vertex, there are 14 other vertices (17 total vertices - 1 fixed vertex - 2 adjacent vertices) to which diagonals can be drawn. Each diagonal creates a triangle with the fixed vertex and two of the vertices connected by the diagonal. Therefore, the number of triangles that can be formed is equal to the number of diagonals, which is 14.
Suppose a polygon has n vertices (and sides). From each vertex, a diagonal can be drawn to all vertices, excluding itself and the two adjacent vertices. So n-3 diagonals can be drawn from each vertex. Multiplying by the full complement of n vertices gives n(n-3). However, as things stand we have counted each diagonal twice: once at both ends. Dividing by two gives the actual number of diagonals. number of diagonals = n(n-3)/2