An octagon has 8 sides. How many diagonals does an octagon have?
Little confused with this question. . can anyone help?
Thanks
Let's take one vertex. It can be linked to 7 other vertices.
Let's take the next vertex. It can also be linked to 7 other vertices but there is one diagonal that is already drawn. Therefore, there is 6 new diagonals.
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$\displaystyle \sum_{i=1}^7 i = \frac{i(i+1)}{2} = 28 $*
Unfortunately, the answer given above is an over count.
The answer is 20. An octagon is a subgraph of complete graph, $\displaystyle K_8$ on eight vertices. $\displaystyle K_8$ has 28 edges but only 20 are diagonals of the octagon.
In general, for an n sided figure there are $\displaystyle {n \choose 2}-n = \frac {n(n-1)}{2}-n = \frac {n(n-3)}{2}$ diagonals.
Originally Posted by Carrick
The above quote is from a PM: please ask such questions in the open. Others may also benefit from the answer.
An octagon is a figure with eight vertices and eight edges. If two edges share a vertex, then the edges are said to be adjacent. But any two vertices will determine a line segment. If you join each vertex with each other vertex we have what is known as a complete graph. In the case of eight vertices the complete graph, $\displaystyle K_8$, has 28 edges. The diagonals are the edges that are in the original octagon. Thus 20 are diagonals.
Thus in $\displaystyle K_n\;,\;n\ge 3$ there are $\displaystyle {n \choose 2} = \frac {n(n-1)}{2}$ edges.
But n of the edges are not diagonals.