Question

How many diagonals can you draw from the vertices of an octagon?

Answer 1

Hint:The given question requires us to find the number of diagonals that can be drawn from the vertices from an octagon. An octagon has eight sides. We can find out the number of diagonals of an octagon using the concepts of Permutations and Combinations. Diagonal of a polygon is a line joining any two vertices of the polygon when those two points are not on the same edge and do not constitute a side of a polygon.

Complete step by step solution:
Total no. of vertices in an octagon$ = 8$
From two given points only one line can pass.
So, number of possible lines that can be drawn from these 8 vertices\[ = {}^8{C_2}\]
We can evaluate \[{}^8{C_2}\] using the combination formula, $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ . So, we get,
$^8{C_2} = \dfrac{{8!}}{{(8 - 2)! \times 2!}}$

Now, factorial represents the product of all the natural numbers from that specific number, say n, till $1$ .
So, Number of possible lines that can be drawn from these 8 vertices$ = \dfrac{{8 \times 7 \times 6!}}{{6! \times 2!}}$
Therefore, number of possible lines that can be drawn from these 8 vertices$ = 28$
Now, out of these 28 lines, 8 lines form the edges of the octagon.
So, number of diagonals $ = 28 - 8$
Thus, the number of diagonals = 20.

Note: The direct formula for finding the number of diagonals of a polygon of n sides is: $\left( {^n{C_2} - n} \right)$. In the given formula, we have the number of possible lines that can be drawn from the n vertices of the polygon as $\left( {^n{C_2}} \right)$ and n as the number of sides of the polygon.