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How many diagonals can be drawn by joining the angular points of an octagon?

1. $14$
2. $20$
3. $21$
4. $28$
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If we have a polynomial having n vertices then corresponding to each vertices we will have total (n-3) diagonals. ( By joining a vertex to every other vertices except its 2 neighbours. )

So total number of diagonal  = n *(n-3)/2  ( Dividing by 2 because of duplicate diagonals ).

Here n= 8 for octagon,

No of diagonal = (8-3)*8/2 = 20

Ans- B
by Boss (26.6k points)
+5

8c2 - 8 = 20

Option B

Draw an octagon, select one vertex and construct each diagonal from this vertex.You will see there are 5 such diagonals. Thus for each of the 8 vertices you can draw 5 diagonals and hence you have constructed $5 * 8 = 40$ diagonals. But you have constructed each diagonal twice, once from each of its ends. Thus there are $20$ diagonals in a octagon.

by Active (2k points)
Number of digonals = (n-3) + Sum of first (n-3) numbers

n = 8

Number of diagonals = 5 + (1+2+3+4+5) = 20
by (299 points)

The formula to find the number of sides of a polygon is n(n-3)/2 where “n” equals the number of sides of the polygon. Using the distributive property this can be rewritten as (n2 - 3n)/2. You may see it either way, both equations are identical.

• This equation can be used to find the number of diagonals of any polygon.
• Note that the triangle is an exception to this rule. Due to the shape of the triangle, it does not have any diagonals
by Active (4.9k points)
+1 vote
it is 20 ( you can draw and find ) For each vertex have a line with all other vertex ( a direct edge )

so option b
by Loyal (9.9k points)
+1 vote
for any n vertex polygon no of diagonal=no of total edge -no of sides(non diagonal) =nC2-n=20
by Active (3.6k points)