Number of a diagonals of a polygon of side n = n*(n-3)/2.

We are told that number of diagonals here is 20. Hence n(n-3)/2 = 20.
Hence n*(n-3)=40 = 8*5. Hence n = 8.

Hence E is the right answer.

Let me know if this helps

let usassume polygon has n sides. hence, it has n vertices.

no of ways in which any two points can be selected from these n points= nC2
(please note this gives us the total number of lines that can be formed from the given n points).

nC2 = no of sides of the ploygon + no of diagonals of the polygon.

here,no od diagonals= 20.
hence, nC2 = n +20
solve for n .

Hope this helps!

Soumita Ghosh wrote:A POLYGON HAS 20 DIAGONALS, How Many Sides Does It Have?


A) 12

B) 11

C) 10

D) 9

E) 8

A diagonal is a COMBINATION of 2 vertices.
Any combination of 2 vertices that do not form a side of the polygon can serve to form a diagonal.
We can plug in the answers, which represent the number of vertices needed to form 20 diagonals.

Answer choice C: 10
Number of combinations of 2 that can be formed from 10 options = 10C2 = 45.
But this result includes those combinations of 2 that form the sides of the polygon -- combinations that do not yield diagonals.
These bad combinations -- in other words, the number of sides of the polygon -- must be subtracted from the total:
45-10 = 35.
Too big.
Eliminate A, B and C.

Answer choice D: 9
Number of combinations of 2 that can be formed from 9 options = 9C2 = 36.
Subtracting the number of sides, we get:
36-9 = 27.
Eliminate D.

The correct answer is E.

Answer choice E: 8
Number of combinations of 2 that can be formed from 8 options = 8C2 = 28.
Subtracting the number of sides, we get:
28-8 = 20.
Success!