Question

5. A polygon has 27 diagonals. How many sides does it have?
(a) 7
(b) 8
(c) 9
(d) 12​

Answer 1

we we know that,

number number of polygons =n(n-3)/2

according to the question,

n(n-3)/2=27

n(n-3)=27x2

n(n-3)=54

n²-3n-54=0

n²9n+6(n9)=0

(n+6) (n-9) =0

n=9, -6 ( neglecting negative value)

n=9

hence, sides of polygon are 9

hope it's helpful

Answer 2

Answer:

Option (c) The Polygon have 9 sides.

Step-by-step explanation:

An n sided polygon have n number of vertices.

Each of them can connect to the rest of the vertices to form straight lines.  And amongst those straight lines, n are the sides of that polygon.  That leaves remaining straight lines as diagonals.

Number of straight lines formed = ⁿ$C$r = ⁿ$C$₂ $=\frac{n!}{2!(n - 2)!} = \frac{n(n-1)}{2}$

Number of sides = n

Therefore, number of diagonals $= \frac{n(n-1)}{2} - n = \frac{n^2 - 3n}{2}$

According to the problem:

$\frac{n^2 - 3n}{2} = 27$

$=> n^2 - 3n - 54 = 0$

$=> (n - 9)(n + 6) = 0$

n cannot be negative, hence n = 9.

Option (c) The Polygon have 9 sides.