Question

A certain polygon has a total of five diagonals that can be drawn from all vertices. How many sides would that polygon have? Select one: a. 2 b. 7 c. 5 d. 3

Answer 1

C. 5

Explanation:

We are given that the polygon has total 5 diagonals.

It is known that the formula for the number of diagonals in a polygon is,

Number of diagonals =
`(n(n-3))/(2)`, where n is the number of sides.

As, the total number of diagonals in the given polygon are 5.

So, we get,

5 =
`(n(n-3))/(2)`

i.e.
`10=n^(2)-3n`

i.e.
`n^(2)-3n-10=0`

Since, the solution of a quadratic equation
`ax^2+bx+c=0` is given by,
`x=\frac{-b\pm \sqrt{b^(2)-4ac}}{2a}`

We have,

`n^(2)-3n-10=0` implies a= 1, b= -3 and c= -10.

So, the solution is,

`n=\frac{3\pm \sqrt{(-3)^(2)-4* 1* (-10)}}{2* 1}`

i.e.
`n=(3\pm √(9+40))/(2)`

i.e.
`n=(3\pm √(49))/(2)`

i.e.
`n=(3\pm 7)/(2)`

i.e.
`n=(3+7)/(2)` and i.e.
`x=(3-7)/(2)`

i.e.
`n=(10)/(2)` and i.e.
`x=(-4)/(2)`

i.e. n= 5 and n= -2.

As, the number of sides cannot be negative.

So, n= 5.

Thus, the polygon will have 5 sides.

Answer 2

5 vertices

Explanation:

We know that the total number of diagonals of n vertices of a polygon is given by

`d=(n(n-3))/(2)`

Number of diagonal = 5

Hence, d = 5

Plugging this value in above mentioned formula, we get

`5=(n(n-3))/(2)\\\\n(n-3)=10\\n^2-3n-10=0\\\\n^2-5n+2n-10=0\\n(n-5)+2(n-5)=0\\(n-5)(n+2)=0\\n=-2,5`

n can't be negative because it represents the vertices.

Hence, n = 5

c is the correct option.