How many sides of a polygon of 20 diagonals?

M4maths Previous Puzzles

28 Sep 2010 Find the sides?

How many sides of a polygon of 20 diagonals?

OPtion

1) 7
2) 8
3) 9
4) 6
5) 10
6) 11
7) 12
8) 13
9) 14
10) 5

Solution

if number of sides are s,
The number of diagonals D = s(s-3)/2
by solve this equation 40 = s^2-3s
By solve this we get s = 8, -5
so the number of sides are 8

Correct Option: 2 Best Solution (16)

SUDESHNA SEN   14 years AGO

Formula for the number of diagonals in a regular polygon:
D = (n(n-3)/2)
Here D = number of diagonals which is 20
n = number of sides which is not known

20=(n^2-3n)/2

40=n^2-3n
n^2-3n-40= 0
By factorizing the equation you will get

n^2-8n+5n-40=0
n(n-8)+5(n-8)=0
(n-8)(n+5)= 0

So either
n-8=0 or n+5=0
n=8 or n=-5
n can not be negative therefore
n = 8 sides or regular Octagon...... Option 2)

Like? Yes (5) | No   

YadeeshKumar   14 years AGO

formula is
D = (n(n-3)/2)
Here D = number of diagonals which is 20
n = number of sides which is not known
so
20=n(n-3)/2
solve this we get 8

Like? Yes (3) | No   

HARMEET KAUR   14 years AGO

If there are D diagonals and n sides in a polygon, then D = n(n-3)/2 means 20 = n(n-3)/2 means n=8

Like? Yes (2) | No   

Karthik Pandiyan   14 years AGO

D = (n(n-3)/2)
D = no.of diagonals n = no. of sides
20=n^2-3n/2
n^2-3n-40=0
(n-8)(n+5)
hence n=8

Like? Yes (2) | No (1)  

Dr.Dipin Singh   14 years AGO

No of sides = x and D is no of diagonals,

then x*(x-3)/2 = D; Here D = 20 and so we get x=8...option 2

Like? Yes (1) | No   

Mayur   14 years AGO

If there are n sides/ vertices in a polygon, no of diagonals starting from a point ( vertex) can be (n-3) only. Each diagonal is shared between 2 points ( vertices) .

So no of diagonals are n(n-3)/2 =20 and n=8

Like? Yes (1) | No   

Shikha   14 years AGO

n* (n-3)/2 =20 ...or .. n=8

Like? Yes (1) | No   

Brij   14 years AGO

if number of sides is n,
then No of diagonals = 20 = n(n-3)/2 and hence n=8

Like? Yes (1) | No   

Sulekha   14 years AGO

N*(N-3)/2=20 or N=8

Like? Yes (1) | No   

Narayanan   14 years AGO

formula : diagonals for a regular polygon= side(side-3)/2
-> d=n(n-3)/2 -> n=8 for d=20.option :2

Like? Yes (1) | No   

CT   14 years AGO

formula : diagonals for a regular polygon= side(side-3)/2
-> d=n(n-3)/2 -> n=8 for d=20.option :2
8 sided polygon

Like? Yes (1) | No   

S. Singh   14 years AGO

d= n(n-3)/2 , here d=20 , so n=8

Like? Yes (1) | No (1)  

~Sandeep~   14 years AGO

(n(n-3))/2 =20 => n=8

Like? Yes (1) | No (1)  

sanjay   14 years AGO

Formula for the number of diagonals in a regular polygon:
D = (n(n-3)/2)
Here D = number of diagonals which is 20
n = number of sides which is not known

20=(n^2-3n)/2

40=n^2-3n
n^2-3n-40= 0
By factorizing the equation you will get

n^2-8n+5n-40=0
n(n-8)+5(n-8)=0
(n-8)(n+5)= 0

So either
n-8=0 or n+5=0
n=8 or n=-5
n can not be negative therefore
n = 8 sides or regular Octagon...... Option 2)

Like? Yes (1) | No (1)  

Gunaseelan   14 years AGO

Formula:(x-3)*x/2=no' of diagonals;
Given Diagonals=20;
(x-3)*x/2=20;
x^2-3x=40;
x^2-8x+5x-40=0;
x=8

Like? Yes (1) | No (1)  

Sn. Babu   14 years AGO

d = (n(n-3)/2)
20=n^2-3n/2
n^2-3n-40=0
hence n=8 and n=-5(nt possible)

Like? Yes (1) | No (3)