Polygon -Geometry

Here is the solution
Formula to find diagonals in a polygon:
D = (n(n-3)/2)
D = number of diagonals that is 20
n = ?

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

40=n^2-3n
n^2-3n-40= 0
Splitting the middle term,

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

So there are two values for n
n-8=0 or n+5=0
n = 8 or n = -5
n can't be negative
So n = 8

5th grade math

Consider a polygon with n vertices. Each vertex can be connected to the n-1 other vertices but two of those connections (to the two adjacent vertices) form sides of the polygon, not diagonals. So there are n-3 diagonals emanating from each of the n vertices giving n(n-3) diagonals. This formula overcounts because it counts each diagonal twice: once for each vertex on the diagonal. Divide by two to compensate for the overcounting giving n(n-3)/2 diagonals.

You can set n(n-3)/2 = 20, expand and solve the quadratic equation by factorization but this is time consuming and inefficient. Way faster is to simply plug in the answer choices for n and see which one satisfies the equation: n(n-3) = 40. Ans: n=8.

Thank You Tutorphd for the explanation but do you think that this question can be a part of GMAT. till date in case of polygons, i have seen questions only abt internal angles which has a basic formula.Moreover, the guides also teach only abt the angles formula.

@poster, what is the source of the question.??

tutorphd wrote:Consider a polygon with n vertices. Each vertex can be connected to the n-1 other vertices but two of those connections (to the two adjacent vertices) form sides of the polygon, not diagonals. So there are n-3 diagonals emanating from each of the n vertices giving n(n-3) diagonals. This formula overcounts because it counts each diagonal twice: once for each vertex on the diagonal. Divide by two to compensate for the overcounting giving n(n-3)/2 diagonals.

You can set n(n-3)/2 = 20, expand and solve the quadratic equation by factorization but this is time consuming and inefficient. Way faster is to simply plug in the answer choices for n and see which one satisfies the equation: n(n-3) = 40. Ans: n=8.

shekhar.kataria wrote:Thank You Tutorphd for the explanation but do you think that this question can be a part of GMAT. till date in case of polygons, i have seen questions only abt internal angles which has a basic formula.Moreover, the guides also teach only abt the angles formula.

This is really a combinatorics question in disguise; it can be solved without knowing the formula for the number of diagonals.
Check my post here:

https://www.beatthegmat.com/polygon-geom ... tml#482906

That being said, questions about polygons are rare.

By doing Excel with formula n*(n-3) / 2, you can have an idea of the different outcomes

sides diagonals
1 -1
2 -1
3 0
4 2
5 5
6 9
7 14
8 20
9 27
10 35
11 44
12 54
13 65
14 77
15 90
16 104
17 119
18 135
19 152
20 170
21 189
22 209
23 230
24 252
25 275
26 299

Yes as a previous poster said, this question has very little geometry in it, it's a combinatorics in disguise. Questions like that do happen on GMAT but not as frequent as more obvious combinatorics questions. I can think of about two problems like that in OG12.

If question changes to 44 , then:


A polygon has 44 diagonals, How Many Sides Does It Have?


A) 12

B) 11

C) 10

D) 9

E) 8

Consider a polygon with n vertices. Each vertex can be connected to the n-1 other vertices but two of those connections (to the two adjacent vertices) form sides of the polygon, not diagonals. So there are n-3 diagonals emanating from each of the n vertices giving n(n-3) diagonals. This formula overcounts because it counts each diagonal twice: once for each vertex on the diagonal. Divide by two to compensate for the overcounting giving n(n-3)/2 diagonals.

You can set n(n-3)/2 = 44, expand and solve the quadratic equation by factorization but this is time consuming and inefficient. Way faster is to simply plug in the answer choices for n and see which one satisfies the equation: n(n-3) = 88. Ans: n=11.
The equation a*x^2 + b*x + c = 0 has as solutions: x = ( -b + sqrt(b^2-4*a*c) ) / (2*a) or x = ( -b - sqrt(b^2-4*a*c) )

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

88=n^2-3n

So there are two values for n
n = 11 or n = -8
n can't be negative
So n = 11


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