How many diagonals does a 63-sided convex polygon have?

Question

A. 1890
B. 1953
C. 3780
D. 3843
E. 3906

BrentGMATPrepNow · Accepted Answer

A 63-sided convex polygon has 63 vertices. 
If we examine a single vertex, we can see that we can connect it with 60 other vertices to create a diagonal. NOTE: there are 60 options because we can't connect the vertex to ITSELF, and we can't connect it to its ADJACENT vertices, since this would not create a diagonal.

If each of the 63 vertices can be connected with 60 vertices to create a diagonal then...
...the total number of diagonals = (63)(60) = 3780
 HOWEVER , before we select answer choice C, we must recognize that we have counted every diagonal TWICE. 
For example, we might connect vertex A with vertex F and count that as 1 diagonal, and at the same time we connect vertex F with vertex A and count that as 1 diagonal. Of course these diagonals are the SAME. 
To account for counting each diagonal twice, we must divide 3780 by 2 to get: 1890
Answer:  A

Cheers,
Brent

chetan2u · Answer

Hi,

the method to find number of diagonals in a polygon is nC2 - n...
 nC2 is the number of selecting two vertex, but these will also include the adjoining vertex, which will make the EDGES and not diagonal..
so we subtract EDGES from the total .....

ans =  nC2-n = 63C2 - 63 =\frac{63!}{61!2!}- 63 = 63*31 -63 = 63(31-1) = 63 * 30 = 1890 
A

subhamgarg91 · Answer

63 sided polygon will have 63 vertices.

No. of lines connecting any two vertices = 63C2

So, no. of diagonals = 63C2 - 63(no. of sides) = 1890

Abhishek009 · Answer

No fo diagonals of an n sided polygon is -  \frac{n(n − 3)}{2}  =  \frac{(63*60)}{2}  => 1890

Hence answer will be (A) 1890

shashankism · Answer

The number of diagonals of a convex polygon = (No. of combination of 2 points of polygon ) - (no. of sides)
63C2 - 63 
63 * 62/2 - 63 = 1890

Answer A

jaoliveraaravena · Answer

There is a rule for this:
For any geometric figure, the number of sides is equal to the number of vertexes. Let "n" be that number.

At first, you may think that n times n is the number of diagonals but that is not the case for two reasons.

First, with such procedure, you are counting each line joining two vertexes, including the edges of the figure, which are not diagonals. So, for each vertex, you need to discard two lateral lines. That is n-2. In addition, you need to discard the same vertex you are focusing on. That is n-2-1.

So far, we have n vertexes, each with n-1-2 diagonals coming from it, but that is not the end for a second reason. With n(n-2-1) you are counting each diagonal twice; so divide the number by 2.

The formula is then: n(n-3)/2.

For this question, we have that n=63. Let's operate.

63(63-3)/2 ---> 63*60/2 ---> 63*30 ---> 1890. So A is the answer.

GMATinsight · Answer

Method 1

A diagonal is aline joining two vertices and two vertices out of 63 may be chosen in 63C2 ways

BUT If two vertices chosen are adjacent then they will NOT form a diagonal instead they will for a side of the polygon hence we need to remove those cases.
Since the polygon has 63 sides therefore there will 63 such undesired cases

Favourable cases = 63C2 - 63 = 1953-63 = 1890

Method 2

From any one vertex 60 diagonals can be drawn (It can't join with itself and can't be joined with adjacent vertices to form a diagonal)

Total Such line that may be drawn from all 63 vertices = 60*63

But every line has been drawn twice because point joined with point 3 and then point 3 is joined with point 1 when it's 60 cases are considered and likewise)

Therefore total unique cases = (60*63)/2 = 1890

Answer: Option A

Kinshook · Answer

The number of diagonals = n(n-3)/2 = 63*62/2 = 63*31 = 1953

IMO B

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How many diagonals does a 63-sided convex polygon have?

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How many diagonals does a 63-sided convex polygon have?

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