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

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

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

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

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.

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

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

IMO B

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