Question

Task 4- Application
Draw a Quadrilateral, a Pentagon and a Hexagon
Draw all the diagonals for each polygon and count the number of diagonals.
Based on above, form an algebraic expression for the number of diagonals a polygon of n number of sides will have.​

Answer 1

Answer:

Hello!

I am a maths nut and I love solving problems that deal with non routine mathematics. So , to find the generalized formula for finding number of diagonals of a n sided polygon , we can use the concept of permutation and combination.

In this answer , I am going to explain the concept of permutation and combination in detail and in the end how we can apply this concept to find the formula.

Permuatation And Combination

1)Permutation is the number of ways of selecting and arranging suppose n number of articles out of a total of m number of articles. The total permutations of r number of objects from n number of objects is denoted by n(sub)P(super sub)r

2)Combination is the number of ways by which you can select n number of articles out of a total of m number of articles. The total combinations of r number of objects from n number of objects is denoted by n(sub)C(super sub)r

There is a formula to find n(sub)P(super sub)r and n(sub)C(super sub)r . It is as follows-

n(sub)P(super sub)r = (n)!/(n-r)! and

n(sub)C(super sub)r = (n)!/[{(r)!}*{(n-r)!}]

Now that we have understood the concept of Permutation and combination, let's get back to the proving.

Let's consider an n sided polygon. A diagonal has teo end points isn't it? And those two points are the vertices of the n sided polygon. So what if we take n(sub)C(super sub)2? Won't it be the number of diagonals of the polygon. But if we write n(sub)C(super sub)2 then in this set , it will also consider the lines which are nothing but the sides of the polygon. So we have to subtract n from it as n is the number of sides of the n sided polygon. So the number of diagonals will be -

n(sub)C(super sub)2 - n

Let's solve this

{n! / 2!*(n-2)!} - n

{(n-1)(n)}/2 - n

{(n-1)(n) - 2n}/2

Taking n common

n(n-2-1)/2

n(n-3)/2

You can verify it as well.

Suppose you want to find the number of diagonals of a quadrilateral. It is a 4 sided polygon right? Put 4 in place of n. You get 2. And 2 indeed is the number of diagonals a quadrilateral has.

Hope it is clear to all

(Note: n(sub)P(super sub)r means n

P

r and

n(sub)C(super sub)r means n

C

r

and r can be any positive integer smaller than or equal to n)

Hope you all have understood

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