Question

# A polygon has 27 diagonals. How many sides does it have??

Solution

## The number of diagonals in any polygon is = n(n−3)/2  Given that this is equal to 27, we have that n(n−3)=2⋅27=54 , => n2−3n−54=0. Now you could just try values of n until you find one that works (shouldn’t take too long) or you could solve the quadratic equation if you know how to do that. n2−3n−54=0 (n-9)(n+6)=0 ,This is not solve the quadratic equation by the method you know. The solutions are: n=9 or n=−6. As n is no.of sides ,cannot be negative so we have n=No.of sides=9 Explanation for no.of diagonals: If you want to know more about the above equation ,refer below Let’s assume the polygon has n sides. We’ll try to work out what nn is from the information given. First, let’s consider just 1 vertex of the polygon. How many diagonals are there in that point? Well, a diagonal is a line connecting non-adjacent vertices. So, the point itself and its 2 neighbors do not have a diagonal arriving in this point, but all other (n−3) vertices do. So, we have (n−3) diagonals arriving in this point. Now, we have n such points. We might be tempted to say that n(n−3) is the number of diagonals, but that would be wrong! Note that every diagonal arrives in 2 vertices, one for each end-point. So, in n(n−3), we counted all diagonals twice. The number of diagonals in any polygon is = n(n−3)/2

Suggest corrections