Question

The number of intersection points of diagonals of 2009 sides regular polygon with lie inside the polygon​

Answer 1

$\huge\star{\tt{\underline{\underline{\red{Answer}}}}}\star$

Each intersection. engage four vertices of polygon so number of intersection points will be

$\large{\boxed{\boxed{^{2004}C_{4}}}}$

Answer 2

2015027 diagnols can be made from a regular polygon of sides 2009.

Explanation:

As we know for a regular polygon, no. of diagonals can be find by the formula:

              No. of diagonals = $\frac{n(n-3)}{2}$

No. of diagonals are the ratio of product of no. of sides and the no. of sides minus 3 to 2.

Here is the no. of sides of a regular polygon.

So, according to this formula,

                               ⇒   No. of diagonals = $\frac{2009 (2009 - 3)}{2}$

                               ⇒ No. of diagonals = $\frac{2009\times2006}{2}$

                               ⇒  No. of diagonals = 2015027

Diagonals are those lines that connect two vertices in a polygon or regular polygon from inside.