Answer :

i. We know that the decagon has 10 vertices and each side and diagonal can be formed by joining two vertices of a hexagon,

We know that decagon has 10 sides,

Let us assume the no. of diagonals of the hexagon are N,

⇒ N = (no. of lines formed on joining any two vertices) – (no. of sides of the hexagon)

⇒ N = ^{10}C_{2} – 10

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 45 – 10

⇒ N = 35

∴ The total no. of diagonals formed is 35.

ii. Given that we need to find the no. of triangles that can be drawn in a decagon.

We know that 3 points are required to draw a triangle.

We know that decagon has 10 sides

Let us assume the no. of triangles formed be N_{1},

⇒ N_{1} = (total no. of triangles formed by all 10 points)

⇒ N_{1} = ^{10}C_{3}

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{1} = 120

∴ The total no. of ways of different lines formed are 120.

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