Question

How many triangles can be formed by joining the vertices of an n-sided polygon? How many of these triangles have 

(i) exactly one side common with that of the polygon?

(ii) exactly two sides common with that of the polygon? 

(iii) no sides common with that of the polygon? 

Answer 1

Number of triangles formed by joining the vertices of the polygon = Number of selections of 3 points from n points

= ⁿC₃ = n(n - 1)(n - 2)/3.2.1

Let the vertices of the polygon be marked as A₁ , A₂ , A₃ , …, An .

(i) Select two consecutive vertices A₁ , A₂ of the polygon. For the required triangle, we can select the third vertex from the points A₄ , A₅ , …, A_n–1. This can be done in ^n–4C₁ ways. Also two consecutive points (end points of a side of polygon) can be selected in n ways. Hence, the total number of required triangles  = n ^n–4C₁ = n(n – 4).

(ii)  For the required triangle, we have to select three consecutive vertices of the polygon, that is, (A₁ A₂ A₃ ), (A₂ A₃ A₄), (A₃ A₄ A₅), …, (A_n A₁ A₂). This can be done in n ways.

(iii) Triangles having no side common + Triangles having exactly one side common + Triangles having exactly two sides common (with those of the polygon)

= Total number of triangles formed

⇒ Triangles having no side common with those of the polygon