Number of triangles formed by diagonals in a polygon

Consider a regular polygon with $n$ number of vertices $\mathrm{A_1, \ A_2,\ A_3, \ A_3, \ldots , A_{n-1}}$ & $\mathrm{A_{n}}$

Total number of triangles formed by joining the vertices of n-sided regular polygon $$N=\text{number of ways of selecting 3 vertices out of n}=\color{}{\binom{n}{3}}$$ $$N=\color{red}{\frac{n(n-1)(n-2)}{6}}$$ $\forall \ \ \color{blue}{n\geq 3}$

Consider a side $\mathrm{A_1A_2}$ of regular n-polygon. To get a triangle with only one side $A_1A_2$ common (As shown in figure-1 below)

(figure-1)

Join the vertices $A_1$ & $A_2$ to any of $(n-4)$ vertices i.e. $A_4, \ A_5,\ A_6, \ \ldots \ A_{n-1}$ to get triangles with only one side common. Thus there are $(n-4)$ different triangles with only one side $A_1A_2$ common. Similarly, there are $(n-4)$ different triangles with only one side $A_2A_3$ common & so on. Thus there are $(n-4)$ different triangles with each of $n$ sides common. Therefore, number of triangles $N_1$ having only one side common with that of the polygon $$N_1=\text{(No. of triangles corresponding to one side)}\text{(No. of sides)}=\color{blue}{(n-4)n}$$

(figure-2)

Now, join the alternate vertices $A_1$ & $A_3$ by a straight (blue) line to get a triangle $A_1A_2A_3$ with two sides $A_1A_2$ & $A_2A_3$ common. Similarly, join alternate vertices $A_2$ & $A_4$ to get another triangle $A_2A_3A_4$ with two sides $A_2A_3$ & $A_3A_4$ common & so on (as shown in above figure-2). Thus there are $n$ pairs of alternate & consecutive vertices to get $n$ different triangles with two sides common (Above fig-2 shows $n$ st. lines of different colors to join alternate & consecutive vertices). Therefore, number of triangles $N_2$ having two sides common with that of the polygon $$N_2=\color{blue}{n}$$ If $N_0$ is the number of triangles having no side common with that of the polygon then we have $$N=N_0+N_1+N_2$$ $$N_0=N-N_1-N_2$$ $$=\binom{n}{3}-(n-4)n-n$$ $$=\color{}{\frac{n(n-1)(n-2)}{6}-n^2+3n}$$ $$N_0=\color{red}{\frac{n(n-4)(n-5)}{6}}$$ The above formula $(N_0)$ is valid for polygon having $n$ no. of the sides such that $ \ \ \color{blue}{n\geq 6}$

We will learn how to find the number of triangles contained in a polygon.

If the polygon has ‘n’ sides, then the number of triangle in a polygon is (n – 2).

In a triangle there are three sides. In the adjoining figure of a triangle ABC we can observe that the number of triangles contained = 3 – 2 = 1.

In a quadrilateral there are four sides. Number of triangles contained in a quadrilateral = 4 – 2 = 2.

In the adjoining figure of a quadrilateral ABCD, if diagonal BD is drawn, the quadrilateral will be divided into two triangles i.e. ∆ABD and ∆BDC.

In a pentagon there are five sides. Number of triangles contained in a pentagon = 5 – 2 = 3.

In the adjoining figure of a pentagon ABCDE, on joining AC and AD, the given pentagon is divided into three triangles i.e. ∆ABC, ∆ACD and ∆ADE.

In a hexagon there are six sides. Number of triangles contained in a hexagon = 6 – 2 = 4.

In the adjoining figure of a hexagon ABCDEF, on joining AC, AD and AE, the given hexagon is divided into four triangles i.e. ∆ABC, ∆ACD, ∆ADE and ∆AEF.

● Polygons

Polygon and its Classification

Terms Related to Polygons

Interior and Exterior of the Polygon

Convex and Concave Polygons

Regular and Irregular Polygon

Number of Triangles Contained in a Polygon

Angle Sum Property of a Polygon

Problems on Angle Sum Property of a Polygon

Sum of the Interior Angles of a Polygon

Sum of the Exterior Angles of a Polygon

7th Grade Math Problems