Question

In an even $n$-sided polygon, the sides are painted blue and red alternatively. The vertices of the polygon are joined to form triangles with one side in common with the polygon. Find the number of such triangles with one blue side.

• A. $n(n-4)$
• B. $2n(n-3)$
• C. $\frac{n}{2}(n-4)$
• D. $(n-3)(n-4)$

Answer 1

The correct option is C $\frac{n}{2}(n-4)$

No. of ways to select one side of a $n$-sided polygon is $n$.

No. of ways to select the third vertex to form a triangle would be $n-4$ as two vertices are already selected when we selected one side of the polygon and we can not select the two vertices adjacent to the side selected, since only one side is common with the polygon.

Therefore, total no. of triangles formed $=n(n-4)$
Since, only half of the sides are blue therefore, total no. of triangles with one blue side $=\frac{n}{2}(n-4)$