Question
Question: The number of triangles that can be formed by using the vertices of a regular polygon of \(\left( n+...
The number of triangles that can be formed by using the vertices of a regular polygon of (n+3) sides is 220. Then n is equal to:
1. 8
2. 9
3. 10
4. 11
5. 12
Solution
We first describe how triangles are formed and how we can make triangles by joining the vertices of a n- sided polygon. Then we will find the formula of choosing r things out of n things. If Tn be the number of triangles made, then we will use the formula to find the number of ways 3 points can be chosen from 10 points. We will put the values in the equation and solve it to find the solution of the problem.
Complete step-by-step solution:
According to the question it is asked to find n if the number of triangles that can be formed by using the vertices of a regular polygon of (n+3) sides is 220. A n- sided polygon has n vertices and n sides. We know that in order to form a triangle we need three points which are non-linear. We also have that no three points out of those n vertices are non-linear. As we are trying to create triangles by joining the vertices of a n- sided polygon, we have to choose 3 points out of those (n+3) points. So,
The number of triangles = n+3C3=220
Therefore, 3!.n!(n+3)!=220
So, it means,
(n+1)(n+2)(n+3)=1320⇒(n+1)(n+2)(n+3)=12×10×11⇒(n+1)(n+2)(n+3)=(9+1)(9+2)(9+3)
By the comparison of both sides it can be written that n=9.
Hence the correct answer is option 2.
Note: We have to remember that in the case of triangles, we don’t have to worry, but if we are asked about diagonal, where we need to find any 2 points, then at the end, we have to subtract the number of sides from the option as they will be considered as diagonals. So, the answer will be nC2−n.