In a regular polygon with 10 sides, find the number of triangles that can be formed with the vertices of the polygon, such that the triangles formed have at least one side in common with the polygon?
Explanation:
Exactly one common side: We can chose one common side in 10 ways (any one of the 10 sides of the polygon)
If we chose two adjacent vertices of the polygon, we will get one side which will be common to polygon and the triangle. Now the next vertex should be chose such that we don’t get anymore common sides. Hence, the 3rd vertex should not be adjacent to the ones already chose.
∴ The 3rd vertex can be chose in 6 ways.
∴ Total number of such triangles = 6 × 10 = 60
Exactly one common side: For two sides to be common we need to choose 3 adjacent vertices of the polygon. This can be done in 10 ways. (i.e., ABC, BCD, CDE, …, JAB)
∴ Total number of such triangles = 10
⇒ Total number of triangles = 60 + 10 = 70.
Hence, option (d).
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