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Question

If the sides AB, BC and CA of a ABC have a, b and c points lying on them respectively, then the number of triangles that can be constructed using these points as vertices, if
  1. two vertices lie on same side =a+b+cC3(aC3+bC3+cC3+abc)  
  2. two vertices lie on the same side =a+b+cC3abc  
  3. all the vertices lie on different sides =abc  
  4. all the vertices lie on different sides =a+b+cC3(aC3+bC3+cC3)


Solution

The correct options are
A two vertices lie on same side =a+b+cC3(aC3+bC3+cC3+abc)  
C all the vertices lie on different sides =abc  
Total number of triangles = total number of ways of choosing 3 points number of ways of choosing all the 3 points from AB or BC or CA
=a+b+cC3(aC3+bC3+cC3)
Number of triangle formed such that all the vertices lies on different sides =aC1bC1cC1=abc

Number of triangles formed such that two of the vertices lies on same side =a+b+cC3(aC3+bC3+cC3+abc)

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