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Q.

Consider a convex polygon which has 44 diagonals then the number of triangles joining the vertices of polygon in which exactly one side is common in triangle and polygon is


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a

77

b

70

c

76

d

75 

answer is A.

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Detailed Solution

https://www.vedantu.com/question-sets/0fef2c04-8fb0-4005-9605-82569018f86f6336035680054669771.png
  Now let us suppose there is a hexagon ABCDEF and from the point A the number of possible diagonals will be 3
And there are a total of six points so the total diagonal = (6)(3) = 18 but in this way we counted one diagonal twice.
For example from point A we have the diagonal AC and from point C also we have the same diagonal AC but these both are the same so need to be counted once only.
So the number of diagonals = (3)(6)2 = 9
Similarly for the square
Total number of diagonals = 1(4)2 Similarly for polygon of side n we can write
Total number of diagonals = n(n-3)2
Where n is the sides of the polygon.
So here we are given that there are 44 diagonals in the polygon.
Then n(n-3)2 = 44
⇒ n2 − 3n – 88 = 0
Upon solving we get
⇒ n2 − 11n + 8n – 88 = 0
⇒ n (n − 11) + 8 (n − 11) = 0
⇒ (n − 11) (n + 8) = 0  
So n = 11, −8  
n cannot be negative so the polygon has 11 sides
So the number of triangles can be formed having one side common to the polygon is given as
= n (n − 4)
= 11 (11 − 4)
= 11 (7)
= 77    So 77 possible triangles can be formed.
Option (1) is correct.
 
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