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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answer is A.

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Detailed Solution

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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 =

\frac{(3) (6)}{2}

= 9
Similarly for the square
Total number of diagonals =

\frac{1 (4)}{2}

Similarly for polygon of side n we can write
Total number of diagonals =

\frac{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

\frac{n (n - 3)}{2}

= 44
⇒

n^{2}

− 3n – 88 = 0
Upon solving we get
⇒

n^{2}

− 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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