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**Hint:**As we know for a polygon of the side $n$, the total number of diagonals is given by $\dfrac{{n(n - 3)}}{2}$

As we are given that the number of diagonals is $44$, then we can get the number of sides of the polygon. Now we can find the number of sides of the triangles.

**Complete step-by-step answer:**

We need to find the number of diagonals if the polygon contains \[n\] sides. Now let us suppose there is a hexagon $ABCDEF$ the 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 $ = \dfrac{{(3)(6)}}{2} = 9$

Similarly for the square

Total number of diagonals $ = \dfrac{{1\left( 4 \right)}}{2}$

Similarly for polygon of side n we can write

Total number of diagonals $ = \dfrac{{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 $\dfrac{{n(n - 3)}}{2} = 44$

$\Rightarrow$${n^2} - 3n - 88 = 0$

Upon solving we get

$\Rightarrow$${n^2} - 11n + 8n - 88 = 0$

$\Rightarrow$$n(n - 11) + 8(n - 11) = 0$

$\Rightarrow$$(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.**

**Note:**Number of triangles formed by joining the vertices of n sided polygon which has no side in common with that of the polygon is given by $\dfrac{{n(n - 4)(n - 5)}}{{3!}}$ where n is the side of the polygon.

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