$A$ polygon has $35$ diagonals,then the number of its sides is

The number of diagonals in a polygon of $m$ sides is

There are $10$ points in a plane,out of which $6$ are collinear. If $N$ is the number of triangles formed by joining these points,then . . . .

Suppose $t_n$ is the number of triangles formed using the vertices of a regular polygon of $n$ sides. If $t_{n+1} = t_n + 28$,then $n =$

Consider a rectangle $ABCD$ having $5, 6, 7, 9$ points in the interior of the line segments $AB, BC, CD, DA$ respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then $(\beta-\alpha)$ is equal to:

If the number of diagonals of a regular polygon is $35$,then the number of sides of the polygon is