Q: The point at which the diagonals of a parallelogram intersect is. equdistant from the four vertices?

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No because the diagonals of a parallelogram are of different lengths

The given vertices when plotted on the Cartesian plane forms a rectangle with diagonals of square root of 50 in lengths and they both intersect at (3.5, 4.5)

They are diagonals. In a rhombus, diagonals join opposite vertices.

A parallelogram has 4 vertices

They are straight joining vertices to non-adjacent vertices.

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No because the diagonals of a parallelogram are of different lengths

sometimes

sometimes

Suppose that the parallelogram is a rhombus (a parallelogram with equal sides). If we draw the diagonals, isosceles triangles are formed (where the median is also an angle bisector and perpendicular to the base). Since the diagonals of a parallelogram bisect each other, and the diagonals don't bisect the vertex angles where they are drawn, then the parallelogram is not a rhombus.

The three major diagonals in an ordinary hexagon do not intersect at the same point. Therefore, in such a hexagon, the diagonals form 111 triangles.

The given vertices when plotted on the Cartesian plane forms a rectangle with diagonals of square root of 50 in lengths and they both intersect at (3.5, 4.5)

The vertices of a rhombus have no right angles but its diagonals intersect each other at right angles.

They are diagonals. In a rhombus, diagonals join opposite vertices.

This is false for all rhomboids (a distinct parallelogram such that 4 sides are equal, and has non-right angles), since by congruency, a parallelogram can be flipped on its axis (with 2 closer vertices), producing 2 unequal length diagonals.

Only a square and a rhombus will have all its diagonals bisecting vertices. In other shapes some - but not all - diagonals can bisect vertices.

A parallelogram has 4 vertices

They are straight joining vertices to non-adjacent vertices.