You could prove this by congruent triangles, but here are two simpler arguments: --------------- Since a square is a rhombus, and the diagonals of a rhombus are perpendicular bisectors of each other, then the diagonals of a square must be perpendicular bisectors of each other -------------------- A square has four-fold rotational symmetry - as you rotate it around the point where the diagonals cross, there are four positions in which it looks the same. This means that the four angles at the centre must be equal. They will each measure 360/4 = 90 degrees, so the diagonals are perpendicular. Also. the four segments joining the centre to a vertex are all equal, so the diagonals bisect each other.
The diagonals of a square are congruent, bisect each other, perpendicular, and either diagonal's length is sqrt(2) times any side length.
A square, a rhombus and a kite are three examples of quadrilaterals that have perpendicular diagonals that intersect each other at right angles.
Only for a square or rhombus (diamond shape). The diagonals of a rectangle bisect each other, but are not perpendicular and do not bisect the opposite angles they join.
It is a kite or a rhombus both of which have unequal diagonals that are perpendicular to each other creating right angles.
If the diagonals are congruent and are perpendicular bisectors of each other then the parallelogram is a square. If the diagonals are not congruent but are perpendicular bisectors of each other then the figure would be a rhombus.
Yes
square
Perpendicular bisectors of each other.
You could prove this by congruent triangles, but here are two simpler arguments: --------------- Since a square is a rhombus, and the diagonals of a rhombus are perpendicular bisectors of each other, then the diagonals of a square must be perpendicular bisectors of each other -------------------- A square has four-fold rotational symmetry - as you rotate it around the point where the diagonals cross, there are four positions in which it looks the same. This means that the four angles at the centre must be equal. They will each measure 360/4 = 90 degrees, so the diagonals are perpendicular. Also. the four segments joining the centre to a vertex are all equal, so the diagonals bisect each other.
Not necessarily - the diagonals of a rhombus bisect each other (they are perpendicular bisectors of each other), but are not equal.
It is a rhombus
No but the diagonals of a square, rhombus and a kite are perpendicular to each other
It is a square because its diagonals are equal in length and they bisect each other at right angles which is 90 degrees The diagonals of a rhombus are not equal in length but they meet at right angles.
If you are talking about the diagonals of a quadrilateral, the only quadrilateral that have diagonals that are perpendicular and bisect each other is a square, because a rectangle has bisecting diagonals, while a rhombus has perpendicular diagonals. And a square fits in both of these categories.
Not always but they are perpendicular in a square, a rhombus and a kite in that the diagonals intersect each other at 90 degrees
Yes they do.