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The diagonals of a square are perpendicular bisectors of each other?
Yes
Are diagonals on a parallelogram perpendicular bisectors?
No, they are just bisectors. The angle between them is not (usually) the 90o required to be perpendicular.
Why are the diagonals of a 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.
Are the diagonals of a rectangle sometimes always or never perpendicular?
The diagonals of a rectangle are never perpendicular but the diagonals of a square are perpendicular
If the diagonals of a quadrilateral are perpendicular can the quadrilateral be a rectangle?
Sure, a square is a rectangle and the diagonals of a square are perpendicular.
What kind of parellogram is it If the diagonals of a parallelogram are congruent and are perpendicular bisectors of each other?
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.
The diagonals of a square are perpendicular bisectors of each other?
Yes
Are the diagonals of a square perpendicular bisectors to each other?
Yes
What quadrilateral has the diagonals of a perpendicular bisectors of equal length?
A square.
The diagonals of a rhombus are perpendicular bisectors of one another?
True, the diagonals of a rhombus are perpendicular bisectors of one another.
What is the center of rotation for square?
Where the diagonals meet. Also where the perpendicular bisectors of the sides meet.
What type of parallelogram always has diagonals that are congruent and are perpendicular bisectors of each other?
square
Are diagonals on a parallelogram perpendicular bisectors?
No, they are just bisectors. The angle between them is not (usually) the 90o required to be perpendicular.
Diagonals of a rhombus are?
Perpendicular bisectors of each other.
Why are the diagonals of a 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.
A shape that has 4 lines of symmerty?
A square is on example. The perpendicular bisectors of the sides and the two diagonals comprie four lines of symmetry.
What diagonals are always perpendicular?
The diagonals of a square are always perpendicular.
