The diagonals of a square are always perpendicular.
Proof: Without loss of generality, assume the square has side length 1 and one vertex is at the origin. The square ABCD is given by:
A = (0,0) , B = (1,0) , C = (1,1) , D = (0,1)
The diagonals are d1=AC and d2=BD. Finding equations for each of them yields
d1 = x
d2 = 1-x (you can double check this)
So, the relative slopes are 1 and -1. Since their product is -1, they are perpendicular.
No but the diagonals of a square, rhombus and a kite are perpendicular to each other
No.
Not always but they are perpendicular in a square, a rhombus and a kite in that the diagonals intersect each other at 90 degrees
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.
none, no
The diagonals of a rectangle are never perpendicular but the diagonals of a square are perpendicular
The diagonals of a square are always perpendicular.
Sure, a square is a rectangle and the diagonals of a square are perpendicular.
The diagonals of a square are perpendicular whereas the diagonals of a rectangle are not perpendicular.
The diagonals of a square are perpendicular whereas the diagonals of a rectangle are not perpendicular.
No. The diagonals of a rhombus are perpendicular only if the rhombus is a square.
No but the diagonals of a square, rhombus and a kite are perpendicular to each other
No, not necessarily. It would have to also be a square or a kite in order to have perpendicular diagonals.
Yes
square
No.
No but the diagonals of a square intersect at right angles