They are diagonals. In a rhombus, diagonals join opposite vertices.
Any type of rhombus has perpendicular diagonals. Please note that squares are a type of rhombus.
Two - the two diagonals.
Two - its diagonals.
The diagonals of a rhombus bisect one another at right angles. So you can use Pythagoras on half the lengths of the diagonals. If the two diagonals ore of lengths a and b, then side2 = (a/2)2 + (b/2)2 or, equivalently, side = 1/2*sqrt(a2 + b2)
Since the diagonals of a rhombus are perpendicular between them, then in one forth part of the rhombus they form a right triangle where hypotenuse is the side of the rhombus, the base and the height are one half part of its diagonals. Let's take a look at this right triangle.The base and the height lengths could be congruent if and only if the angles opposite to them have a measure of 45⁰, which is impossible to a rhombus because these angles have different measures as they are one half of the two adjacent angles of the rhombus (the diagonals of a rhombus bisect the vertex angles from where they are drawn), which also have different measures (their sum is 180⁰ ).Therefore, the diagonals of a rhombus are not congruent as their one half are not (the diagonals of a rhombus bisect each other).
They are diagonals. In a rhombus, diagonals join opposite vertices.
It has two diagonals, and they are perpendicular to each other.
Any type of rhombus has perpendicular diagonals. Please note that squares are a type of rhombus.
The diagonals of a rhombus (not rombhus) bisect one another at right angles. The sides of the rhombus form the hypotenuses of triangles whose other sides are half the diagonals. So use Pythagoras.
Two - the two diagonals.
I'm awesome! I have like 3 diagonals...
The diagonals of a rhombus are lines drawn from one corner, or vertex, to the opposite one. They have two important properties. 1. Diagonals bisects a pair of opposite angles. 2. Diagonals are !!perpendicular!!
Two - its diagonals.
The diagonals of a rhombus bisect one another at right angles. So you can use Pythagoras on half the lengths of the diagonals. If the two diagonals ore of lengths a and b, then side2 = (a/2)2 + (b/2)2 or, equivalently, side = 1/2*sqrt(a2 + b2)
yes
No but its diagonals are perpendicular to each other