That will depend on the lengths of the diagonals of the rhombus which are of different lengths and intersect each other at right angles but knowing the lengths of the diagonals of the rhombus it is then possible to work out its perimeter and area.
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...
Oh, dude, it's like the diagonals in a rhombus are totally those lines that connect the opposite corners. They're like the fancy crossroads of the rhombus world, making all those right angles and stuff. So yeah, if you ever need to find them, just look for those lines that cut the rhombus in half diagonally.
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