The height and longer diagonal do not provide enough information to calculate the sides.
The diagonals are perpendicular to one another. The shorter diagonal is bisected by the longer diagonal. The kite is symmetrical about the longer diagonal. The longer diagonal bisects the angles at each end of the diagonal.
Yes, but only when the shape is a rectangle (or square). Other paralleograms will have one diagonal longer than the other. And yes, rectangles and squares ARE parallelograms.
Yes. Starting with a square, shear it by sliding one edge parallel to the opposite edge; this creates a parallelogram with its height the length of the side of the square which is two of the sides; the other two sides will be longer.
Kites, arrowheads.
It can be any length you like greater than (or equal to) 4 x √60 ≈ 30.98 A parallelogram is a sheared rectangle; the more the rectangle shears, the longer the originally "vertical" sides (the height of the parallelogram) become. (Shearing does not change the area of the shape.)
The diagonals are perpendicular to one another. The shorter diagonal is bisected by the longer diagonal. The kite is symmetrical about the longer diagonal. The longer diagonal bisects the angles at each end of the diagonal.
Yes, but only when the shape is a rectangle (or square). Other paralleograms will have one diagonal longer than the other. And yes, rectangles and squares ARE parallelograms.
Yes. Starting with a square, shear it by sliding one edge parallel to the opposite edge; this creates a parallelogram with its height the length of the side of the square which is two of the sides; the other two sides will be longer.
The longer diagonal bisects the shorter diagonal.
Kites, arrowheads.
It can be any length you like greater than (or equal to) 4 x √60 ≈ 30.98 A parallelogram is a sheared rectangle; the more the rectangle shears, the longer the originally "vertical" sides (the height of the parallelogram) become. (Shearing does not change the area of the shape.)
I guess the diagonal length given is from one corner of the box to the opposite corner reached by traversing one length side, one edge side and one height side. Using Pythagoras, the length of the diagonal of the base (length by width) can be found. Using this diagonal and the height of the box, the diagonal from corner-to-opposite-corner of the box can be found using Pythagoras. However, as this [longer] diagonal is know, the height can be found by rearranging this last use of Pythagoras: Diagonal_base2 = length2 + width2 Diagonal_box2 = diagonal_base2 + height2 ⇒ height = √(diagonal_box2 - diagonal_base2 ) = √(diagonal_box2 - (length2 + width2)) = √(diagonal_box2 - length2 - width2) Now that the formula has been derived, plugging in (substituting) the various lengths will allow the height to be calculated.
If you look ate the parallelogram you'll see two kinds of triangles. Two that have longer diagonal and bigger angle, and two sides of parallelogram. Then, you have two triangles that have two sides of parallelogram, shorter diagonal and smaller angle. This triangles obviously have two sides that are the same (sides of parallelogram). If this two triangles had been congruent diagonals would have been congruent too, since these triangles would have been congruent. But this is not true unless angles of parallelogram are the same, therefore diagonals cannot be the same length. Of course, there are parallelograms that have same angles, and those are square and rectangle, which do have the same angles. I hope I made this more clear, and I'm sorry for my bad English.
No, they do not. Only the longer diagonal bisects the shorter diagonal.
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Yes.