Parallelogram

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A parallelogram.

A parallelogram is a four-sided plane figure that has two sets of opposite parallel sides. Every parallelogram is a polygon, and more specifically a quadrilateral. Special cases of a parallelogram are the rhombus, in which all four sides are of equal length, the rectangle, in which the two sets of opposing, parallel sides are perpendicular to each other, and the square, in which all four sides are of equal length and the two sets of opposing, parallel sides are perpendicular to each other. In any parallelogram, the diagonals bisect each other, i.e, they cut each other in half. The parallelogram law distinguishes Hilbert spaces from other Banach spaces.

It is possible to create a tessellation with any parallelogram.

The three-dimensional counterpart of a parallelogram is a parallelepiped.

The area of a parallelogram can be seen as twice the area of a triangle created by one of its diagonals. The area of a parallelogram can be found by using the formula $B\times H=A$ (Base multiplied by Height equals area). The area can also be computed as the magnitude of the vector cross product of two of its non-parallel sides.

1 Proof that diagonals bisect each other
2 Area
- 2.1 Alternate method
3 See also
4 External links

[edit] Proof that diagonals bisect each other

Prove that the diagonals of a parallelogram bisect each other.

(Prove that $E/\overrightarrow{A C}=1:1$ and $E/\overrightarrow{B D}=1:1$ )

Proof:

$\Rightarrow \overrightarrow{A E}=k(\overrightarrow{A D}+\overrightarrow{A B})$ since $\overrightarrow{D C}=\overrightarrow{A B}$

$\overrightarrow{A E}=k\overrightarrow{A D}+k\overrightarrow{A B}$

since E,D,B are collinear, by the division-point theorem,

k + k = 1

2k = 1

k = 0.5

sub k = 0.5

$\overrightarrow{A E}=k\overrightarrow{A C}$

$\overrightarrow{A E}=(0.5)\overrightarrow{A C}$

$\overrightarrow{A E} / \overrightarrow{A C}=0.5$ (the ratio of AE to AC is 1:2)

also sub k = 0.5 into:

$\overrightarrow{A E}=k\overrightarrow{A D}+k\overrightarrow{A B}$

$\overrightarrow{A E}=0.5\overrightarrow{A D}+0.5\overrightarrow{A B}$

by the division-point theorem,

$E/\overrightarrow{D B}=1:1$

by adding the division ratios to the parallelogram, we see that E divides both diagonals in the ratio 1:1, and E bisects AC and BD.

Therefore, the diagonals of a parallelogram bisect each other.

[edit] Area

Area of the parallelogram is in blue

The area of a parallelogram can be derived in the following way. The area of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is

A r e c t = (B + A) H

and the area of one orange triangle is:

$A_{tri} = \frac{1}{2} A H$

Therefore, two triangles has the area of

A t r i s = A H

which is equivalent to the area of a rectangle of dimensions $A \times H$ . The area of the parallelogram is

$A_{parallelogram} = A_{rect} - 2 A_{tri} = A_{rect} - A_{tris} = \left( (B+A) H \right) - \left( A H \right) = B H$

[edit] Alternate method

Step one: ends of parallelogram are chopped off

Step two: pieces are rearranged

An alternative, less mathematically sophisticated method, to show the area is by rearrangement of the area. First, take the two ends of the parallelogram and chop them off to form two more triangles. Each of these two new triangles are equal in every way with the orange triangles. This first step is shown to the right.

The second step is merely swap the left orange triangle with the right blue triangle. Clearly, the two blue triangles plus the blue rectangle have an area equivalent to $B H$ .

To further demonstrate this, the first image on the right could be printed off and cut up along the lines:

Cut along the lines between the orange triangles and the blue parallelogram
Cut along the vertical lines on the end to form the two blue triangles and the blue rectangle
Rearrange all five pieces as shown in the second image