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Parallelogram

In Euclidean geometry, a parallelogram is a simple (non self-intersecting) quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. The congruence of opposite sides and opposite angles is a direct consequence of the Euclidean Parallel Postulate and neither condition can be proven without appealing to the Euclidean Parallel Postulate or one of its equivalent formulations. The three-dimensional counterpart of a parallelogram is a parallelepiped.

The etymology (in Greek - , a shape "of parallel lines") reflects the definition.

Contents


Characterizations

A simple (non self-intersecting) quadrilateral is a parallelogram if and only if any one of the following statements is true:[1][2]

Properties

  • Opposite sides of a parallelogram are parallel (by definition) and so will never intersect.
  • The area of a parallelogram is twice the area of a triangle created by one of its diagonals.
  • The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides.
  • Any line through the midpoint of a parallelogram bisects the area.[3]
  • Any non-degenerate affine transformation takes a parallelogram to another parallelogram.
  • A parallelogram has rotational symmetry of order 2 (through 180 ). If it also has two lines of reflectional symmetry then it must be a rhombus or an oblong.
  • The perimeter of a parallelogram is 2(a + b) where a and b are the lengths of adjacent sides.
  • The sum of the distances from any interior point of a parallelogram to the sides is independent of the location of the point. (This is an extension of Viviani's theorem). The converse also holds: If the sum of the distances from a point in the interior of a quadrilateral to the sides is independent of the location of the point, then the quadrilateral is a parallelogram.[4]

Types of parallelogram

  • Rhomboid A quadrilateral whose opposite sides are parallel and adjacent sides are unequal, and whose angles are not right angles
  • Rectangle A parallelogram with four angles of equal size
  • Rhombus A parallelogram with four sides of equal length.
  • Square A parallelogram with four sides of equal length and four angles of equal size (right angles).

Area formulas

The area of the parallelogram is the area of the blue region, which is the interior of the parallelogram

  • The area K 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_\text{rect} = (B+A) \times H\,
and the area of a single orange triangle is
A_\text{tri} = \frac{1}{2} A \times H. \,
Therefore, the area of the parallelogram is
\begin{align} K &= A_\text{rect} - 2 \times A_\text{tri} \\ &= \left( (B+A) \times H \right) - \left( A \times H \right) \\ &= B \times H \\ \end{align}
  • Another area formula, for two sides B and C and angle , is
K = B \cdot C \cdot \sin \theta.\,
  • The area of a parallelogram with sides B and C (B C) and angle \gamma at the intersection of the diagonals is given by[5]
K = \frac{|\tan \gamma|}{2} \cdot \left| B^2 - C^2 \right|.

The area on coordinate system

Let vectors \mathbf{a},\mathbf{b}\in\R^2 and let V = \begin{bmatrix} a_1 & a_2 \\ b_1 & b_2 \end{bmatrix} \in\R^{2 \times 2} denote the matrix with elements of a and b. Then the area of the parallelogram generated by a and b is equal to |\det(V)| = |a_1b_2 - a_2b_1|\,.

Let vectors \mathbf{a},\mathbf{b}\in\R^n and let V = \begin{bmatrix} a_1 & a_2 & \dots & a_n \\ b_1 & b_2 & \dots & b_n \end{bmatrix} \in\R^{2 \times n} Then the area of the parallelogram generated by a and b is equal to \sqrt{\det(V V^\mathrm{T})}.

Let points a,b,c\in\R^2. Then the area of the parallelogram with vertices at a, b and c is equivalent to the absolute value of the determinant of a matrix built using a, b and c as rows with the last column padded using ones as follows:

K = \left| \det \begin{bmatrix} a_1 & a_2 & 1 \\ b_1 & b_2 & 1 \\ c_1 & c_2 & 1 \end{bmatrix} \right|.

Proof that diagonals bisect each other

Parallelogram ABCD To prove that the diagonals of a parallelogram bisect each other, we will use congruent triangles:

\angle ABE \cong \angle CDE (alternate interior angles are equal in measure)
\angle BAE \cong \angle DCE (alternate interior angles are equal in measure).

(since these are angles that a transversal makes with parallel lines AB and DC).

Also, side AB is equal in length to side DC, since opposite sides of a parallelogram are equal in length.

Therefore triangles ABE and CDE are congruent (ASA postulate, two corresponding angles and the included side).

Therefore,

AE = CE
BE = DE.

Since the diagonals AC and BD divide each other into segments of equal length, the diagonals bisect each other.

Separately, since the diagonals AC and BD bisect each other at point E, point E is the midpoint of each diagonal.

See also

References

External links

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