Movatterモバイル変換

[0]ホーム
URL:

Jump to content
WikipediaThe Free Encyclopedia
Search

Parallelogram

From Wikipedia, the free encyclopedia
Quadrilateral with two pairs of parallel sides
This article is about the quadrilateral. For the music album, seeParallelograms (album).
Parallelogram
This parallelogram is a rhomboid as it has unequal sides and no right angles.
TypeQuadrilateral,Trapezium
Edges andvertices4
Symmetry groupC2, [2]+,
Areabh (base × height);
ab sinθ (product of adjacent sides and sine of the vertex angle determined by them)
PropertiesConvex polygon

InEuclidean geometry, aparallelogram is asimple (non-self-intersecting)quadrilateral with two pairs ofparallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. Thecongruence of opposite sides and opposite angles is a direct consequence of the Euclideanparallel postulate and neither condition can be proven without appealing to the Euclidean parallel postulate or one of its equivalent formulations.

By comparison, a quadrilateral with at least one pair of parallel sides is atrapezoid in American English or a trapezium in British English.

The three-dimensional counterpart of a parallelogram is aparallelepiped.

The word "parallelogram" comes from the Greek παραλληλό-γραμμον,parallēló-grammon, which means "a shape of parallel lines".

Special cases

[edit]
  • Rectangle – A parallelogram with four right angles.
  • Rhombus – A parallelogram with four sides of equal length.
  • Square – A parallelogram with four sides of equal length and four right angles.
  • Rhomboid – A parallelogram with adjacent sides that are of unequal lengths andnon-right angles (and thus is neither a rectangle nor a rhombus). This term is not used in modern mathematics but it does survive in some contexts in biology in names like therhomboid muscles orrhomboid leaf shapes.[1]

Characterizations

[edit]

Asimple (non-self-intersecting)quadrilateral is a parallelogramif and only if any one of the following statements is true:[2][3]

  • Two pairs of opposite sides are parallel (by definition).
  • Two pairs of opposite sides are equal in length.
  • Two pairs of opposite angles are equal in measure.
  • Thediagonals bisect each other.
  • One pair of opposite sides isparallel and equal in length.
  • Adjacent angles aresupplementary.
  • Each diagonal divides the quadrilateral into twocongruenttriangles.
  • The sum of thesquares of the sides equals the sum of the squares of the diagonals. (This is theparallelogram law.)
  • It hasrotational symmetry of order 2.
  • The sum of the distances from any interior point to the sides is independent of the location of the point.[4] (This is an extension ofViviani's theorem.)
  • There is a pointX in the plane of the quadrilateral with the property that every straight line throughX divides the quadrilateral into two regions of equal area.[5]

Thus, all parallelograms have all the properties listed above, andconversely, if just any one of these statements is true in a simple quadrilateral, then it is considered a parallelogram.

Other properties

[edit]
  • 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 thevector cross product of twoadjacent sides.
  • Any line through the midpoint of a parallelogram bisects the area.[6]
  • Any non-degenerateaffine transformation takes a parallelogram to another parallelogram.
  • A parallelogram hasrotational symmetry of order 2 (through 180°) (or order 4 if a square). If it also has exactly two lines ofreflectional symmetry then it must be a rhombus or anoblong (a non-square rectangle). If it has four lines of reflectional symmetry, it is asquare.
  • The perimeter of a parallelogram is 2(a +b) wherea andb are the lengths of adjacent sides.
  • Unlike any other convex polygon, a parallelogram cannot be inscribed in any triangle with less than twice its area.[7]
  • The centers of four squares all constructed either internally or externally on the sides of a parallelogram are the vertices of a square.[8]
  • If two lines parallel to sides of a parallelogram are constructedconcurrent to a diagonal, then the parallelograms formed on opposite sides of that diagonal are equal in area.[8]
  • The diagonals of a parallelogram divide it into four triangles of equal area.

Area formula

[edit]
A diagram showing how a parallelogram can be re-arranged into the shape of a rectangle
A parallelogram can be rearranged into a rectangle with the same area.
Animation for the area formulaK=bh{\displaystyle K=bh}.

All of thearea formulas for general convex quadrilaterals apply to parallelograms. Further formulas are specific to parallelograms:

A parallelogram with baseb and heighth can be divided into atrapezoid and aright triangle, and rearranged into arectangle, as shown in the figure to the left. This means that thearea of a parallelogram is the same as that of a rectangle with the same base and height:

K=bh.{\displaystyle K=bh.}
The area of the parallelogram is the area of the blue region, which is the interior of the parallelogram

The base × height area formula can also be derived using the figure to the right. The areaK 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

Krect=(B+A)×H{\displaystyle K_{\text{rect}}=(B+A)\times H\,}

and the area of a single triangle is

Ktri=A2×H.{\displaystyle K_{\text{tri}}={\frac {A}{2}}\times H.\,}

Therefore, the area of the parallelogram is

K=Krect−2×Ktri=((B+A)×H)−(A×H)=B×H.{\displaystyle K=K_{\text{rect}}-2\times K_{\text{tri}}=((B+A)\times H)-(A\times H)=B\times H.}

Another area formula, for two sidesB andC and angle θ, is

K=B⋅C⋅sin⁡θ.{\displaystyle K=B\cdot C\cdot \sin \theta .\,}

Provided that the parallelogram is a rhombus, the area can be expressed using sidesB andC and angleγ{\displaystyle \gamma } at the intersection of the diagonals:[9]

K=|tan⁡γ|2⋅|B2−C2|.{\displaystyle K={\frac {|\tan \gamma |}{2}}\cdot \left|B^{2}-C^{2}\right|.}


When the parallelogram is specified from the lengthsB andC of two adjacent sides together with the lengthD1 of either diagonal, then the area can be found fromHeron's formula. Specifically it is

K=2S(S−B)(S−C)(S−D1)=12(B+C+D1)(−B+C+D1)(B−C+D1)(B+C−D1),{\displaystyle K=2{\sqrt {S(S-B)(S-C)(S-D_{1})}}={\frac {1}{2}}{\sqrt {(B+C+D_{1})(-B+C+D_{1})(B-C+D_{1})(B+C-D_{1})}},}

whereS=(B+C+D1)/2{\displaystyle S=(B+C+D_{1})/2} and the leading factor 2 comes from the fact that the chosen diagonal divides the parallelogram intotwo congruent triangles.

From vertex coordinates

[edit]

Let vectorsa,b∈R2{\displaystyle \mathbf {a} ,\mathbf {b} \in \mathbb {R} ^{2}} and letV=[a1a2b1b2]∈R2×2{\displaystyle V={\begin{bmatrix}a_{1}&a_{2}\\b_{1}&b_{2}\end{bmatrix}}\in \mathbb {R} ^{2\times 2}} denote the matrix with elements ofa andb. Then the area of the parallelogram generated bya andb is equal to|det(V)|=|a1b2−a2b1|{\displaystyle |\det(V)|=|a_{1}b_{2}-a_{2}b_{1}|\,}.

Let vectorsa,b∈Rn{\displaystyle \mathbf {a} ,\mathbf {b} \in \mathbb {R} ^{n}} and letV=[a1a2…anb1b2…bn]∈R2×n{\displaystyle V={\begin{bmatrix}a_{1}&a_{2}&\dots &a_{n}\\b_{1}&b_{2}&\dots &b_{n}\end{bmatrix}}\in \mathbb {R} ^{2\times n}}. Then the area of the parallelogram generated bya andb is equal todet(VVT){\displaystyle {\sqrt {\det(VV^{\mathrm {T} })}}}.

Let pointsa,b,c∈R2{\displaystyle a,b,c\in \mathbb {R} ^{2}}. Then thesigned area of the parallelogram with vertices ata,b andc is equivalent to the determinant of a matrix built usinga,b andc as rows with the last column padded using ones as follows:

K=|a1a21b1b21c1c21|.{\displaystyle K=\left|{\begin{matrix}a_{1}&a_{2}&1\\b_{1}&b_{2}&1\\c_{1}&c_{2}&1\end{matrix}}\right|.}

Proof that diagonals bisect each other

[edit]
Parallelogram ABCD
Parallelogram ABCD

To prove that the diagonals of a parallelogram bisect each other, we will usecongruenttriangles:

∠ABE≅∠CDE{\displaystyle \angle ABE\cong \angle CDE}(alternate interior angles are equal in measure)
∠BAE≅∠DCE{\displaystyle \angle BAE\cong \angle DCE}(alternate interior angles are equal in measure).

(since these are angles that a transversal makes withparallel linesAB andDC).

Also, sideAB is equal in length to sideDC, since opposite sides of a parallelogram are equal in length.

Therefore, trianglesABE andCDE are congruent (ASA postulate,two corresponding angles and the included side).

Therefore,

AE=CE{\displaystyle AE=CE}
BE=DE.{\displaystyle BE=DE.}

Since the diagonalsAC andBD divide each other into segments of equal length, the diagonals bisect each other.

Separately, since the diagonalsAC andBD bisect each other at pointE, pointE is the midpoint of each diagonal.

Lattice of parallelograms

[edit]

Parallelograms can tile the plane by translation. If edges are equal, or angles are right, the symmetry of the lattice is higher. These represent the fourBravais lattices in 2 dimensions.

Lattices
FormSquareRectangleRhombusRhomboid
SystemSquare
(tetragonal)		Rectangular
(orthorhombic)		Centered rectangular
(orthorhombic)		Oblique
(monoclinic)
Constraintsα=90°, a=bα=90°a=bNone
Symmetryp4m, [4,4], order 8npmm, [∞,2,∞], order 4np1, [∞+,2,∞+], order 2n
Form

Parallelograms arising from other figures

[edit]

Automedian triangle

[edit]

Anautomedian triangle is one whosemedians are in the same proportions as its sides (though in a different order). IfABC is an automedian triangle in which vertexA stands opposite the sidea,G is thecentroid (where the three medians ofABC intersect), andAL is one of the extended medians ofABC withL lying on the circumcircle ofABC, thenBGCL is a parallelogram.

Varignon parallelogram

[edit]
Main article:Varignon's theorem
Proof without words of Varignon's theorem

Varignon's theorem holds that themidpoints of the sides of an arbitrary quadrilateral are the vertices of a parallelogram, called itsVarignon parallelogram. If the quadrilateral isconvex orconcave (that is, not self-intersecting), then the area of the Varignon parallelogram is half the area of the quadrilateral.

Proof without words (see figure):

  1. An arbitrary quadrilateral and its diagonals.
  2. Bases of similar triangles are parallel to the blue diagonal.
  3. Ditto for the red diagonal.
  4. The base pairs form a parallelogram with half the area of the quadrilateral,Aq, as the sum of the areas of the four large triangles,Al is 2Aq (each of the two pairs reconstructs the quadrilateral) while that of the small triangles,As is a quarter ofAl (half linear dimensions yields quarter area), and the area of the parallelogram isAq minusAs.

Tangent parallelogram of an ellipse

[edit]

For anellipse, two diameters are said to beconjugate if and only if thetangent line to the ellipse at an endpoint of one diameter is parallel to the other diameter. Each pair of conjugate diameters of an ellipse has a correspondingtangent parallelogram, sometimes called a bounding parallelogram, formed by the tangent lines to the ellipse at the four endpoints of the conjugate diameters. All tangent parallelograms for a given ellipse have the same area.

It is possible toreconstruct an ellipse from any pair of conjugate diameters, or from any tangent parallelogram.

Faces of a parallelepiped

[edit]

Aparallelepiped is a three-dimensional figure whose sixfaces are parallelograms.

See also

[edit]

References

[edit]
  1. ^"CIMT - Page no longer available at Plymouth University servers"(PDF).www.cimt.plymouth.ac.uk. Archived fromthe original(PDF) on 2014-05-14.
  2. ^Owen Byer, Felix Lazebnik andDeirdre Smeltzer,Methods for Euclidean Geometry, Mathematical Association of America, 2010, pp. 51-52.
  3. ^Zalman Usiskin and Jennifer Griffin, "The Classification of Quadrilaterals. A Study of Definition", Information Age Publishing, 2008, p. 22.
  4. ^Chen, Zhibo, and Liang, Tian. "The converse of Viviani's theorem",The College Mathematics Journal 37(5), 2006, pp. 390–391.
  5. ^Problem 5,2006 British Mathematical Olympiad,[1].
  6. ^Dunn, J.A., and J.E. Pretty, "Halving a triangle",Mathematical Gazette 56, May 1972, p. 105.
  7. ^Weisstein, Eric W."Triangle Circumscribing".Wolfram Math World.
  8. ^abWeisstein, Eric W. "Parallelogram." From MathWorld--A Wolfram Web Resource.http://mathworld.wolfram.com/Parallelogram.html
  9. ^Mitchell, Douglas W., "The area of a quadrilateral",Mathematical Gazette, July 2009.

External links

[edit]
Wikimedia Commons has media related toParallelograms.
Triangles
Quadrilaterals
By number
of sides
1–10 sides
11–20 sides
>20 sides
Star polygons
Classes
Retrieved from "https://en.wikipedia.org/w/index.php?title=Parallelogram&oldid=1322990934"
Categories:
Hidden categories:
[8]ページ先頭
©2009-2025 Movatter.jp