Parallelograms

 Table of Contents

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A parallelogram is a special kind of quadrilateral. 

Rectangle, square, and rhombus are parallelogram examples.

Parallelogram: Definition

A parallelogram is defined as a quadrilateral in which both pairs of opposite sides are parallel and equal.

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A parallelogram is a convex polygon with \(4\) edges and \(4\) vertices.

It is a type of quadrilateral where the opposite sides are parallel and equal.

Use the interactive tool below to learn more about parallelograms.

Drag the vertices to understand the relations between different elements of a parallelogram.

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Rhombuses, rectangles, and squares are common parallelogram examples.

The parallelogram properties for each are listed below.

Rectangle

A rectangle has:

• two pairs of parallel sides\[A B\|D C \text{ and } A D\| B C\]
• four right angles\[\angle A=\angle B=\angle C=\angle D=90^{\circ} \]
• opposite sides of equal lengths\[A B=D C \text{ and } A D= B C\]
• two equal diagonals\[AC=BD\]
• diagonals that bisect each other i.e., one diagonal divides the other diagonal into exactly two halves.

Square

A square has:

• four equal sides\[AB=BC=CD=DA\]
• four right angles\[\angle A=\angle B=\angle C=\angle D=90^{\circ} \]
• two pairs of parallel sides\[A B\|D C \text { and } A D\| B C\]
• two equal diagonals\[AC=BD\]
• diagonals that are perpendicular to each other\[AC \perp BD\]
• diagonals that bisect each other. i.e., one diagonal divides the other diagonal into exactly two halves.

Rhombus

A rhombus has:

• two pairs of parallel sides\[E H\|F G \text{ and } E F\| H G\]
• four equal sides\[EH=HG=GF=FE\]
• opposite angles are equal\[\angle E = \angle G \text{ and } \angle H = \angle F\]
• diagonals that are perpendicular to each other\[EG \perp HF\]
• diagonals that bisect each other. i.e., one diagonal divides the other diagonal into exactly two halves.

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Consider the parallelogram \(\text{PQRT}\).

Let us understand the various parallelogram properties using \(\text{PQRT}\).

• The opposites sides of a parallelogram are parallel.
  \[P Q\|R T \text{ and } PR\| Q T \]
• Opposite sides of a parallelogram are equal.
  \[P Q=R T \text{ and } PR= Q T \]
• Opposite angles of a parallelogram are equal.
  \[\angle P = \angle T \text{ and }\angle Q = \angle R\]
• Diagonals of a parallelogram bisect each other. i.e., one diagonal divides the other diagonal into exactly two halves.
  \[R E=E Q \text{ and } PE= ET \]
• Same-side interior angles supplements each other.
  \[\angle PRT +  \angle RTQ =180 ^\circ\]
  \[\angle RTQ +  \angle TQP =180 ^\circ\]
  \[\angle TQP +  \angle QPR =180 ^\circ\]
  \[\angle QPR +  \angle PRT =180 ^\circ\]
• Diagonals divide the parallelogram into two congruent triangles.
  \[\Delta \text{RPQ} \: \text{congruent to} \:\Delta \text{QTR} \]
  \[\Delta \text{RPT} \: \text{congruent to}\: \Delta \text{QTP} \]

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Theorem 1

In a parallelogram, opposite sides are equal. Conversely, if the opposite sides in a quadrilateral are equal, then it is a parallelogram.

Consider the following figure:

Proof:

In \(\Delta ABC\) and \(\Delta CDA\),

\[\begin{align}  
AC&=AC \;(\text{common sides}) \\
\angle 1&=\angle 4 \;(\text{alternate interior angles})
\\ \angle 2&=\angle 3 \;(\text{alternate interior angles})
\end{align}\]

Thus, by the ASA criterion, the two triangles are congruent, which means that the corresponding sides must be equal.

Thus,

\[\begin{align}\boxed{AB=DC\;\text{and}\;AD=BC} \end{align}\]

This proves that opposite sides are equal in a parallelogram.

Theorem 2

In a parallelogram, opposite angles are equal. Conversely, if the opposite angles in a quadrilateral are equal, then it is a parallelogram.

Consider the following figure:

Proof:

In \(\Delta ABC\) and \(\Delta CDA\),

\[\begin{align}
 AC&=AC\: ( \text{common sides} )\\  \angle 1&=\angle 4 \:(\text{alternate interior angles}) \\    
\angle 2&=\angle 3 \: (\text{alternate interior angles} )
\end{align}\]

Thus, the two triangles are congruent, which means that

\[\begin{align}\boxed{\angle B=\angle D} \end{align}\]

Similarly, we can show that

\[\begin{align}\boxed{\angle A=\angle C} \end{align}\]

This proves that opposite angles in any parallelogram are equal.

Theorem 3

In a parallelogram, the diagonals bisect each other. Conversely, if the diagonals in a quadrilateral bisect each other, then it is a parallelogram.

Consider the following figure:

Proof:

In \(\Delta AEB\) and \(\Delta DEC\), we have:

\[\begin{align}   
AB&=CD \:( \text{opposite sides of a parallelogram})\\   
 \angle 1&=\angle 3\:( \text{alternate interior angles}) \\
\angle 2&=\angle 4\:( \text{alternate interior angles})
\end{align}\]

By the ASA criterion, the two triangles are congruent, which means that

\[\begin{align}\boxed{AE=EC\;\text{and}\;BE=ED}\end{align}\]

Thus, the two diagonals bisect each other.

Theorem 4

In a quadrilateral, if one pair of opposite sides are equal and parallel, then it is a parallelogram.

Consider the following figure:

It is given that \(AB=CD\) and \(AB || CD\).

We have to prove that ABCD is a parallelogram.

Proof:

In \(\Delta AEB\) and \(\Delta DEC\), we have:

\[\begin{align}   \text{AB}&=\text{CD} \: \text{given} \\
 \angle 1&=\angle 3\: \text{alternate interior angles} \\
 \angle 2&=\angle 4\: \text{alternate interior angles}
\end{align}\]

Thus, the two triangles are congruent, which means that:

\[\begin{align}\boxed{AE=EC\;\text{and}\;BE=ED}\end{align}\]

Therefore, the diagonals AC and BD bisect each other, and this further means that ABCD is a parallelogram.

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Parallelogram area is given by the formula: 

\(\text{Area of Parallelogram} \!=\!\text{Base} \!\times\! \text{Height}\)

Input the base and height in the parallelogram area calculator shown below to calculate the area of a parallelogram.

Note: Parallelogram perimeter formula: 

\(2 \times\) (Sum of length of adjacent sides)

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Example 1

 

 

In the given parallelogram\( \text{ABCD}\), \(\angle \text{A} =  90^\circ\).

Show that all its angles are equal to \(90^\circ\).

Solution:

Applying the parallelogram properties we have learnt, we know that:  

• the opposite angles are equal

\[\begin{align}
\therefore \angle \text{A} = \angle \text{C} =  90^\circ
\end{align}\]

• the adjacent angles are supplementary

\[\begin{align}
\angle \text{A}  +\angle \text{D} &=  180^\circ \\
\angle\text{D} &= 180^\circ -90^\circ (\because \angle\text{A} = 90^\circ \: \text{given} )\\
\therefore \angle\text{D} &= 90^\circ\ \\
\Rightarrow \angle \text{B} &=90^\circ
\end{align}\]

\(\begin{align}\angle\text{A}=\angle\text{B}=\angle\text{C}=\angle\text{D}= 90^\circ\end{align}\)

Example 2

 

 

Prove that when the parallelogram diagonals bisect each other at \( 90^\circ\), it is a rhombus.

Solution:

Consider the parallelogram \(\text{ABCD}\).

\[\begin{align}
 \Delta \text{AEB}\:\text{and}\: \Delta \text{AED} \\
AE&=AE\: (\text{common})\\ BE&=ED \:(\text{given})\\  \angle AEB&=\angle AED=\,90^\circ\   (\text{given}  )
\end{align}\]

Therefore, by SAS Congruency, \(\Delta AEB\) and \(\Delta AED\) are congruent.

\[\begin{align}
\Rightarrow  \text{AB = AD}
\end{align}\]

Similarly,

Considering  \(\Delta \text{AED}\:\text{and}\: \Delta \text{CED} \)

\[\begin{align}
\Rightarrow  \text{AD = DC}
\end{align}\]

Which further implies,

\[\begin{align}\boxed{ AB=BC=CD=AD} \end{align}\]

\(\therefore\)The given parallelogram is a rhombus.

Example 3

 

 

 If in a parallelogram \(\text{ABCD}\), \(\text{AC}\) bisects  \(\angle\text{A}\) and \(\angle\text{C}\), show that \(\text{AC} \perp \text{BD} \).

Solution:

\[\begin{align}
 \Delta \text{ABC}\:and\: \Delta \text{ADC}: \\
AC&=AC  (\text{common}) \\ \angle 1&=\angle 2 \:( \text{given)}  \\
 \angle 3 &=\angle 4 \:( \text{given})
 \end{align}\]

By the ASA criterion, the two triangles are congruent. This means that

\[\begin{align}
\Rightarrow  \text{AB = AD}
\end{align}\]

\[\begin{align}
 \Delta \text{AEB}\:and\: \Delta \text{AED}: \\
AE&=AE  (\text{common})\\
AB&=AD  (\text{proved above})
 \\ \angle 2&=\angle 1 \:( \text{given)}  \\
  \end{align}\]

By the SAS criterion, the two triangles are congruent.

Therefore,

\[\begin{align}
\angle AEB = &\angle AED\\
= &\frac{1}{2} \times 180^\circ  \\= &\,90^\circ
\end{align}\]

Hence,

\[\begin{align}\boxed{\angle AEB=\angle AED = 90^\circ} \end{align}\]

\(\therefore \;\text{AC} \perp\ \text{BD}\)

Example 4

 

 

Find the area of a parallelogram whose base is \(7\:\text{cm}\) and height is \(10\:\text{cm}\).

Solution:

Given

Base \( =7\:\text{cm}\)

Height\( =10\:\text{cm}\)

Area of a parallelogram

\[\begin{align}
&=\text{base}  \times  \text{height}  \\
&= 7 \times 10 \\
&=70\: \text{cm}^2
\end{align}\]

\(\therefore\) Area of the parallelogram \(=70\: \text{cm}^2\)

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