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Properties of Parallelograms

  • Last Updated : 19 Jan, 2021

A quadrilateral having both the pairs of opposite sides equal is a parallelogram. A parallelogram is a two-dimensional geometrical shape, whose sides are parallel to each other. Below are some simple facts about parallelogram:

  1. Number of sides in Parallelogram = 4
  2. Number of vertices in Parallelogram = 4
  3. Area = Base x Height
  4. Perimeter = 2 (Sum of adjacent sides length)
  5. Type of polygon = Quadrilateral

Below is the representation of a parallelogram:

Proofs: Parallelograms

Proof 1: Opposite sides of a parallelogram is equal.

Given: ABCD is a parallelogram
 



To Prove: AB = CD & DA = BC

Firstly, Join AC

As given ABCD is a parallelogram. Therefore, 

AB || DC  &  AD || BC

Now,  AD || BC and AC is intersecting A and C respectively.

\angle  DAC = \angle  BCA                   …(i)                  [Alternate Interior Angles]

Now, AB || DC and AC is intersecting A and C respectively.

\angle  BAC = \angle  DCA                 …(ii)                      [Alternate Interior Angles]



Now, In \triangle  ADC & \triangleCBA

\angle  DAC = \angle  BCA                    [ From (i) ]

AC = AC                                   [ Common Side ]

\angle  DCA = \angle  BAC                 [ From (ii) ]

So, by ASA(Angle-Side-Angle) criterion of congurence

\triangle  ADC  \cong  \triangle CBA

AB = CD & DA = BC [ Corresponding part of congurent triangles are equal ]

Hence Proved !

Proof 2: Opposte angles of a parallelogram are equal.

Given: ABCD is a parallelogram



To Prove:  \angle  A = \angle  C  and \angle  B = \angle  D

As given ABCD is a parallelogram. Therefore, 

AB || DC  &  AD || BC

Now, AB || DC and AD is Intersecting them at A and D respectively.

\angle  A + \angle  D = 180\degree                 …(i)             [ Sum of consecutive interior angles is 180\degree  ]

Now, AD || BC and DC is Intersecting them at D and C respectively.

\angle  D + \angle  C = 180\degree                …(ii)            [ Sum of consecutive interior angles is 180\degree]

From (i) and (ii) , we get

\angle  A + \angle  D = \angle  D  +  \angle  C

So,  \angle  A = \angle  C



Similarly, \angle  B = \angle  D

\angle  A = \angle  C and  \angle  B = \angle  D

Hence Proved !

Proof 3: Diagonals of a parallelogram bisect each other.

Given: ABCD is a parallelogram

To Prove: OA = OC & OB = OD

As given ABCD is a parallelogram. Therefore,

AB || DC  &  AD || BC

Now, AB || DC and AC is intersecting A and C respectively.

\angle  BAC = \angle  DCA                               [ Alternate Interior Angles are equal ]



So, \angle  BAO = \angle  DCO

Now,  AB || DC and BD is intersecting B and D respectively.

\angle  ABD = \angle  CDB                               [ Alternate Interior Angles are equal ]

So, \angle  ABO = \angle  CDO

Now, in  \triangle  AOB &  \triangle  COD we have, 

\angle  BAO = \angle  DCO                               [ Opposite sides of a parallelogram are equal ]

AB = CD

\angle  ABO = \angle  CDO

So, by ASA(Angle-Side-Angle) congurence criterion 

\triangle  AOB  \cong    \triangle  COD

OA = OC and OB = OD

Hence Proved !

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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