E Board: Suggested Books     ICSE Board:  Foundation Mathematics 
Class 8: Reference Books     Class 8: NTSE Preparation 


Types of Quadrilaterals


Definition: Parallelogram is a quadrilateral where both pairs of the opposite sx1ides are parallel to each other.

In the figure shown, AB\parallel DC \ and AD\parallel BC  .

Note that AC is a transversal to AB \ and\  DC  . Similarly, AC is also a transversal to AB \ and\  DC  .

Properties of a Parallelogram (please refer to the above diagram):

1) Opposite sides are equal and parallel to each other.

m4xAB\parallel DC \ and \ AB = DC  

AD\parallel BC \ and \ AD = BC  

2) Opposite angles are equal

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

3) Adjacent angles are supplementary (consecutive interior angles)

\angle A+\angle D=180^{\circ}  

\angle C +\angle B = 180^{\circ}  

\angle C +\angle D = 180^{\circ}  

\angle A +\angle B = 180^{\circ}  

4) The diagonals (AC and BD) bisect each other

AO = OC 

DO = OB 

5) Each diagonal bisect the parallelogram in two congruent triangles

AC bisects the parallelogram ABCD in \Delta ACD\ and\ \Delta ACB such that in \Delta ACD \cong \Delta ACB

BD bisects the parallelogram ABCD in \Delta ABD\ and\ \Delta BCD such that in \Delta ABD \cong \Delta BCD


Theorem: In a parallelogram,x2

Opposite sides are equal

Opposite angles are equal

Each diagonal bisects the parallelogram


Given: Parallelogram ABCD  such that AB \parallel DC\ and\ AD \parallel BC  

To Prove:

AB = DC \ and\  AD = BC

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

\Delta ABD \cong \Delta BCD  i.e. area of \Delta ABD is equal to area of \Delta BCD

Similarly, \Delta ACD \cong \Delta ACB  i.e. area of \Delta ACD  is equal to area of \Delta ACB


\angle 1 =\angle 2 (alternate angles because AB \parallel DC \ and\  DB is a transversal)

\angle 4 =\angle 3  (alternate angles because AD \parallel BC \ and\ DB is a transversal)

DB is common

Therefore, \Delta ABD \cong \Delta BCD    [A.S.A axiom]

i) Since    \Delta ABD \cong \Delta BCD

AB = DC \ and\  AD = BC

ii) Since \Delta ACD \cong \Delta ACB 

\angle C = \angle A

\angle D = \angle B (since \angle 1 = \angle 2 \ and\  \angle 4 = \angle 3)  

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

Since \Delta ABD \cong \Delta BCD m6x

Area of   \Delta ABD = \ Area \ of \Delta BCD     (since congruent triangles are equal in area)

Similarly, if we were to join AC, we could use the same logic to prove \Delta ACD \cong \Delta ACB .



Definition: A parallelogram where all four sides are equal is called a Rhombus.

AB \parallel DC \ and\ AD \parallel BC

AB = DC = AD = BC


Properties of a Rhombus:

1) Opposite sides are parallel to each other.

AB \parallel DC

AD \parallel BC

2) All sides are equal

AB = DC = AD = BC


3) Diagonals AC and BD bisect each other at right angles

\angle AOB = \angle AOD = \angle BOC = \angle COD = 90^{\circ}


4) Diagonal BD \ bisect\ \angle A   (i.e. \angle ADO = \angle ODC ). SImilarly, diagonal AC \ bisects\ \angle A \ and \ \angle C.

Similarly, diagonal\ AC\ bisects\ \angle A \ and\ \angle C .

Theorem: The diagonals of a Rhombus bisect each other at right angles.

Given: Rhombus ABCD, AC \ and\ BD  are diagonals.

To Prove: \angle BOC=\angle COD=90^{\circ}


 \angle ODC =\angle OBA     (alternate angles)

  \angle OCD =\angle OAB     (alternate angles)

 AB = DC         (Side of a Rhombus)

Therefore \Delta AOB \cong \Delta COD    (A.S.A Axiom)

Hence  OD = OB \ and\  OA = OC    

Again in \Delta ODC\ and\ \Delta OBC m8x

 DC = BC        (Side of a Rhombus)

 OD = OB        (Proved above) And

OC   is common

Hence \Delta ODB \cong \Delta OBC    (S.S.S axiom)


\angle DOC =\angle COB

Now  \angle DOC +\angle COB =180^{\circ}  

Or   \angle DOC =\angle COB =90^{\circ}

E Board: Suggested Books     ICSE Board:  Foundation Mathematics 
Class 8: Reference Books     Class 8: NTSE Preparation