Quadrilaterals Class 9 Notes - Chapter 8

CBSE Class 9 Maths Quadrilaterals Notes:-Download PDF Here

Get the complete notes on quadrilaterals class 9 here. A quadrilateral is a shape which has four sides. In this article, we are going to discuss the different types of quadrilaterals such as square, rectangle, parallelogram properties with proofs.

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Parallelogram: Opposite sides of a parallelogram are equal

In $\Delta ABCand\Delta CDA$

$AC=AC$ [Common / transversal]

$\angle BCA=\angle DAC$ [alternate angles]

$\angle BAC=\angle DCA$ [alternate angles]

$\Delta ABC\cong \Delta CDA$  [ASA rule]

Hence,

$AB=DC$ and $AD=BC$ [ C.P.C.T.C]

Opposite angles in a parallelogram are equal

In parallelogram $ABCD$

$AB\Vert CD$; and $AC$ is the transversal

Hence, $\angle 1=\angle 3$….(1) (alternate interior angles)

$BC\Vert DA$; and $AC$ is the transversal

Hence, $\angle 2=\angle 4$….(2) (alternate interior angles)

Adding (1) and (2)

$\angle 1+\angle 2=\angle 3+\angle 4$

$\angle BAD=\angle BCD$

Similarly,
$\angle ADC=\angle ABC$

Properties of diagonal of a parallelogram

Diagonals of a parallelogram bisect each other.

In  $\Delta AOBand\Delta COD,$

$\angle 3=\angle 5$ [alternate interior angles]

$\angle 1=\angle 2$  [vertically opposite angles]

$AB=CD$ [opp. Sides of parallelogram]

$\Delta AOB\cong \Delta COD$ [AAS rule]

$OB=OD$ and $OA=OC$ [C.P.C.T]

Hence, proved

Conversely,
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Diagonal of a parallelogram divides it into two congruent triangles.

In $\Delta ABC$ and $\Delta CDA,$

$AB=CD$ [Opposite sides of parallelogram]

$BC=AD$ [Opposite sides of parallelogram]

$AC=AC$ [Common side]

$\Delta ABC\cong \Delta CDA$ [by SSS rule]

Hence, proved.

For more information on Properties Of Parallelogram, watch the below video

Diagonals of a rhombus bisect each other at right angles

Diagonals of a rhombus bisect each – other at right angles

In $\Delta AODand\Delta COD,$

$OA=OC$ [Diagonals of parallelogram bisect each other]

$OD=OD$ [Common side]

$AD=CD$ [Adjacent sides of a rhombus]

$\Delta AOD\cong \Delta COD$ [SSS rule]

$\angle AOD=\angle DOC$ [C.P.C.T]

$\angle AOD+\angle DOC=180$ [$\because$ AOC is a straight line]

Hence, $\angle AOD=\angle DOC=90$

Hence proved.

For more information on Properties Of Rhombus, watch the below video

Diagonals of a rectangle bisect each other and are equal

Rectangle ABCD

In $\Delta ABCand\Delta BAD,$

$AB=BA$ [Common side]

$BC=AD$ [Opposite sides of a rectangle]

$\angle ABC=\angle BAD$ [Each = ${90}^0\because$ ABCD is a Rectangle]

$\Delta ABC\cong \Delta BAD$ [SAS rule]

$\therefore AC=BD$  [C.P.C.T]

Consider $\Delta OADand\Delta OCB,$

$AD=CB$  [Opposite sides of a rectangle]

$\angle OAD=\angle OCB$  [$\because$ AD||BC and transversal AC intersects them]

$\angle ODA=\angle OBC$  [$\because$ AD||BC and transversal BD intersects them]

$\Delta OAD\cong \Delta OCB$ [ASA rule]

$\therefore OA=OC$  [C.P.C.T]

Similarly we can prove $OB=OD$

For more information on Properties Of Rectangle, watch the below video

Diagonals of a square bisect each other at right angles and are equal

Square ABCD

In $\Delta ABCand\Delta BAD,$

$AB=BA$ [Common side]

$BC=AD$ [Opposite sides of a Square]

$\angle ABC=\angle BAD$ [Each = ${90}^0\because$ ABCD is a Square]

$\Delta ABC\cong \Delta BAD$ [SAS rule]

$\therefore AC=BD$  [C.P.C.T]

Consider $\Delta OADand\Delta OCB,$

$AD=CB$  [Opposite sides of a Square]

$\angle OAD=\angle OCB$  [$\because$ AD||BC and transversal AC intersects them]

$\angle ODA=\angle OBC$  [$\because$ AD||BC and transversal BD intersects them]

$\Delta OAD\cong \Delta OCB$ [ASA rule]

$\therefore OA=OC$  [C.P.C.T]

Similarly we can prove $OB=OD$

In $\Delta OBAand\Delta ODA,$

$OB=OD$   [ proved above]

$BA=DA$  [Sides of a Square]

$OA=OA$  [ Common side]

$\Delta OBA\cong \Delta ODA,$  [ SSS rule]

$\therefore \angle AOB=\angle AOD$  [ C.P.C.T]

But, $\angle AOB+\angle AOD={180}^0$  [ Linear pair]

$\therefore \angle AOB=\angle AOD={90}^0$

Important results related to parallelograms

Parallelogram ABCD

Opposite sides of a parallelogram are parallel and equal.

$AB||CD,AD||BC,AB=CD,AD=BC$

Opposite angles of a parallelogram are equal adjacent angels are supplementary.

$\angle A=\angle C,\angle B=\angle D,$

$\angle A+\angle B={180}^0,\angle B+\angle C={180}^0,\angle C+\angle D={180}^0,\angle D+\angle A={180}^0$

A diagonal of parallelogram divides it into two congruent triangles.

$\Delta ABC\cong \Delta CDA$ [With respect to AC as diagonal]

$\Delta ADB\cong \Delta CBD$   [With respect to BD as diagonal]

The diagonals of a parallelogram bisect each other.

$AE=CE,BE=DE$

$\angle 1=\angle 5$ (alternate interior angles)

$\angle 2=\angle 6$ (alternate interior angles)

$\angle 3=\angle 7$ (alternate interior angles)

$\angle 4=\angle 8$ (alternate interior angles)

$\angle 9=\angle 11$ (vertically opp. angles)

$\angle 10=\angle 12$ (vertically opp. angles)

The Mid-Point Theorem

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of the third side

In $\Delta ABC,E$ – the midpoint of $AB;F$ – the midpoint of $AC$

Construction: Produce $EF$ to $D$ such that $EF=DF.$

In $\Delta AEF$ and $\Delta CDF$,

$AF=CF$  [ F is the midpoint of AC]

$\angle AFE=\angle CFD$  [ V.O.A]

$EF=DF$ [ Construction]

$\therefore \Delta AEF\cong \Delta CDF$ [SAS rule]

Hence,

$\angle EAF=\angle DCF$….(1)

$DC=EA=EB$  [ E is the midpoint of AB]

$DC\Vert EA\Vert AB$ [Since, (1), alternate interior angles]

$DC\Vert EB$

So $EBCD$ is a parallelogram

Therefore, $BC=ED$ and $BC\Vert ED$

Since, $ED=EF+FD=2EF=BC$  [ $\because$ EF=FD]

We have,$EF=\frac{1}{2}BC$ and $EF||BC$

Hence proved.

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Introduction to Quadrilaterals

For more information on Quadrilaterals, watch the below videos

Quadrilaterals

Any four points in a plane, of which three are non-collinear are joined in order results into a four-sided closed figure called ‘quadrilateral’

Quadrilateral

For more information on Elementary Shapes – Quadrilaterals, watch the below video

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Angle sum property of a quadrilateral

Angle sum property – Sum of angles in a quadrilateral is 360

In $△ADC,$

$\angle 1+\angle 2+\angle 4=180$ (Angle sum property of triangle)…………….(1)

In $△ABC,$

$\angle 3+\angle 5+\angle 6=180$ (Angle sum property of triangle)………………(2)

(1) + (2):

$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6=360$

I.e, $\angle A+\angle B+\angle C+\angle D=360$

Hence proved

Types of Quadrilaterals

Trapezium

A trapezium is a quadrilateral with any one pair of opposite sides parallel.

Trapezium

$PQRS$ is a trapezium in which $PQ||RS$

Parallelogram

A parallelogram is a quadrilateral, with both pair of opposite sides parallel and equal. In a parallelogram, diagonals bisect each other.

Parallelogram ABCD

Parallelogram ABCD in which $AB||CD,BC||AD$  and $AB=CD,BC=AD$

Rhombus

A rhombus is a parallelogram with all sides equal. In a rhombus, diagonals bisect each other perpendicularly

Rhombus ABCD

A rhombus $ABCD$ in which $AB=BC=CD=AD$ and  $AC\perp BD$

Rectangle

A rectangle is a parallelogram with all angles as right angles.

Rectangle ABCD

A rectangle $ABCD$ in which, $\angle A=\angle B=\angle C=\angle D={90}^0$

Square

A square is a special case of a parallelogram with all angles as right angles and all sides equal.

Square ABCD

A square $ABCD$ in which $\angle A=\angle B=\angle C=\angle D={90}^0$ and $AB=BC=CD=AD$

Kite

A kite is a quadrilateral with adjacent sides equal.

Kite ABCD

A kite $ABCD$ in which $AB=BC$ and $AD=CD$

For more information on Types Of Quadrilaterals, watch the below videos

To know more about Different Types of Quadrilateral, visit here.

Venn diagram for different types of quadrilaterals