Type of Quadrilaterals…continued

Rectangle

Definition: A parallelogram where all the four angles are right angles is called a rectangle.

Properties:a1

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

AB \parallel DC \ and \ AB=DC

AD \parallel BC \ and \ AD=BC

2) Each angle measures 90^{\circ} \ i.e.\  \angle A=\angle B=\angle C=\angle D=90^{\circ}

3) The diagonals are equal i.e. AC = BC

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

m10xAO = OC

DO = OB

Theorem: The diagonals of a rectangle are equal.

Given: ABCD is a rectangle and the diagonals are AC \ and\  BD

To prove: AC = BD 

Consider \Delta BCD \ and\ \Delta ACD

AD = BC  (given that it is a rectangle)

DC  is common

\angle ADC =\angle BCD  (right angles)

Hence \Delta ACD \cong \Delta BCD (S.A.S axiom)

Therefore AC = BD . Hence Proved

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Square

A parallelogram which has all sides equal and all four angles are right angle is called a square.a2 Or we can also say that a square is a rectangle that has all four sides equal.

Properties of square:

1) All sides are equal i.e. AB = BC = CD = DA 

2) Each angle measures 90^{\circ} i.e. \angle A =\angle B =\angle C =\angle D = 90^{\circ} 

3) The diagonals are equal i.e. AC = BC 

4) The diagonals (AC \ and\  BD)   bisect each other  AO = OC\  \& \ DO = OB 

The diagonals intersect at right angles i.e.  AC \perp  BD 

m11xTheorem: The diagonals of a square are equal and perpendicular to each other.

To prove:  AC = BD \ and\  AC \perp BD 

Given: Square  ABCD , Diagonals  AC \ and\  BD  \ intersect \ at \ O 

Proof:

Consider \Delta BCD \ and\ \Delta ACD

 AD = BC  (given that it is a square)

DC  is common

\angle ADC = \angle BCD \ (right angles)

Hence \Delta ACD \cong \Delta BCD (S.A.S axiom)

Therefore  AC = BD . Hence Proved

Consider \Delta COD \ and\ \Delta AOD

OC = OA   (diagonals of a parallelogram bisect each other)

DO  is common

D=DC (Sides of a triangle)

Hence \Delta COD \cong \Delta AOD (S.S.S axiom)

Hence \angle DOC=\angle DOA

We know that \angle DOC+\angle DOA=180^{\circ}

Therefore \angle DOC=\angle DOA=90^{\circ}

Hence  DO\perp AC \ i.e.\  AC\perp BD

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