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Maths / Quadrilaterals / Diagonals of a Rectangle Are of Equal Length
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Diagonals of a Rectangle Are of Equal Length

Theorem 2: The diagonals of a rectangle are of equal length.

Given : PQRS is a rectangle.

To Prove : PR = QS

Proof:As each rectangle is a parallelogram and PQRS is a rectangle

Therefore PQRS is a parallelogram 

PS = QR  '................(1) [ Opposite sides of a parallelogram]

As each angle of a rectangle is a right angle

angle P = angle Q=angle R=angle S=90 ^0

In Delta PRS ;and;Delta QRS

   PS = QR         [ From Equation 1 ]

  angle S = angle R      [ Each right angle ]

  RS = RS          [ Common ]

therefore ;Delta PRS ;cong;Delta QSR   [By SAS Congruence Criteria]

Rightarrow PR = QS   [ CPCT]

Hence Proved

Converse of Theorem 2: If the diagonals of a parallelogram are of equal length, it is a rectangle.

Given : PQRS is a parallelogram such that  PR  = QS.

To Prove : PQRS is a rectangle

Proof: In Delta PRS ;and;Delta QRS

   PR = QS         [ Given ]

  PS = QR          [ Opposite sides of parallelogram ]

  RS = RS          [ Common ]

therefore ;Delta PRS ;cong;Delta QSR   [By SSS Congruence Criteria]

Rightarrowangle S = angle R    --------(1)   [ CPCT]

As PQRS is a parallelogram, PS || QR

Now PS || QR and RS is the transversal

    angle S + angle R = 180^0      [ Co interior angles are supplementary ]

Rightarrow angle S + angle S = 180^0     [ Replacing R from equation 1]

Rightarrow 2angle S = 180^0 Rightarrow ;angle S = 90 ^0

Now PQRS is a parallelogram in which one angle is a right angle.

Therefore PQRS is a rectangle

Hence Proved

Illustration: The diagonals of a rectangle PQRS intersect at O, If angle QOR= 58^0;find ;angle QSP.

Solution: PQRS is a rectangle and we know that diagonals of a rectangle are equal

Each rectangle is aparallelogram and we know that diagonals of a parallelogram bisect each other

Therefore  OS = OR    [ Because when diagonals are equal halves are equal ]

In Delta SOR, As OS = OR

angle 1 = angle 2         [Angles opposite equal side are equal ]

Now angle QOR is the exterior angle of Delta SOR

angle 1 +angle 2= angle QOR

Rightarrow ; angle 1 +angle 1= angle QOR      [ because  angle 1 = angle 2]

Rightarrow ; 2angle 1 = 58^0

Rightarrow ; angle 1 = 29^0

Now each angle of a rectangle is a right angle.

angle S = 90^0

angle QSP + angle1 = 90^0

Rightarrow angle QSP + 29^0 = 90^0Rightarrow angle QSP = 90-29 = 61^0

 

 


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