Diagonals of Square Are Equal and Perpendicular to Each Other.


 
 
Concept Explanation
 

Diagonals of Square Are Equal and Perpendicular to Each Other.

Theorem: In a square the diagonals are equal and perpendicular to each other.

Given: A square ABCD.

To Prove: AC = BD and AC and BC intersect perpendicularly.

Proof:  In Delta ADB and Delta BCA

       AD  =  BC           [ because Sides of square are equal]

  angle DAB = angleCBA       [Each angle is 90^0]

       AB = AB             [ Common]

Therfore Delta DAB cong DeltaCBA     [SAS Criteria of Congruence]

  Rightarrow BD = AC   

As ABCD is a square Therefore it is a parallelogram. Also diagonals of a parallelogram bisect each other.

Rightarrow AO = OC and BO = OD              ...................(1)

In Delta BOC and Delta COD

  BO = DO       [ From Equation (1)]

  BC = CD       [ Sides of a square]

 OC = OC       [ Common ]

Therfore Delta BOC cong Delta DOC     [SSS Criteria of Congruence]

Rightarrow angle COB = angle COD     [ By CPCT ]    ............(2)

But angle COB + angle COD = 180^0               [Linear Pair]

angle COB + angle COB = 180^0                     [Using Eq 2]

  2 angle COB = 180^0Rightarrow angle COB = 90^0 Rightarrow angle COB = angle COD = 90^0

Similarly we can prove that angle AOB = angle AOD = 90^0

angle AOB = angle AOD = angle COB = angle COD =90^0

Hence Proved that In a square the diagonals are equal and perpendicular to each other.

Theorem : If the diagonals of a parallelogram are equal and intersect at right angles, then the parallelogram is a square.

Given : ABCD is a parallelogram, such that AC = BD and ACperp BD

To Prove: ABCD is a square

Proof: In Delta AOB and DeltaCOB

         AO = OC      [ Diagonals of parallelogram bisect each other]

         BO = OB      [ Common]

   angle AOB = angleCOB  [ Each 90^0 as AC perp BD]

Therefore Delta AOB cong DeltaCOB     [ SAS Criteria of Congurence]

     AB = BC     [CPCT]

As ABCD is a IIgm and the adjacent sides are equal AB = BC Therefore AB = BC = CD= DA.

In Delta ADB and Delta BCA

       AD  =  BC           [ Proved above]

       BD =  AC           [Given]

       AB = AB             [ Common]

Therfore Delta DAB cong DeltaCBA     [SSS Criteria of Congruence]

Rightarrowangle A = angle B    --------(1)   [ CPCT]

As ABCD is a parallelogram, AD || BC

Now AD || BC  and AB is the transversal

    angle A + angle B = 180^0      [ Co interior angles are supplementary ]

Rightarrow angle A + angle A = 180^0     [ Replacing B from equation 1]

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

Now ABCD is a IIgm in which all sides are equal and one angle is 90 ^0.Thus ABCD is a square

Illustration: If the side of a square ABCD whose diagonals intersect at O is 6cm find the length of OA and OB;

Solution: As the diagonals of a square are equal and bisect each other perpendicularly

We can say that OA = OB = x(say)            ..............(1)

and angleAOB = 90^0

AB = 6cm   ...................(2)         [Given]

Now Delta AOB is aright angled at O

AB^2= AO^2+ OB^2                         [By Pythagorous Theorem]

Substituting the values from Eq (1) and (2), we get

6^2= x^2+ x^2

2 x^2= 36

x^2= 18

x= sqrt{18}= 3sqrt2cm

Hence OA = OB = 3sqrt2 cm

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  • Sample Questions
    (More Questions for each concept available in Login)
    Question : 1

    PQRS is a square. PR and SQ intersects at O. State the measure of angle POQ.

    Right Option : B
    View Explanation
    Explanation
    Question : 2

    If the side of a square ABCD whose diagonals intersect at O is 10sqrt{2} cm find the length of OA and OB.

    Right Option : A
    View Explanation
    Explanation
    Question : 3

    ABCD is a square. AC and BD intersect at O. State the measure of angle AOB.

    Right Option : A
    View Explanation
    Explanation
     
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