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Diagonals of Square Are Equal and Perpendicular to Each Other.

Quadrilaterals II - Concepts

Properties of Rectangle

Diagonals of a Rectangle Are of Equal Length

Properties of Rhombus

The Diagonal of a Rhombus are Perpendicular to Each Other

Properties of Square

Diagonals of Square Are Equal and Perpendicular to Each Other.

Mid Point Theorem

Converse Mid Point Theorem

Class - 9th Foundation NTSE Subjects

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 = $angle$ CBA [Each angle is $90^0$ ]

AB = AB [ Common]

Therfore $Delta$ DAB $cong$ $Delta$ CBA [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 AC $perp$ BD

To Prove: ABCD is a square

Proof: In $Delta$ AOB and $Delta$ COB

AO = OC [ Diagonals of parallelogram bisect each other]

BO = OB [ Common]

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

Therefore $Delta$ AOB $cong$ $Delta$ COB [ 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 $cong$ $Delta$ CBA [SSS Criteria of Congruence]

$Rightarrow$ $angle A = angle B$ --------(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 $angle$ AOB = $90^0$

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

$6^2= x^2+ x^2$

$2 x^2= 36$

$x^2= 18$

$x= sqrt{18}= 3sqrt2$ cm

Hence OA = OB = $3sqrt2$ cm

Sample Questions

(More Questions for each concept available in Login)

Question : 1

If the side of a square ABCD whose diagonals intersect at O is $5sqrt{2}$ cm find the length of OA and OB.
A. 6	B. 8	C. 10	D. 5
Right Option : D
View Explanation

Question : 2

PQRS is a square. PR and SQ intersects at O. State the measure of $angle POQ$ .
A. 45	B. 90	C. 180	D. 360
Right Option : B
View Explanation

Question : 3

If the diagonals of a parallelogram are equal and intersect at right angles, then the parallelogram is a _____________ .
A. Rhombus	B. Rectangle	C. Kite	D. Square
Right Option : D
View Explanation