Question

show that quadrilateral formed by joining the mid points of sides of a square is also a square ​

Answer 1

Answer:

In a square ABCD, P,Q,R and S are the mid-points of AB,BC,CD and DA respectively.

⇒ AB=BC=CD=AD [ Sides of square are equal ]

In △ADC,

SR∥AC and SR=

2

1

AC [ By mid-point theorem ] ---- ( 1 )

In △ABC,

PQ∥AC and PQ=

2

1

AC [ By mid-point theorem ] ---- ( 2 )

From equation ( 1 ) and ( 2 ),

SR∥PQ and SR=PQ=

2

1

AC ---- ( 3 )

Similarly, SP∥BD and BD∥RQ

∴ SP∥RQ and SP=

2

1

BD

and RQ=

2

1

BD

∴ SP=RQ=

2

1

BD

Since, diagonals of a square bisect each other at right angle.

∴ AC=BD

⇒ SP=RQ=

2

1

AC ----- ( 4 )

From ( 3 ) and ( 4 )

SR=PQ=SP=RQ

We know that the diagonals of a square bisect each other at right angles.

∠EOF=90

o

.

Now, RQ∥DB

RE∥FO

Also, SR∥AC

⇒ FR∥OE

∴ OERF is a parallelogram.

So, ∠FRE=∠EOF=90

o

(Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R=90

o

and SR=PQ=SP=RQ

∴ PQRS is a square.

Step-by-step explanation:

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Answer 2

In a square ABCD, P,Q,R and S are the mid-points of AB,BC,CD and DA respectively.

⇒  AB=BC=CD=AD           [ Sides of square are equal ]

In △ADC,

SR∥AC and SR=21AC       [ By mid-point theorem ]  ---- ( 1 )

In △ABC,

PQ∥AC and PQ=21AC     [ By mid-point theorem ]  ---- ( 2 )

From equation ( 1 ) and ( 2 ),

SR∥PQ and SR=PQ=21AC    ---- ( 3 )

Similarly, SP∥BD and BD∥RQ

∴  SP∥RQ and SP=21BD

and RQ=21BD

∴  SP=RQ=21BD

Since, diagonals of a square bisect each other at right angle.

∴  AC=BD

⇒  SP=RQ=21AC          ----- ( 4 )

From ( 3 ) and ( 4 )

SR=PQ=SP=RQ

We know that the diagonals of a square bisect each other at right angles.

∠EOF=90o.

Now, RQ∥DB

RE∥FO

Also, SR∥AC

⇒ FR∥OE

∴  OERF is a parallelogram.

So, ∠FRE=∠EOF=90o  (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R=90o and SR=PQ=SP=RQ

∴  PQRS is a square.