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Question

The figure formed by joining the mid-points of the adjacent sides of a square is a

A

rhombus

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B

square

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C

rectangle

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D

parallelogram

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Solution

The correct option is A. square


Let ABCD be a square. Let P,Q,R and S be the mid points of the side AB,BC,CD and AD respectively.

Consider ΔADC, since S and R are the mis points of AD and DC, therefore, by Mid Point Theorem,

SRAC and SR=AC2...(i)

Similarly, in ΔABC, since P and Q are the mis points of AB and BC, therefore, by Mid Point Theorem,

PQAC and PQ=AC2...(ii)

From (i) and (ii)

SRPQ and SR=PQ ...(iii)

Therefore, PQRS is a parallelogram.

Now, consider ΔAPS and ΔPBQ

AS=BQ [As AS and BQ are the sides of the square]

AP=BP [P is the mid point of AB]

SAP=QBP=90 [Each angle is 90 of a square]

Therefore, ΔSAP and ΔBQP by SAS congruence rule.

Hence, SP=PQ (By CPCT) ...(iv)

From (iii) and (iv),

PQ=QR=RS=PS...(v)

Now consider, quadrilateral ERFO,

Since, EORF and EROF, therefore,

Quadrilateral ERFO is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at 90.

EOF=90

Hence, SRQ=ERF=EOF=90 [Opposite angles of parallelogram are equal]...(vi)

From (v) and (vi), PQRS is a square.


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