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Question

ABCD is a rectangle and P,Q,R and S are mid-points of the sides AB,BC,CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.


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Solution

Step 1: Drawing the diagram:

ABCD is a rectangle,

P,Q,R and S are mid-points of the sides AB,BC,CD and DA respectively.

Join diagonals AC and BD which intersect at O.

Step 2: Proving PQRS is a parallelogram:

In, ADC,

S is the midpoint of AD and R is the midpoint of CD.

Therefore, by mid point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half ot it.

Therefore, SRAC and SR=12AC…………………..(i)

In, ABC,

P is the midpoint of AB and Q is the midpoint of BC.

Therefore, by mid point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half ot it.

Therefore, PQAC and PQ=12AC……………..(ii)

From (i) and (ii), we get

Therefore, PQ=SR

We have, SRAC and PQAC

SRPQ (Lines parallel to same line are parallel to each other)

And, also PQ=SR.

PQRS is a parallelogram because a pair of opposite side of quadrilateral PQRS is equal and parallel.

Step 3:Proving PQRS is a rhombus:

In QBP and QCR,

QB=QCQisthemidpointofBCQBP=QCREach90°BP=CROppositesidesareequal,hencehalflengthisalsoequal

Therefore, By SAS congruency criteria

QBPQCR

QR=QP (By C.P.C.T.)

As PQRS is a parallelogram and having adjacent sides are equal

PQ=QR=RS=SP

Hence, PQRS is a rhombus.

Hence, proved


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