Q. 113.7( 6 Votes )

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,
A. ABCD is a rhombus

B. diagonals of ABCD are equal

C. diagonals of ABCD are equal and perpendicular

D. diagonals of ABCD are perpendicular.

Answer :

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.


PQRS is a square


PQ = QR = RS = SP


Also, PR = SQ [Diagonals of a square]


And,


BC = PR


SQ = AB


BC = PR = SQ = AB


i.e, BC = AB


Thus all sides of ABCD are equal


ABCD is a square or a rhombus.


In ∆ABD,


S and P are midpoints of AD and AB


By midpoint theorem,


SP DB and SP = 1/2DB


Similarly,


In ∆ABC,


Q and P are midpoints of BC and AB


By midpoint theorem,


PQ AC and PQ = 1/2AC


PQRS is a square


SP = PQ


1/2DB = 1/2AC


DB = AC


i.e, Diagonals of ABCD are equal


it cannot be a square


Hence, it is a rhombus

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