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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