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

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. 

Answers (1)

Answer:    [C]

Solution.

Given : ABCD is a quadrilateral and P, Q, R and S are the midpoints of sides of AB, BC, CD and DA. Then, PQRS is a square
\therefore PQ=QR=RS=PS              …..(1)
PR=SQ
Also,   PR=BC and SQ=AB
\therefore AB=BC
Thus all sides are equal. 
Hence ABCD is either a square or a rhombus.
In \triangle ADB by mid-point theorem
SP\parallel DB
SP=\frac{1}{2}DB                       …..(2)
Similarly in  \triangle ABC,AQ=\frac{1}{2}AC           …..(3)
From equation 1
PS=PQ
\frac{1}{2}DB=\frac{1}{2}AC
Thus ABCD is a square so diagonals of a quadrilateral are also perpendicular.
Therefore option (C) is correct.

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