Question

Question 11
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 1

Given, ABCD is a quadrilateral and P,Q,R and S are the mid-points of sides of AB , BC , CD and DA, respectively, then, PQRS is a square.

$\therefore PQ=QR=RS=PS$

And PR = SQ

But PR = BC and SQ = AB

$\therefore AB=BC$

Thus all the sides of quadrilateral ABCD are equal.

Hence, Quadrilateral ABCD is either a square of a rhombus.

Now, in $\Delta ADB$ use-mid – point theorem

SP ∥ DB
$AndSP=\frac{1}{2}DB$
Similarly in $\Delta ABC$ ( by mid-point theorem) PQ ∥ AC and PQ = AC ….(iii)
From Eq. (i) PS = PQ
$\Rightarrow \frac{1}{2}DB=\frac{1}{2}AC$ [ From Eqs. (ii) and (iii)]
$\Rightarrow DB=AC$
Thus, diagonals of ABCD are equal and therefore quadrilateral ABCD is a square not rhombus, so, Diagonals of quadrilateral are also perpendicular.