The figure formed by joining the mid-points of the sides of a quadrilateral $PQRS$ taken in order is a square if and only if, (a) $P Q R S$ is a rhombus. (b) diagonals of $P Q R S$ are perpendicular. (c) diagonals of $P Q R S$ are equal and perpendicular. (d) diagonals of $P Q R S$ are equal.
Given :
The figure is formed by joining the mid-points of the sides of a quadrilateral $PQRS$
To do :
We have to find a suitable option if the figure formed is a square.
Solution :
In the given figure, $A, B, C, D$ are the midpoints of the quadrilateral $PQ, QR, RS,$ and $PS$ respectively.
It is given that $ABCD$ is a square.
In a square, all four sides are equal and diagonals are congruent.
So, $AB = BC = CD = AD$
$AC=BD$..........(i)
Here, $AC = PS = QR$ and $BD=PQ=RS$...........(ii)
From (i) and (ii),
$PS = QR =PQ=RS$
Therefore, $PQRS$ can be a square or a rhombus.
In $\triangle PQS$,
$AD \parallel QS$ and $AD = \frac{1}{2} QS$.............(iii) [By midpoint theorem]
In $\triangle PQR$,
$AB \parallel PR$ and $AB = \frac{1}{2} PR$.............(iv) [By midpoint theorem]
We already know that $AB=AD$
From (iii) and (iv),
$\frac{1}{2} PR = \frac{1}{2} QS$
So, $PR = QS$
If $PR = QS$, it should be a square. [Diagonals are equal in the square]
In a square, diagonals are equal and perpendicular.
Therefore, (c) diagonals of $P Q R S$ are equal and perpendicular is the correct option.
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