# 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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Updated on: 10-Oct-2022

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