Question

Question 5
The quadrilateral formed by joining the mid – points of the side of quadrilateral PQRS, taken in order, is a rhombus, if
(A) PQRS is a rhombus
(B) PQRS is a parallelogram
(C) diagonals of PQRS are perpendicular
(D) diagonals of PQRS are equal.

Answer 1

Given, the quadrilateral ABCD is a rhombus.

So, sides AB, BC, CD and AD are equal.

Now, in $\Delta PQS$, we have

D and C are the mid – points of PQ and PS.
So, $DC=\frac{1}{2}QS$ [by mid –d point theorem] . . . . . (i)
Similarly, in $\Delta PSR$ $BC=\frac{1}{2}PR$ [by mid – point theorem] . . . . . .(ii)
As BC = DC [since, ABCD is a rhombus]
$\therefore \frac{1}{2}QS=\frac{1}{2}PR$ [from Eqs. (i) and (ii) ]
$\Rightarrow QS=PR$
Hence, diagonals of PQRS are equal.