Question

The figure formed by joining the midpoints of the adjacent sides of a rhombus is a

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

Answer 1

Correct option is C. rectangle.

Let $ABCD$ be a rhombus such that $P,Q,R$ and $S$ are the mid points of the respective sides $AB,BC,CD$ and $DR$.

Consider $\Delta ADC$, since, $S$ and $R$ are the mid points of the sides $AD$ and $DC$, therefore, by Mid Point Theorem,

$\Rightarrow SR\parallel AC$ and $SR=\frac{AC}{2}$ ...(i)

Similarly, in $\Delta ABC$, since, $P$ and $Q$ are the mid points of the sides $AB$ and $BC$, therefore, by Mid Point Theorem,

$\Rightarrow PQ\parallel AC$ and $PQ=\frac{AC}{2}$ ...(ii)

From (i) and (ii),

$\Rightarrow SR\parallel PQ$ and $SR=PQ$ ...(iii)

Hence, $PQRS$ is a parallelogram.

Similarly we can prove that $SP\parallel RQ$ and $SP=RQ$ ...(iv)

From (iii) and (iv), the opposite sides of $PQRS$ are equal. ...(v)

Now consider, quadrilateral $ERFO$,

Since, $EO\parallel RF$ and $ER\parallel OF$, therefore,

$\Rightarrow$ Quadrilateral $ERFO$ is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at ${90}^{\circ }$.

$\therefore \angle EOF={90}^{\circ }$

Hence, $\angle SRQ=\angle ERF=\angle EOF={90}^{\circ }$ [Opposite angles of parallelogram are equal]...(iv)

From (v) and (iv), $PQRS$ is a rectangle.