Question

Prove that the midpoints of the adjacent sides of a rectangle will form a rhombus. [4 MARKS]

Answer 1

Diagram : 1 Mark
Concept : 1 Mark
Proof : 2 Marks


Consider a rectangle $ABCD$.

Let $PQRS$ be the quadrilateral formed by the midpoint of the adjacent sides of the rectangle.

Let the length of the rectangle be $2a$ and let the breadth be $2b.$

$\Rightarrow AP=PB=DR=RC=a.$ [Since $P$ and $R$ are midpoints of $AB$ and $CD$]

and $AS=SD=BQ=QC=b.$ [Since $S$ and $Q$ are midpoints of $AD$ and $BC$]

Consider triangles $\Delta PAS$ and $\Delta RDS$

$PA=RD$ [Proved above]

$AS=DS$ [Proved above]

$\angle PAS=\angle RDS={90}^{\circ }$ [Angles of a rectangle – included angle]

$\Delta PAS\cong \Delta RDS$ [SAS congruency]

$\Rightarrow PS=RS$......(i) [CPCTC]

Similarly $\Delta PAS\cong \Delta PBQ$

$\Rightarrow PS=PQ$.....(ii) [CPCTC]

$\Delta PBQ\cong \Delta RCQ$

$\Rightarrow PQ=RQ$.....(iii) [CPCTC]

$\Delta RCQ\cong \Delta RDS$

$\Rightarrow RQ=RS$.....(iv) [CPCTC]

By (i), (ii), (iii) and (iv) $PS=RS=PQ=RQ$

$\Rightarrow PQRS$ is a rhombus.