Question

In the given figure, if $PQRS$ is a rectangle which one is true ?

• A. $PA=RB$
• B. $Ar\Delta \left(APS\right)=Ar\Delta \left(QRB\right)$
• C. $Ar\left(PQS\right)=Ar\left(QRS\right)$
• D. All of the above

Answer 1

Answer: (D)

In $△APS$ and $△QRB$

$\Rightarrow$ $\angle PAS=\angle QBR={90}^o$ [ Given ]

$\Rightarrow$ $\angle PSA=\angle RQB$ [ Alternate angles ]

$\Rightarrow$ $PS=QR$ [ Opposite sides of rectangle are equal ]

$\therefore$ $△APS\cong △QRB$ [ By $AAS$ congruence rule ]

$\Rightarrow$ $Ar\left(△APS\right)=Ar\left(△QRB\right)$ [ Area of two congruent triangles are equal ]

$\Rightarrow$ $PA=RB$ [ By C.P.C.T. ]

In $△PQS$ and $△QRS$

$\Rightarrow$ $PS=QR$ [ Opposite sides of rectangle are equal ]

$\Rightarrow$ $QS=QS$ [ Common sides ]

$\Rightarrow$ $PQ=RS$ [ Opposite sides of rectangle are equal ]

$\therefore$ $△PQS\cong △QRS$ [ By $SSS$ rule ]

The area of two congruent triangles are equal.

$\therefore$ $Ar\left(△PQS\right)=Ar\left(△QRS\right)$