Question

State true or false:

In the given figure, diagonals $PR$ and $QS$ of the parallelogram $PQRS$ intersect at point $O$ and $LM$ is parallel to $PS$, that

Area $\left(\Delta POS\right)+$ Area $\left(\Delta QOR\right)=$ Area $\left(\Delta POQ\right)+$ Area $\left(\Delta SOR\right)$

• A. True
• B. False

Answer 1

The correct option is A True

In $\left(\Delta POS\right)$ and $\left(\Delta SOR\right)$
$PO=OR$ (diagonal bisects eachother)
$\angle POS=\angle SOR$ (${90}^0$)
$OS=OS$ (common side)
$△POS\cong \left(\Delta SOR\right)$ (SAS congrience)
Area of $△POS$ = Area $△SOR$
In $\left(\Delta QOR\right)$ and $\left(\Delta POQ\right)$
$OP=OR$ (diagonal bisects eachother)
$\angle POQ=\angle QOR$ (${90}^0$)
$OQ=OQ$ (common side)
$△QOS\cong \left(\Delta QOR\right)$ (SAS congrience)
Area of $△QOS$ = Area $△QOR$
Area of parallelogram PQRS=Area of $△POS$+Area of $△POQ$+Area of $△QOR$+Area of $△SOR$
=Area of $△SOR$+Area of $△SOR$+Area of $△POQ$+Area of $△POQ$
=$2△SOR$+$2Areaof△POQ$
$\therefore 2△POS$+$2△QOR$=$2△SOR$+$2△POQ$
$Area\left(POS\right)+Area\left(QOR\right)=Area\left(POQ\right)+Area\left(SOR\right)$