In the given figure, ^@ABCD^@ is a parallelogram whose diagonals | Math Question and Answer

Answer:

Step by Step Explanation:

We know that diagonal $AC$ of $||gm \space ABCD$ divides it into two triangles of equal area. $\begin{aligned} \therefore ar(\Delta ACD) = \dfrac { 1 } { 2 } ar(||gm \space ABCD) &&\ldots \text{(i)} \end{aligned}$
Now, In $\Delta OAP$ and $\Delta OCQ$ , we have: $\begin{aligned} &OA = OC &&[\text{Diagonals of a ||gm bisect each other}] \\ &\angle AOP = \angle COQ &&[\text{Vertically opposite angles}]\\ &\angle PAO = \angle QCO &&[\text{Alternate interior angles}]\\ \therefore &\Delta OAP \cong \Delta OCQ \end{aligned}$
We know that if two triangles are congurrent then their respective areas are equal. $\begin{aligned} \therefore& \space ar(\Delta OAP) = ar(\Delta OCQ) \\ \implies& ar(\Delta OAP) + ar( \square AOQD) = ar(\Delta OCQ) + ar(\square AOQD) \\ \implies& ar(\square APQD) = ar(\Delta ACD) \\ &\space\space\space\space\space\space\space\space\space\space\space\space\space \space\space\space\space\space\space\space\space\space = \dfrac { 1 } { 2 } ar(||gm\space ABCD) &&[\text{Using eq (i)}] \\ \therefore&\space ar(\square APQD) = \dfrac { 1 } { 2 } ar(||gm\space ABCD) \end{aligned}$

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