Question

$ABCD$ and $AEFD$ are two ${\parallel }^{gms}$. then

$\left(i\right)$ $PE=FQ$
$\left(ii\right)$ $Ar\left(△APE\right):Ar\left(△PFA\right)=Ar\left(△QFD\right):Ar\left(△PFD\right)$.
$\left(iii\right)$ $Ar\left(△PEA\right)=Ar\left(△QFD\right)$.

statements are ?

• A. True
• B. False

Answer 1

The correct option is A True

$ABCD$ and $AEFD$ are two parallelograms.

$\left(i\right)$

In $△EPA$ and $△FQD$

$\Rightarrow$ $\angle PEA=\angle QFD$ [ Corresponding angles ]

$\Rightarrow$ $\angle EPA=\angle FQD$ [ Corresponding angles ]

$\Rightarrow$ $PA=QD$ [ Opposite sides of parallelogram ]

$\Rightarrow$ $△EPA\cong △FQD$ [ By $AAS$ condition ]

$\therefore$ $EP=FQ$ [ C.P.C.T ]

$\left(ii\right)$

Since, $△PEA$ and $△QFD$ stand on the same base $PE$ and $FQ$ lie between the same parallels $EQ$ and $AD$.

$\therefore$ $ar\left(△PEA\right)ar\left(△QFD\right)$ ---- ( 1 )

And $ar\left(△PFA\right)=ar\left(△PFD\right)$ ----- ( 2 )

Divide the equation ( 1 ) by equation ( 2 )

$\Rightarrow$ $\frac{ar\left(△PEA\right)}{ar\left(△PFA\right)}=\frac{ar\left(△QFD\right)}{ar\left(△PFD\right)}$

$\left(iii\right)$

From part $\left(i\right)$ $△EPA\cong △FQD$

Then, $ar\left(△EPA\right)=ar\left(△FQD\right)$

$i.e.$ $ar\left(△PEA\right)=ar\left(△QFD\right)$

Thus, we can see all three statements are correct.