Pl slove this

Answer :

From given information we form our diagram , As  :



Here , we know

AP  =  BP  =  $\frac{1}{2}$AB
And
BQ  =  CQ  =  $\frac{1}{2}$BC
And
AR  =  CR  =  $\frac{1}{2}$AC

And

APQR is a parallelogram

And from converse of mid-point theorem we know if we join mid point of two sides , that the joining line is parallel to the third line .

Here we get $∆$ ABC , $∆$ PBQ and $∆$ RQC are similar to each other .

So , we get

$$\begin{gathered} \frac{Area of paralle\log ram APQR}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}} = \frac{Area of ∆ ABC - Area of ∆ PBQ - Area of ∆ RQC}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}} \\ \Rightarrow \frac{Area of ∆ ABC - Area of ∆ PBQ - Area of ∆ RQC}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}} = \frac{B{C}^2 - B{Q}^2 - Q{C}^2}{B{C}^2\hspace{0.17em}\hspace{0.17em}} \\ \Rightarrow \frac{Area of paralle\log ram APQR}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}}= \frac{B{C}^2 - B{Q}^2 - Q{C}^2}{B{C}^2\hspace{0.17em}\hspace{0.17em}} \\ \Rightarrow \frac{Area of paralle\log ram APQR}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}}= \frac{B{C}^2 - {\left({\displaystyle \frac{BC}{2}}\right)}^2 - {\left({\displaystyle \frac{BC}{2}}\right)}^2}{B{C}^2\hspace{0.17em}\hspace{0.17em}} \\ \Rightarrow \frac{Area of paralle\log ram APQR}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}}= \frac{B{C}^2 - {{\displaystyle \frac{BC}{4}}}^2 - {{\displaystyle \frac{BC}{4}}}^2}{B{C}^2\hspace{0.17em}\hspace{0.17em}} \\ \Rightarrow \frac{Area of paralle\log ram APQR}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}}= \frac{ {\displaystyle }{\displaystyle \frac{4B{C}^2 - 2B{C}^2 }{4}}}{B{C}^2\hspace{0.17em}\hspace{0.17em}} \\ \Rightarrow \frac{Area of paralle\log ram APQR}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}}= \frac{ {\displaystyle }{\displaystyle \frac{2B{C}^2 }{4}}}{B{C}^2\hspace{0.17em}\hspace{0.17em}} \\ \Rightarrow \frac{Area of paralle\log ram APQR}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}}= \frac{ {\displaystyle }{\displaystyle 2B{C}^2 }}{4B{C}^2\hspace{0.17em}\hspace{0.17em}} \\ \Rightarrow \frac{Area of paralle\log ram APQR}{Area of ∆ ABC\hspace{0.17em}\hspace{0.17em}}= \frac{ {\displaystyle }1{\displaystyle }}{2\hspace{0.17em}} \\ \Rightarrow Area of paralle\log ram APQR= \frac{ {\displaystyle }{\displaystyle 1 }}{2\hspace{0.17em}\hspace{0.17em}}\times Area of ∆ ABC\hspace{0.17em}\hspace{0.17em} \end{gathered}$$

So, we can say that

Area of parallelogram APQR is $\frac{1}{2}$ of the $∆$ ABC( answer)
Regards