In $\triangle QAP , l ( AP )=30 m , l ( QA )=50 m$
$A (\triangle QAP )$
$=\frac{1}{2} \times$ product of sides forming the right angle
$=\frac{1}{2} \times I(A P) \times I(Q A)$
$=\frac{1}{2} \times 30 \times 50$
$=750 sq \cdot m$
In $\square QACR , l ( QA )=50 m , l ( RC )=25 m$,
$I(A C)=I(A B)+I(B C)$
$=30+30=60 m$
$A (\square QACR )$
$=\frac{1}{2} \times$ sum of lengths of parallel sides $x$ height
$=\frac{1}{2} \times[ I ( QA )+ I ( RC )] \times I ( AC )$
$=\frac{1}{2} \times(50+25) \times 60$
$=\frac{1}{2} \times 75 \times 60$
$=2250$ sq.m
In $\triangle RCS , I ( CS )=60 m , I ( RC )=25 m A (\triangle RCS )$
$=\frac{1}{2} \times$ product of sides forming the right angle
$=\frac{1}{2} \times I ( CS ) \times I ( RC )$
$=\frac{1}{2} \times 60 \times 25$
$=750$ sq. $m$
\begin{aligned}
& \text { In } \triangle PTS , I ( TB )=30 m \text {, } \\
& I(P S)=I(P A)+I(A B)+I(B C)+I(C S) \\
& =30+30+30+60 \\
& =150 m \\
& A (\triangle PTS )=\frac{1}{2} \times { base } \times { height } \\
& =\frac{1}{2} \times I ( PS ) \times I ( TB ) \\
& =\frac{1}{2} \times 150 \times 30 \\
& =2250 sq . m \\
& \therefore \text { Area of plot } QPTSR = A (\triangle QAP )+ A (\square QACR )+ A (\triangle RCS )+ \\
& A (\triangle PTS ) \\
& =750+2250+750+2250 \\
& =6000 sq \cdot m \\
& \therefore \text { The area of the given plot is } 6000 \text { sq.m. } \\
\end{aligned}
\begin{aligned}
& \text { ii. In } \triangle A B E, m \angle B A E=90^{\circ}, I ( AB )=24 m , I ( BE )=30 m \\
& \therefore[ I ( BE )]^2=[ I ( AB )]^2+[ I ( AE )]^2 \\
& \ldots[\text { Pythagoras theorem }] \\
& \therefore(30)^2=(24)^2+[ I ( AE )]^2 \\
& \therefore 900=576+[ I ( AE )]^2 \\
& \therefore[ I ( AE )]^2=900-576 \\
& \therefore[ I ( AE )]^2=324 \\
& \therefore I ( AE )=\sqrt{ } 324=18 m \\
& \ldots[\text { Taking square root of both sides }] \\
& A(\triangle A B E) \\
& =\frac{1}{2} \times \text { product of sides forming the right angle } \\
& =\frac{1}{2} \times I ( AE ) \times I ( AB ) \\
& =\frac{1}{2} \times 18 \times 24 \\
& =216 \text { sq. } m \\
& \text { In } \triangle B C E, a=30 m , b=28 m , C =26 m
\end{aligned}
\begin{aligned}
& \text { Semiperimeter of } \triangle BCE = s =\frac{1}{2}(a+b+c) \\
& =\frac{30+28+26}{2} \\
& =\frac{84}{2} \\
& =42 m \\
& A(\Delta B C E)=\sqrt{s(s-a)(s-b)(s-c)} \\
& =\sqrt{42(42-30)(42-28)(42-26)} \\
& =\sqrt{42 \times 12 \times 14 \times 16} \\
& =\sqrt{2 \times 3 \times 7 \times 2 \times 2 \times 3 \times 2 \times 7 \times 4 \times 4} \\
& =\sqrt{2^2 \times 2^2 \times 3^2 \times 4^2 \times 7^2} \\
& =2 \times 2 \times 3 \times 4 \times 7 \\
& =336 sq . m \\
&
\end{aligned}
In $\triangle EDC , l( CE )=28 m , l( DF )=16 m$
$
\begin{aligned}
A (\triangle EDC ) & =\frac{1}{2} \times \text { base } \times \text { height } \\
& =\frac{1}{2} \times l( CE ) \times l( DF ) \\
& =\frac{1}{2} \times 28 \times 16 \\
& =224 \text { sq. } m
\end{aligned}
$
$\therefore$ Area of plot $A B C D E$
$= A (\triangle ABE )+ A (\triangle BCE )+ A (\triangle EDC )$
$=216+336+224$
$=776$ sq. $m$
$\therefore$ The area of the given plot is 776 sq.m.
[Note: In the given figure, we have taken $I(D F)=16 m$ ]
