Answer


$\begin{aligned} & \text { Length of rectangle } \Rightarrow 2 x+y+8=4 x-y \\
& \Rightarrow 2 x-4 x+y+y=-8 \\
& \Rightarrow-2 x+2 y=-8 \\
& \Rightarrow-x+y=-4 \ldots \ldots(I)
\end{aligned}$
$\begin{aligned}
& \text { Breadth of the rectangle }=2 y=x+4 \\
& \Rightarrow-x+2 y=4 \ldots . . \text { (II) }
\end{aligned}$
Equating Eq. I and II and change sign of Eq. II
$\begin{gathered}
-x+y=-4 \\
x-2 y=-4 \\
\hline -y=-8 \\
y=8
\end{gathered}$
Substituting $y=8$ in Eq.I
$\begin{aligned}
& -x+8=-4 \\
& -x=-4-8 \\
& -x=-12 \\
& x=12
\end{aligned}$
$\begin{aligned} & \text { Length }=2 \times 12+8+8=40 \\ & \text { Breadth }=2 \times 8=16 \\ & \text { Area }=\text { Length } \times \text { breadth }=40 \times 16=640 \text { sq. unit } \\ & \text { Perimeter }=2 \text { (Length }+ \text { Breadth) }=2(40+16)=2(56)=112 \text { unit. }\end{aligned}$

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