Answer
The dimensions are 5 meter by 8 meter.
Work Step by Step
We know that: Area $=\text{ }40\text{ square meters}$
Perimeter= $26\text{ meter}$
Consider, L is length, W is width
Then: Area of rectangle= $ L\times W=40,L=\frac{40}{W}$
Perimeter $2(L+W)=26$
Put L
$\begin{align}
& 2\left( \frac{40}{W}+W \right)=26 \\
& 2\left( \frac{40+{{W}^{2}}}{W} \right)=26 \\
& 40+{{W}^{2}}=26\times \frac{W}{2} \\
& {{W}^{2}}+40=13W
\end{align}$
Solving further as given below:
$\begin{align}
& {{W}^{2}}-13W+40=0 \\
& W\left( W-8 \right)-5\left( W-8 \right)=0 \\
& \left( W-8 \right)\left( W-5 \right)=0
\end{align}$
Thus, the rectangle dimensions are $8\text{ meter}$ and $5\text{ meter}$.
