## Precalculus (6th Edition) Blitzer

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 $5\text{ meter}$.