# Molly's rectangular backyard has 300 ft of fencing. one side is 40 ft, find the area of the...

## Question:

Molly's rectangular backyard has 300 ft of fencing. one side is 40 ft, find the area of the backyard.

## Area and Perimeter of a Rectangle

In this scenario, we are going to find the width of the backyard first by using the perimeter and length, then solve for the area by multiplying the length and width.

{eq}\underline{\text{Given:}}\\ \bullet \text{Perimeter (p): }300 \text{ ft.}\\ \bullet \text{Length (l): }40 \text{ ft.}\\ \bullet \text{Width (w): ?}\\ \bullet \text{Area (A): ?}\\ \bullet \text{Perimeter of rectangle: } 2l + 2w = p\\ \bullet \text{Area of rectangle: } l \cdot w = A\\~\\ \text{Before solving for the area, we need to find the width.}\\ \text{To find the width, plugin } l \text{ and } p \text{ into the perimeter equation and solve for }w.\\ \require{cancel} \begin{align} 2l + 2w &= p\\ 2(40 \text{ ft.}) + 2w &= (300 \text{ ft.})\\ 80 \text{ ft.} + 2w &= 300 \text{ ft.}\\ 2w &= 300 \text{ ft.} - 80 \text{ ft.}\\ 2w &= 220 \text{ ft.}\\ w &= \frac{220 \text{ ft.}}{2}\\ w &= \frac{2 \cdot 2 \cdot 5 \cdot 11 \cdot \text{ ft.}}{2}\\ w &= \cancel{\frac{2}{2}} \cdot \frac{2 \cdot 5 \cdot 11 \cdot \text{ ft.}}{1}\\ w &= 1 \cdot 110 \text{ ft.}\\ w &\mathbf{= 110 }\text{ ft.}\\ \end{align}\\ {/eq}

{eq}\text{With the width defined, we can solve for the area.}\\ \text{To solve for the area, plugin }l \text{ and } w \text{ into the area equation and solve for } A.\\ \begin{align} A &= l \cdot w\\ &= (40 \text{ ft.}) \cdot (110 \text{ ft.})\\ &\mathbf{= \color{blue}4\color{blue}4\color{blue}0\color{blue}0 \text{ ft.}^{2}}\\ \end{align}\\~\\ \textbf{Therefore, the area of this rectangular backyard is } \mathbf{\color{blue}4\color{blue}4\color{blue}0\color{blue}0 \text{ ft.}^{2}}. {/eq} 