This question **aims** to understand the **geometry** of the **rectangle**Â and how to calculate its **area** and **parameter.** Furthermore, it explains the **quadratic** equations and the **quadratic** formula.

A **rectangle** has four sides. All the internal **angles** in the rectangle are equal to $90 \space degree $. Two sides produce a **right angle** when they meet at any **corner.** In a rectangle, only the **opposite** sides are **equal** in length, making it different from a **square.**

To calculate the **area** of the rectangle, you have to multiply the **length** and the **width.** Area $A$ is $L$ times $W$.

\[Area=L \times W\]

For calculating the **perimeter,** you simply **add** all the **sides** of the rectangle. there are **4** sides and the **opposite** sides are **equal** in length. The **formula** is given as:

\[Perimeter = 2(L+W)\]

The quadratic **formula** enables solving any **quadratic** equation. First, we **convert** the equation in the form $axÂ²+bx+c=0$, where **a**, **b**, and **c** are **coefficients.** Then, we plug these **coefficients** in the formula: $\dfrac{-b \pm \sqrt{4ac}}{2a}$ and find the **unknown** $x$.

## Expert Answer

A **rectangular** playground is to be **fenced** off and split into two by another fence **parallel** to one side. Total **fencing** is $400$ feet.

That means the **4** four sides of the **rectangle** $(2L+2W)$ and the one fence **parallel** to the one side $(**W**)$, a total of $**5$,** $**(2L+3W)$** fences add up to $400$ the total **fencing.** The following equation **represents** the scenario.

\[2L+3W=400\]

\[3W=400-2L\]

\[W= \dfrac{400-2L}{3}\]

\[W= \dfrac{400}{3}- \dfrac{2L}{3}\]

Area of the **rectangle** (in terms of $L$)

\[Area= \space W \times L\]

Inserting $W$ and **calculating** Area in terms of $L$.

\[ Area=W \left(\dfrac{400}{3} – \dfrac{2L}{3}Â \right)\]

\[=\left(\dfrac{400}{3}- \dfrac{2L}{3}Â Â \right)\]

\[=L \left(\dfrac{400}{3} – \dfrac{2L}{3}Â Â \right) \]

\[= -\dfrac{2}{3}L^{2}+ \dfrac{400}{3}L \]

Now solving the **quadratic** equation using the formula:

\[L = \dfrac{-b \pm \sqrt{4ac}}{2a} \]

Here: $a=-\dfrac{2}{3}$, $b = \dfrac{400}{3}$ and $c = 0$

\[= \dfrac{- \dfrac{400}{3} \pm \sqrt{4(\dfrac{2}{3}) (0)}}{- 2 (\dfrac{2}{3})} \]

\[ L = \dfrac{- \dfrac{400}{3}}{- 2 (\dfrac{2}{3})} \]

By solving the expression, $L$ comes out to be:

\[L = 100 \]

Plugging in the $L= 100$ in the Width $W$ equation:

\[ W= \dfrac{400}{3} – \dfrac{2L}{3} \]

\[ W= \dfrac{400}{3} – \dfrac{2(100)}{3} \]

\[ W = 66.6 \]

After calculating the **Length** and **Width,** calculate the **area** of the rectangle:

\[Area = L\times W\]

\[Area = 100 \times 66.6\]

\[Area = 6666.6 \]

## Numerical Answer

The **dimensions** of the rectangular playground that **maximize** the total enclosed **area** is $Length = 100 \space feet$ and $Width= 66.6 \space feet$. The **maximum area** is $6666.6 \space feet^2$.

## Example

The **length** and **width** of the rectangle shape are given as $17 cm$ and $13 cm$ respectively. Find the **area** and the **parameter.**

Given the **Length** $L = 17$ and **Width** $W=13$

The area of the **rectangle** is given by:

\[ Area = L\times W\]

\[ Area = 17 \times13\]

\[ Area = 221 cm^2 \]

The **parameter** of the rectangle is given by:

\[Parameter = 2(L + W)\]

\[= 2(17 + 13)\]

\[= 2(30)\]

\[Parameter = 60 cm\]