**aims**to understand the

**geometry**of the

**rectangle**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 $ax²+bx+c=0$, 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 $L$. \[ Area=W \left(\dfrac{400}{3} – \dfrac{2L}{3} \right)\] \[=\left(\dfrac{400}{3}- \dfrac{2L}{3} \right)\] \[=L \left(\dfrac{400}{3} – \dfrac{2L}{3} \right) \] \[= -\dfrac{2}{3}L^{2}+ \dfrac{400}{3}L \] 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: \[Parameter = 2(L + W)\] \[= 2(17 + 13)\] \[= 2(30)\] \[Parameter = 60 cm\]

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