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A rectangular concrete patio has a perimeter of $36\,\text{meters}$ . Its area is $80\,\text{square meters}$ . What are the dimensions of the patio? $\newline$ $\_\_\_\_$ meters by $\_\_\_\_$ meters

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Area and perimeter: word problems

Full solution

Q. A rectangular concrete patio has a perimeter of

36\,\text{meters}

. Its area is

80\,\text{square meters}

. What are the dimensions of the patio?

\newline

\_\_\_\_

meters by

\_\_\_\_

meters

Rectangle Perimeter Equation: Let $l$ and $w$ be the length and width of the rectangle, respectively. $\newline$ The perimeter of a rectangle is given by $P = 2l + 2w$ . $\newline$ Given $P = 36$ meters, we can write the equation: $\newline$ $2l + 2w = 36$ $\newline$ Simplify by dividing everything by $2$ : $\newline$ $l + w = 18$
Rectangle Area Equation: We also know the area of the rectangle is $A = lw$ and it's given as $80$ square meters. $\newline$ So, we have the equation: $\newline$ $lw = 80$
Solving for Width: From the perimeter equation $l + w = 18$ , solve for one variable in terms of the other. Let's solve for $w$ : $\newline$ $w = 18 - l$
Substitute into Area Equation: Substitute $w = 18 - l$ into the area equation $lw = 80$ : $l(18 - l) = 80$ Expand the equation: $18l - l^2 = 80$ Rearrange to form a quadratic equation: $l^2 - 18l + 80 = 0$
Quadratic Equation Solution: Solve the quadratic equation using the quadratic formula, $l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -18$ , and $c = 80$ : $\newline$ $l = \frac{-(-18) \pm \sqrt{(-18)^2 - 4\cdot1\cdot80}}{2\cdot1}$ $\newline$ $l = \frac{18 \pm \sqrt{324 - 320}}{2}$ $\newline$ $l = \frac{18 \pm \sqrt{4}}{2}$ $\newline$ $l = \frac{18 \pm 2}{2}$ $\newline$ $l = 10$ or $l = 8$

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