# The perimeter of a rectangle is 24 ft. Its length is five times its width Let x be the length and y be the width. What is the area of the rectangle?

Jan 30, 2016

$x = 10 \text{ " y= 2" so area } = 20 f {t}^{2}$

#### Explanation:

$\textcolor{b r o w n}{\text{Build the equation by breaking down the question into parts}}$

Perimeter is $24 f t$

Width $\to y f t$
Length "-> xft

So perimeter is $\to \left(2 y + 2 x\right) f t = 24 f t$ ......................(1)

But Length = $5 \times$ width

So $x = 5 y$ ..................................(2)
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$\textcolor{b l u e}{\text{To find y}}$

Substitute equation (2) into equation (1) giving:

$2 y + 2 \left(5 y\right) = 24$

$12 y = 24$

$\textcolor{b r o w n}{\text{y=2}}$ .....................................(3)
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$\textcolor{b l u e}{\text{To find x}}$

Substitute (3) into (2) giving:

color(brown)("x=5(2) -> x=10)
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Check
2x+2y = 24 " "=>" " 2(10)+2(2)=24 " "color(red)("True")

Jan 31, 2016

$x = 10 , y = 2$

$A r e a = 20 f {t}^{2}$

#### Explanation:

In the statement length is said to be $x$ which is $5$ times the width which is $y$

So,

$5 x = y$

We can assume $y$ as $x$

Then perimeter=$\left(5 x + 5 x\right) + \left(x + x\right) = 24$

$\rightarrow 10 x + 2 x = 24$

$\rightarrow 12 x = 24$

$\rightarrow x = \frac{24}{12} = 2$

So,the width=$2$

Then length=$2 \left(5\right) = 10$

Area of rectangle =$L e n g$$t h \cdot w i \mathrm{dt} h = 10 \left(2\right) = 20 f {t}^{2}$