# The length of a rectangle is twice the width. The area is 578\,yd^{2}.Find the length and width.

## Question:

The length of a rectangle is twice the width. The area is {eq}578\,\text{yd}^{2}. {/eq} Find the length and width.

## Area of a rectangle

A rectangle is a quadrilateral or four-sided figure where the sides are called length and width. The length is longer than the width. The area of a rectangle is the product of the length and the width.

The length, {eq}l {/eq}, is twice the width, {eq}x {/eq}. This means that the length, {eq}l = 2x {/eq}.

Since the area is equal to {eq}\rm 578 \ yd^{2} {/eq}, then:

{eq}Area_{\square} = l \times w {/eq}

{eq}578 = 2x \times x {/eq}

{eq}578 = 2x^{2} {/eq}

{eq}2x^{2} = 578 {/eq}

{eq}\displaystyle \frac{2x^{2}}{2} = \frac{578}{2} {/eq}

{eq}x^{2} = 289 {/eq}

Using the square root to get the value for {eq}x {/eq}:

{eq}\sqrt{x^{2}} = \sqrt{289} {/eq}

{eq}x = \pm 17 {/eq}.

Thus, the value of the width we will use is {eq}\rm 17 \ yards {/eq}. Since the length is twice the width, then the length is {eq}\rm 2 \times 17 \ yards = 34 \ yards {/eq}. The rectangle with an area of {eq}578 \ yd^{2} {/eq} has a width of {eq}\rm 17 \ yd {/eq} and a length of {eq}\rm 34 \ yd {/eq}.