Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Suppose that the width of a certain rectangle is 1 inch more than one-fourth of its length. The perimeter of the rectangle is 52 inches. Find the length and width of the rectangle.

Sagot :

Answer:

Length = 20

Width = 6

Step-by-step explanation:

Let the length of the rectangle be l inch, so, its width

[tex]w= (\frac{1}{4}l+1)[/tex]

So the perimeter equals:

[tex]P= 2l+2w=2l+2(\frac{1}{4}l+1)=2l+\frac{1}{2}l+2=\frac{5}{2}l+2[/tex]

[tex]52= \frac{5}{2}l+2[/tex]

[tex]50 = \frac{5}{2}l[/tex]    [tex]100=5l[/tex]      [tex]20 = l[/tex]

Hence width:

[tex]w = (\frac{1}{4}l+1)[/tex]

[tex]w=5+1 = 6[/tex]

Length = 20

Width = 6

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.