Find the dimension. Then find the area of the rectangle.

Question

The perimeter of a rectangle is 80ft. Find the dimensions if the length is 5ft longer than four times the width. Then find the area of the rectangle.

Answer 1

In this problem, you have two variables–length and width. Let's look at what is given using L for length and W for width.

You know that the perimeter of the rectangle is 80 ft, which is the sum of all sides when added together. For that shape, there are 2 lengths and 2 widths, so 2L+2W= 80.

You also know that the length is 5 ft longer than four times the width, so this means that the length is: L= 4W +5.

Now because there are two variables, substitute L for "4W+5" into the other equation, so it looks like this:

2(4W+5) +2W=80

 

Using the distributive property, "partner up" the 2 with each of the terms inside of the ( ) by multiplication, so the equation becomes

2•4W+2•5 +2W=80 which is 8W+10+2W=80.

 

Combine the like terms

10W+10=80

You want W to be isolated, so we must remove the coefficient (the number in front of the variable) by doing the opposite operation, which is division:

10W ÷10= 70÷10

W=7

 

So now we can go back and replace W with 7. Either equation would work, but since L is isolated in L=4W+5, the math is easier.

L= 4(7) +5 which is 28+5 so L=33.

Check to see if it works in both equations, with L as 33 and W as 7:

2(7)+ 2(33)= 80.

14+66= 80 which is 80=80.

 

(33)= 4(7)+5

33=28+5

33=33