A rectangle has four sides and four right angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, [latex]L[/latex], and the adjacent side as the width, [latex]W[/latex]. See the image below.

A rectangle has four sides, and four right angles. The sides are labeled L for length and W for width.

The perimeter, [latex]P[/latex], of the rectangle is the distance around the rectangle. If you started at one corner and walked around the rectangle, you would walk [latex]L+W+L+W[/latex] units, or two lengths and two widths. The perimeter then is

[latex]\begin{array}{c}P=L+W+L+W\hfill \\ \hfill \text{or}\hfill \\ P=2L+2W\hfill \end{array}[/latex]

What about the area of a rectangle? Remember the rectangular rug from the beginning of this section. It was [latex]2[/latex] feet long by [latex]3[/latex] feet wide, and its area was [latex]6[/latex] square feet. See the image below. Since [latex]A=2\cdot 3[/latex], we see that the area, [latex]A[/latex], is the length, [latex]L[/latex], times the width, [latex]W[/latex], so the area of a rectangle is [latex]A=L\cdot W[/latex].

The area of this rectangular rug is [latex]6[/latex] square feet, its length times its width.

For easy reference as we work the examples in this section, we will restate the Problem Solving Strategy for Geometry Applications here.

example

The length of a rectangle is [latex]32[/latex] meters and the width is [latex]20[/latex] meters.

1. Find the perimeter

2. Find the area

Solution

1.
Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the perimeter of a rectangle
Step 3. Name. Choose a variable to represent it.	Let [latex]P[/latex] = the perimeter
Step 4. Translate. Write the appropriate formula. Substitute.
Step 5. Solve the equation.	[latex]P=64m+40m[/latex] [latex]P=104m[/latex]
Step 6. Check: [latex]p\stackrel{?}{=}104m[/latex] [latex]20m+32m+20m+32m\stackrel{?}{=}104m[/latex] [latex]104m=104m\checkmark[/latex]
Step 7. Answer the question.	The perimeter of the rectangle is [latex]104[/latex] meters.

2.
Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the area of a rectangle
Step 3. Name. Choose a variable to represent it.	Let A = the area
Step 4. Translate. Write the appropriate formula. Substitute.
Step 5. Solve the equation.	[latex]A=640m^2[/latex]
Step 6. Check: [latex]A\stackrel{?}{=}640m^2[/latex] [latex]32m\cdot 20m\stackrel{?}{=}640m^2[/latex] [latex]640m^2=640m^2\checkmark[/latex]
Step 7. Answer the question.	The area of the rectangle is [latex]640[/latex] square meters.

example

Find the length of a rectangle with perimeter [latex]50[/latex] inches and width [latex]10[/latex] inches.

Show Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the length of the rectangle
Step 3. Name. Choose a variable to represent it.	Let [latex]L[/latex] = the length
Step 4. Translate. Write the appropriate formula. Substitute.
Step 5. Solve the equation.	[latex]50\color{red}{- 20}=2L+20\color{red}{- 20}[/latex] [latex]30=2L[/latex] [latex]{\Large\frac{30}{\color{red}{2}}}={\Large\frac{2L}{\color{red}{2}}}[/latex] [latex]15=L[/latex]
Step 6. Check: [latex]p=50[/latex] [latex]15+10+15+10\stackrel{?}{=}50[/latex] [latex]50=50\checkmark[/latex]
Step 7. Answer the question.	The length is [latex]15[/latex] inches.

In the next example, the width is defined in terms of the length. We’ll wait to draw the figure until we write an expression for the width so that we can label one side with that expression.

example

The width of a rectangle is two inches less than the length. The perimeter is [latex]52[/latex] inches. Find the length and width.

Show Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the length and width of the rectangle
Step 3. Name. Choose a variable to represent it. Now we can draw a figure using these expressions for the length and width.	Since the width is defined in terms of the length, we let L = length. The width is two feet less that the length, so we let L − 2 = width
Step 4.Translate. Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information. Substitute in the given information.
Step 5. Solve the equation.	[latex]52=2L+2L - 4[/latex]
Combine like terms.	[latex]52=4L - 4[/latex]
Add [latex]4[/latex] to each side.	[latex]56=4L[/latex]
Divide by [latex]4[/latex].	[latex]{\Large\frac{56}{4}}={\Large\frac{4L}{4}}[/latex]
	[latex]14=L[/latex]
	The length is [latex]14[/latex] inches.
Now we need to find the width.	[latex]L-2[/latex] [latex]\color{red}{14}-2[/latex] [latex]12[/latex]
The width is [latex]L−2[/latex].	The width is [latex]12[/latex] inches.
Step 6. Check: Since [latex]14+12+14+12=52[/latex] , this works!
Step 7. Answer the question.	The length is [latex]14[/latex] feet and the width is [latex]12[/latex] feet.

example

The length of a rectangle is four centimeters more than twice the width. The perimeter is [latex]32[/latex] centimeters. Find the length and width.

Show Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the length and width
Step 3. Name. Choose a variable to represent it.	let W = width The length is four more than twice the width. [latex]2w+4[/latex] = length
Step 4.Translate. Write the appropriate formula and substitute in the given information.
Step 5. Solve the equation.	[latex]32=4w+8+2w[/latex] [latex]32=6w+8[/latex] [latex]24=6w[/latex] [latex]4=w[/latex] (width) [latex]2w+4=[/latex] length [latex]2(\color{red}{4})+4[/latex] [latex]8+4=12[/latex] The length is [latex]12[/latex]cm.
Step 6. Check: [latex]P=2L+2W[/latex] [latex]P\stackrel{?}{=}2\cdot 12+2\cdot 4[/latex] [latex]32=32\quad\checkmark[/latex]
Step 7. Answer the question.	The length is [latex]12[/latex] cm and the width is [latex]4[/latex] cm.

Watch this video to see another similar example of finding area given the relationship between the length and width of a rectangle.

example

The area of a rectangular room is [latex]168[/latex] square feet. The length is [latex]14[/latex] feet. What is the width?

Show Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the width of a rectangular room
Step 3. Name. Choose a variable to represent it.	Let W = width
Step 4.Translate. Write the appropriate formula and substitute in the given information.	[latex]A=LW[/latex] [latex]168=14W[/latex]
Step 5. Solve the equation.	[latex]\Large\frac{168}{14}\normalsize= \Large\frac{14W}{14}[/latex] [latex]12=W[/latex]
Step 6. Check: [latex]A=LW[/latex] [latex]168\stackrel{?}{=}14\cdot 12[/latex] [latex]168=168\checkmark[/latex]
Step 7. Answer the question.	The width of the room is [latex]12[/latex] feet.

In the next example you will see a similar application involving perimeter, length, and width.

example

The perimeter of a rectangular swimming pool is [latex]150[/latex] feet. The length is [latex]15[/latex] feet more than the width. Find the length and width.

Show Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the length and width of the pool
Step 3. Name. Choose a variable to represent it. The length is [latex]15[/latex] feet more than the width.	Let [latex]W=\text{width}[/latex] [latex]W+15=\text{length}[/latex]
Step 4.Translate. Write the appropriate formula and substitute.
Step 5. Solve the equation.	[latex]150=2w+30+2w[/latex] [latex]150=4w+30[/latex] [latex]120=4w[/latex] [latex]30=w[/latex] the width of the pool. [latex]w+15[/latex] the length of the pool. [latex]\color{red}{30}+15[/latex] [latex]45[/latex]
Step 6. Check: [latex]P=2L+2W[/latex] [latex]150\stackrel{?}{=}2(45)+2(30)[/latex] [latex]150=150\checkmark[/latex]
Step 7. Answer the question.	The length of the pool is [latex]45[/latex] feet and the width is [latex]30[/latex] feet.