Find the Perimeter and Area of a Rectangle

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].

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? Below is a rectangular rug. It is [latex]2[/latex] feet long by [latex]3[/latex] feet wide, and its area is [latex]6[/latex] square feet. 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].

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

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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}\color{black}=2L+20\color{red}{- 20}[/latex] [latex]30=2L[/latex] [latex]{\Large\frac{30}{\color{red}{2}}}\color{black}={\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.

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

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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}\color{black}-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.