Using the Properties of Rectangles to Solve Problems

Learning Outcomes

  • Find perimeter and area of a rectangle given length and width
  • Find dimensions of a rectangle given perimeter and a relationship between length and width

 

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.

A rectangle is shown. Each angle is marked with a square. The top and bottom are labeled [latex]L[/latex], the sides are labeled [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? 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.

A rectangle is shown. It is made up of 6 squares. The bottom is 2 squares across and marked as 2, the side is 3 squares long and marked as 3.

Properties of Rectangles

  • Rectangles have four sides and four right [latex]\left(\text{90}^ \circ\right)[/latex] angles.
  • The lengths of opposite sides are equal.
  • The perimeter, [latex]P[/latex], of a rectangle is the sum of twice the length and twice the width. See the first image. [latex]P=2L+2W[/latex]
  • The area, [latex]A[/latex], of a rectangle is the length times the width. [latex]A=L\cdot W[/latex]

 

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

Use a Problem Solving Strategy for Geometry Applications

  1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
  2. Identify what you are looking for.
  3. Name what you are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

 

example

The length of a rectangle is [latex]32[/latex] meters and the width is [latex]20[/latex] meters. Find 1. the perimeter, and 2. 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 P = the perimeter
Step 4. Translate.

Write the appropriate formula.

Substitute.

.
Step 5. Solve the equation. [latex]p=64+40[/latex]

[latex]p=104[/latex]

Step 6. Check:

[latex]p\stackrel{?}{=}104[/latex]

[latex]20+32+20+32\stackrel{?}{=}104[/latex]

[latex]104=104\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=640[/latex]
Step 6. Check:

[latex]A\stackrel{?}{=}640[/latex]

[latex]32\cdot 20\stackrel{?}{=}640[/latex]

[latex]640=640\checkmark[/latex]

Step 7. Answer the question. The area of the rectangle is [latex]60[/latex] square meters.

 

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example

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

 

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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.

 

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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.

 

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example

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

 

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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.

 

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