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Study Guides > ALGEBRA / TRIG I

Using the Properties of Rectangles to Solve Problems

Learning Outcomes

  • Use properties of rectangles
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, LL, and the adjacent side as the width, WW. 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 L, the sides are labeled W. The perimeter, PP, of the rectangle is the distance around the rectangle. If you started at one corner and walked around the rectangle, you would walk L+W+L+WL+W+L+W units, or two lengths and two widths. The perimeter then is

P=L+W+L+WorP=2L+2W\begin{array}{c}P=L+W+L+W\hfill \\ \hfill \text{or}\hfill \\ P=2L+2W\hfill \end{array}

What about the area of a rectangle? Remember the rectangular rug from the beginning of this section. It was 22 feet long by 33 feet wide, and its area was 66 square feet. See the image below. Since A=23A=2\cdot 3, we see that the area, AA, is the length, LL, times the width, WW, so the area of a rectangle is A=LWA=L\cdot W. The area of this rectangular rug is 66 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 (90)\left(\text{90}^ \circ\right) angles.
  • The lengths of opposite sides are equal.
  • The perimeter, PP, of a rectangle is the sum of twice the length and twice the width. See the first image. P=2L+2WP=2L+2W
  • The area, AA, of a rectangle is the length times the width. A=LWA=L\cdot W
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 3232 meters and the width is 2020 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 PP = the perimeter
Step 4. Translate. Write the appropriate formula. Substitute. .
Step 5. Solve the equation. P=64m+40mP=64m+40m P=104mP=104m
Step 6. Check: p=?104mp\stackrel{?}{=}104m 20m+32m+20m+32m=?104m20m+32m+20m+32m\stackrel{?}{=}104m 104m=104m104m=104m\checkmark
Step 7. Answer the question. The perimeter of the rectangle is 104104 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. A=640m2A=640m^2
Step 6. Check: A=?640m2A\stackrel{?}{=}640m^2 32m20m=?640m232m\cdot 20m\stackrel{?}{=}640m^2 640m2=640m2640m^2=640m^2\checkmark
Step 7. Answer the question. The area of the rectangle is 640640 square meters.
 

try it

[ohm_question]146510[/ohm_question]
 

example

Find the length of a rectangle with perimeter 5050 inches and width 1010 inches.

Answer: 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 LL = the length
Step 4. Translate. Write the appropriate formula. Substitute. .
Step 5. Solve the equation. 5020=2L+202050\color{red}{- 20}=2L+20\color{red}{- 20} 30=2L30=2L 302=2L2{\Large\frac{30}{\color{red}{2}}}={\Large\frac{2L}{\color{red}{2}}} 15=L15=L
Step 6. Check: p=50p=50 15+10+15+10=?5015+10+15+10\stackrel{?}{=}50 50=5050=50\checkmark
Step 7. Answer the question. The length is 1515 inches.

 

try it

[ohm_question]146518[/ohm_question]
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 5252 inches. Find the length and width.

Answer: 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. 52=2L+2L452=2L+2L - 4
Combine like terms. 52=4L452=4L - 4
Add 44 to each side. 56=4L56=4L
Divide by 44. 564=4L4{\Large\frac{56}{4}}={\Large\frac{4L}{4}}
14=L14=L
The length is 1414 inches.
Now we need to find the width.  L2L-2 142\color{red}{14}-2 1212
The width is L2L−2.   The width is 1212 inches.
Step 6. Check: Since 14+12+14+12=5214+12+14+12=52 , this works!
Step 7. Answer the question. The length is 1414 feet and the width is 1212 feet.

 

try it

[ohm_question]146504[/ohm_question]
 

example

The length of a rectangle is four centimeters more than twice the width. The perimeter is 3232 centimeters. Find the length and width.

Answer: 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. 2w+42w+4 = length .
Step 4.Translate. Write the appropriate formula and substitute in the given information. .
Step 5. Solve the equation. 32=4w+8+2w32=4w+8+2w 32=6w+832=6w+8 24=6w24=6w 4=w4=w (width) 2w+4=2w+4= length 2(4)+42(\color{red}{4})+4 8+4=128+4=12 The length is 1212cm.
Step 6. Check: P=2L+2WP=2L+2W P=?212+24P\stackrel{?}{=}2\cdot 12+2\cdot 4 32=3232=32\quad\checkmark
Step 7. Answer the question. The length is 1212 cm and the width is 44 cm.

 

try it

[ohm_question]146521[/ohm_question] [ohm_question]146522[/ohm_question]
Watch this video to see another similar example of finding area given the relationship between the length and width of a rectangle. https://youtu.be/zUlU64Umnq4

example

The area of a rectangular room is 168168 square feet. The length is 1414 feet. What is the width?

Answer: 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. A=LWA=LW 168=14W168=14W
Step 5. Solve the equation. 16814=14W14\Large\frac{168}{14}\normalsize= \Large\frac{14W}{14} 12=W12=W
Step 6. Check: A=LWA=LW 168=?1412168\stackrel{?}{=}14\cdot 12 168=168168=168\checkmark
Step 7. Answer the question. The width of the room is 1212 feet.

 

try it

[ohm_question]146523[/ohm_question]
In the next example you will see a similar application involving perimeter, length, and width. https://youtu.be/jlxPgKQfhQs

example

The perimeter of a rectangular swimming pool is 150150 feet. The length is 1515 feet more than the width. Find the length and width.

Answer: 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 1515 feet more than the width. Let W=widthW=\text{width} W+15=lengthW+15=\text{length}
Step 4.Translate. Write the appropriate formula and substitute. .
Step 5. Solve the equation. 150=2w+30+2w150=2w+30+2w 150=4w+30150=4w+30 120=4w120=4w 30=w30=w the width of the pool. w+15w+15 the length of the pool. 30+15\color{red}{30}+15 4545
Step 6. Check: P=2L+2WP=2L+2W 150=?2(45)+2(30)150\stackrel{?}{=}2(45)+2(30) 150=150150=150\checkmark
Step 7. Answer the question. The length of the pool is 4545 feet and the width is 3030 feet.

 

try it

[ohm_question]146524[/ohm_question]

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