Solving Applications Using Rectangle Properties

Solving Applications Using Rectangle Properties

You may already be familiar with the properties of rectangles. Rectangles have four sides and four right \(\left(90\text{°}\right)\) angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, L, and its adjacent side as the width, W.

The distance around this rectangle is \(L+W+L+W,\) or \(2L+2W.\) This is the perimeter, P, of the rectangle.

What about the area of a rectangle? Imagine a rectangular rug that is 2-feet long by 3-feet wide. Its area is 6 square feet. There are six squares in the figure.

The area is the length times the width.

The formula for the area of a rectangle is \(A=LW.\)

Example

The length of a rectangle is 32 meters and the width is 20 meters. What is the perimeter?

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 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.
Step 6. Check. \(\begin{array}{ccc}\hfill P& \stackrel{?}{=}\hfill & 104\hfill \\ \hfill 20+32+20+32& \stackrel{?}{=}\hfill & 104\hfill \\ \hfill 104& =\hfill & 104✓\hfill \end{array}\)
Step 7. Answer the question.	The perimeter of the rectangle is 104 meters.

Example

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

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 width of a rectangular room
Step 3. Name. Choose a variable to represent it.	Let W = the width.
Step 4. Translate.
Write the appropriate formula.	\(\phantom{\rule{0.9em}{0ex}}A=LW\)
Substitute.	\(\phantom{\rule{0.2em}{0ex}}168=14W\)
Step 5. Solve the equation.	\(\frac{168}{14}=\frac{14W}{14}\) \(\phantom{\rule{0.6em}{0ex}}12=W\)
Step 6. Check. \(\begin{array}{ccc}\hfill A& =\hfill & LW\hfill \\ \hfill 168& \stackrel{?}{=}\hfill & 14\cdot 12\hfill \\ \hfill 168& =\hfill & 168✓\hfill \end{array}\)
Step 7. Answer the question.	The width of the room is 12 feet.

Example

Find the length of a rectangle with perimeter 50 inches and width 10 inches.

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 L = the length.
Step 4. Translate.
Write the appropriate formula.
Substitute.
Step 5. Solve the equation.
Step 6. Check. \(\begin{array}{ccc}\hfill P& =\hfill & 50\hfill \\ \hfill 15+10+15+10& \stackrel{?}{=}\hfill & 50\hfill \\ \hfill 50& =\hfill & 50✓\hfill \end{array}\)
Step 7. Answer the question.	The length is 15 inches.

We have solved problems where either the length or width was given, along with the perimeter or area; now we will learn how to solve problems in which the width is defined in terms of the length. We will 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 feet less than the length. The perimeter is 52 feet. Find the length and width.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the length and width of a rectangle
Step 3. Name. Choose a variable to represent it. Since the width is defined in terms of the length, we let L = length. The width is two feet less than the length, so we let L − 2 = width.	\(\phantom{\rule{4em}{0ex}}P=52\) ft
Step 4. Translate.
Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information.	\(P=2L+2W\)
Substitute in the given information.	\(52=2L+2\left(L-2\right)\)
Step 5. Solve the equation.	\(52=2L+2L-4\)
Combine like terms.	\(52=4L-4\)
Add 4 to each side.	\(56=4L\)
Divide by 4.	\(\frac{56}{4}=\frac{4L}{4}\) \(14=L\) The length is 14 feet.
Now we need to find the width.	The width is \(L-2\). The width is 12 feet.
Step 6. Check. Since \(14+12+14+12=52\), this works!
Step 7. Answer the question.	The length is 14 feet and the width is 12 feet.

Example

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

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the length and the width
Step 3. Name. Choose a variable to represent the width.
The length is four more than twice the width.
Step 4. Translate
Write the appropriate formula.
Substitute in the given information.
Step 5. Solve the equation.	12 The length is 12 cm.
Step 6. Check. \(\begin{array}{ccc}\hfill P& =\hfill & 2L+2W\hfill \\ \hfill 32& \stackrel{?}{=}\hfill & 2\cdot 12+2\cdot 4\hfill \\ \hfill 32& =\hfill & 32✓\hfill \end{array}\)
Step 7. Answer the question.	The length is 12 cm and the width is 4 cm.

Example

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

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.	\(\phantom{\rule{3em}{0ex}}P=150\) ft
Step 2. Identify what you are looking for.	the length and the width of the pool
Step 3. Name. Choose a variable to represent the width. The length is 15 feet more than the width.
Step 4. Translate
Write the appropriate formula.
Substitute.
Step 5. Solve the equation.
Step 6. Check. \(\begin{array}{ccc}\hfill P& =\hfill & 2L+2W\hfill \\ \hfill 150& \stackrel{?}{=}\hfill & 2\left(45\right)+2\left(30\right)\hfill \\ \hfill 150& =\hfill & 150✓\hfill \end{array}\)
Step 7. Answer the question.	The length of the pool is 45 feet and the width is 30 feet.