Solving Geometry Applications

Translate to a system of equations and then solve:

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

Solution

\(\begin{array}{cccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem.}\hfill & & & \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & \text{We are looking for the measure of each angle.}\hfill \\ \begin{array}{c}\mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill \\ \end{array}\hfill & & \begin{array}{c}\text{Let}\phantom{\rule{0.2em}{0ex}}x=\text{the measure of the first angle.}\hfill \\ y=\text{the measure of the second angle}\hfill \end{array}\hfill \\ \mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into a system of equations.}\hfill & & \text{The angles are complementary.}\hfill \\ & & \hfill \phantom{\rule{0.2em}{0ex}}x+y=90\hfill \\ & & \text{The difference of the two angles is 26 degrees.}\hfill \\ & & \hfill \phantom{\rule{0.28em}{0ex}}x-y=26\hfill \\ \text{The system is}\hfill & & \hfill \begin{array}{c}x+y=90\hfill \\ x-y=26\hfill \end{array}\hfill \\ \begin{array}{c}\mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the system of equations by elimination.}\hfill \\ \end{array}\hfill & & \hfill \phantom{\rule{0.6em}{0ex}}\begin{array}{c}\underset{\text{_________}}{\begin{array}{l}x+y=90\hfill \\ x-y=26\hfill \end{array}}\hfill \\ 2x\phantom{\rule{1.5em}{0ex}}=116\hfill \end{array}\hfill \\ & & \hfill \phantom{\rule{2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}58\hfill \\ \begin{array}{}\\ \text{Substitute}\phantom{\rule{0.2em}{0ex}}x=58\phantom{\rule{0.2em}{0ex}}\text{into the first equation.}\hfill \end{array}\hfill & & \hfill \begin{array}{ccc}\hfill x+y& =\hfill & 90\hfill \\ \hfill 58+y& =\hfill & 90\hfill \\ \hfill y& =\hfill & 32\hfill \end{array}\hfill \\ \mathbf{\text{Step 6. Check}}\phantom{\rule{0.2em}{0ex}}\text{the answer in the problem.}\hfill & & \\ \begin{array}{c}\hfill 58+32\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}90✓\hfill \\ \hfill 58-32\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}26✓\hfill \end{array}\hfill \\ \mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question.}\hfill & & \text{The angle measures are 58 degrees and 42 degrees.}\hfill \end{array}\)

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for the measure of each angle.
Step 3. Name what we are looking for.	Let \(x=\) the measure of the first angle. \(\phantom{\rule{1.5em}{0ex}}y=\) the measure of the second angle
Step 4. Translate into a system of equations.	The angles are supplementary.
	The larger angle is twelve less than five times the smaller angle
The system is: Step 5. Solve the system of equations substitution.
Substitute 5x − 12 for y in the first equation.
Solve for x.
Substitute 32 for in the second equation, then solve for y.
Step 6. Check the answer in the problem. \(\begin{array}{ccc}\hfill 32+158& =\hfill & 180\phantom{\rule{0.2em}{0ex}}✓\hfill \\ \hfill 5·32-12& =\hfill & 147\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\)
Step 7. Answer the question.	The angle measures are 148 and 32.

Translate to a system of equations and then solve:

Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	We are looking for the length and width.
Step 3. Name what we are looking for.	Let \(L=\) the length of the fenced yard. \(\phantom{\rule{1.3em}{0ex}}W=\) the width of the fenced yard
Step 4. Translate into a system of equations.	One length and two widths equal 125.
	The length will be 5 feet more than four times the width.
The system is: Step 5. Solve the system of equations by substitution.
Substitute L = 4W + 5 into the first equation, then solve for W.
Substitute 20 for W in the second equation, then solve for L.
Step 6. Check the answer in the problem. \(\begin{array}{ccc}\hfill 20+28+20& =\hfill & 125\phantom{\rule{0.2em}{0ex}}✓\hfill \\ \hfill 85& =\hfill & 4·20+5\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\)
Step 7. Answer the equation.	The length is 85 feet and the width is 20 feet.