3.4 Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem

Solve Applications Using Properties of Triangles

In this section we will use some common geometry formulas. We will adapt our problem-solving strategy so that we can solve geometry applications. The geometry formula will name the variables and give us the equation to solve. In addition, since these applications will all involve shapes of some sort, most people find it helpful to draw a figure and label it with the given information. We will include this in the first step of the problem solving strategy for geometry applications.

We will start geometry applications by looking at the properties of triangles. Let’s review some basic facts about triangles. Triangles have three sides and three interior angles. Usually each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.

The plural of the word vertex is vertices. All triangles have three vertices. Triangles are named by their vertices: The triangle in Figure 3.4 is called $\text{△}ABC.$

Figure 3.4 Triangle ABC has vertices A, B, and C. The lengths of the sides are a, b, and c.

The three angles of a triangle are related in a special way. The sum of their measures is $180\text{°}.$ Note that we read $m\text{∠}A$ as “the measure of angle A.” So in $\text{△}ABC$ in Figure 3.4,

Because the perimeter of a figure is the length of its boundary, the perimeter of $\text{△}ABC$ is the sum of the lengths of its three sides.

To find the area of a triangle, we need to know its base and height. The height is a line that connects the base to the opposite vertex and makes a $90\text{°}$ angle with the base. We will draw $\text{△}ABC$ again, and now show the height, h. See Figure 3.5.

Figure 3.5 The formula for the area of $\text{△}ABC$ is $A=\frac{1}{2}bh,$ where b is the base and h is the height.

Triangle Properties

For $\text{△}ABC$

Angle measures:

$$m\text{∠}A+m\text{∠}B+m\text{∠}C=180$$

• The sum of the measures of the angles of a triangle is $180\text{°}.$

Perimeter:

$$P=a+b+c$$

• The perimeter is the sum of the lengths of the sides of the triangle.

Area:

$$A=\frac{1}{2}bh,b=\text{base},h=\text{height}$$

• The area of a triangle is one-half the base times the height.

Example 3.34

The measures of two angles of a triangle are 55 and 82 degrees. Find the measure of the third angle.

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 measure of the third angle in a triangle
Step 3. Name. Choose a variable to represent it.	Let $x=$ the measure of the angle.
Step 4. Translate.
Write the appropriate formula and substitute.	$m\angle A+m\angle B+m\angle C=180$
Step 5. Solve the equation.	$\begin{array}{ccc}\hfill 55+82+x& =\hfill & 180\hfill \\ \hfill 137+x& =\hfill & 180\hfill \\ \hfill x& =\hfill & 43\hfill \end{array}$
Step 6. Check. $\begin{array}{ccc}\hfill 55+82+43& \stackrel{?}{=}\hfill & 180\hfill \\ \hfill 180& =\hfill & 180✓\hfill \end{array}$
Step 7. Answer the question.	The measure of the third angle is 43 degrees.

Example 3.35

The perimeter of a triangular garden is 24 feet. The lengths of two sides are four feet and nine feet. How long is the third side?

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	length of the third side of a triangle
Step 3. Name. Choose a variable to represent it.	Let $c=$ the third side.
Step 4. Translate.
Write the appropriate formula and substitute.
Substitute in the given information.
Step 5. Solve the equation.
Step 6. Check. $\begin{array}{ccc}\hfill \mathrm{P}& =\hfill & a+b+c\hfill \\ \hfill 24& \stackrel{?}{=}\hfill & 4+9+11\hfill \\ \hfill 24& =\hfill & 24✓\hfill \end{array}$
Step 7. Answer the question.	The third side is 11 feet long.

Example 3.36

The area of a triangular church window is 90 square meters. The base of the window is 15 meters. What is the window’s height?

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.	Area $=90m^2$
Step 2. Identify what you are looking for.	height of a triangle
Step 3. Name. Choose a variable to represent it.	Let $h=$ the height.
Step 4. Translate.
Write the appropriate formula.
Substitute in the given information.
Step 5. Solve the equation.
Step 6. Check. $\begin{array}{ccc}\hfill A& =\hfill & \frac{1}{2}bh\hfill \\ \hfill 90& \stackrel{?}{=}\hfill & \frac{1}{2}\cdot 15\cdot 12\hfill \\ \hfill 90& =\hfill & 90✓\hfill \end{array}$
Step 7. Answer the question.	The height of the triangle is 12 meters.

The triangle properties we used so far apply to all triangles. Now we will look at one specific type of triangle—a right triangle. A right triangle has one $90\text{°}$ angle, which we usually mark with a small square in the corner.

Example 3.37

One angle of a right triangle measures $28\text{°}.$ What is the measure of the third angle?

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 measure of an angle
Step 3. Name. Choose a variable to represent it.	Let $x=$ the measure of an angle.
Step 4. Translate.	$m\angle A+m\angle B+m\angle C=180$
Write the appropriate formula and substitute.	$x+90+28=180$
Step 5. Solve the equation.	$\begin{array}{ccc}\hfill x+118& =\hfill & 180\hfill \\ \hfill x& =\hfill & 62\hfill \end{array}$
Step 6. Check. $\begin{array}{ccc}\hfill 180& \stackrel{?}{=}\hfill & 90+28+62\hfill \\ \hfill 180& =\hfill & 180✓\hfill \end{array}$
Step 7. Answer the question.	The measure of the third angle is 62°.

In the examples we have seen so far, we could draw a figure and label it directly after reading the problem. In the next example, we will have to define one angle in terms of another. We will wait to draw the figure until we write expressions for all the angles we are looking for.

Example 3.38

The measure of one angle of a right triangle is 20 degrees more than the measure of the smallest angle. Find the measures of all three angles.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the measures of all three angles
Step 3. Name. Choose a variable to represent it.	$$Let $a=1^{\text{st}}$ angle. $a+20=2^{\text{nd}}$ angle $90=3^{\text{rd}}$ angle (the right angle)
Draw the figure and label it with the given information
Step 4. Translate
Write the appropriate formula. Substitute into the formula.
Step 5. Solve the equation.	55          90 third angle
Step 6. Check. $\begin{array}{ccc}\hfill 35+55+90& \stackrel{?}{=}\hfill & 180\hfill \\ \hfill 180& =\hfill & 180✓\hfill \end{array}$
Step 7. Answer the question.	The three angles measure 35°, 55°, and 90°.

Use the Pythagorean Theorem

We have learned how the measures of the angles of a triangle relate to each other. Now, we will learn how the lengths of the sides relate to each other. An important property that describes the relationship among the lengths of the three sides of a right triangle is called the Pythagorean Theorem. This theorem has been used around the world since ancient times. It is named after the Greek philosopher and mathematician, Pythagoras, who lived around 500 BC.

Before we state the Pythagorean Theorem, we need to introduce some terms for the sides of a triangle. Remember that a right triangle has a $90\text{°}$ angle, marked with a small square in the corner. The side of the triangle opposite the $90\text{°}$ angle is called the hypotenuse and each of the other sides are called legs.

The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. In symbols we say: in any right triangle, $a^2+b^2=c^2,$ where $a\text{and}b$ are the lengths of the legs and $c$ is the length of the hypotenuse.

Writing the formula in every exercise and saying it aloud as you write it, may help you remember the Pythagorean Theorem.

The Pythagorean Theorem

In any right triangle, $a^2+b^2=c^2.$

where a and b are the lengths of the legs, c is the length of the hypotenuse.

To solve exercises that use the Pythagorean Theorem, we will need to find square roots. We have used the notation $\sqrt{m}$ and the definition:

If $m=n^2,$ then $\sqrt{m}=n,$ for $n\ge 0.$

For example, we found that $\sqrt{25}$ is 5 because $25=5^2.$

Because the Pythagorean Theorem contains variables that are squared, to solve for the length of a side in a right triangle, we will have to use square roots.

Example 3.39

Use the Pythagorean Theorem to find the length of the hypotenuse shown below.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the length of the hypotenuse of the triangle
Step 3. Name. Choose a variable to represent it. Label side c on the figure.	Let c = the length of the hypotenuse.
Step 4. Translate.
Write the appropriate formula.	$a^2+b^2=c^2$
Substitute.	$3^2+4^2=c^2$
Step 5. Solve the equation.	$9+16=c^2$
Simplify.	$25=c^2$
Use the definition of square root.	$\sqrt{25}=c$
Simplify.	$5=c$
Step 6. Check.
Step 7. Answer the question.	The length of the hypotenuse is 5.

Try It 3.77

Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.

Try It 3.78

Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.

Example 3.40

Use the Pythagorean Theorem to find the length of the leg shown below.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.		the length of the leg of the triangle
Step 3. Name. Choose a variable to represent it.		Let b = the leg of the triangle.
Lable side b.
Step 4. Translate
Write the appropriate formula.		$a^2+b^2=c^2$
Substitute.		$5^2+b^2={13}^2$
Step 5. Solve the equation.		$25+b^2=169$
Isolate the variable term.		$b^2=144$
Use the definition of square root.		$b^2=\sqrt{144}$
Simplify.		$b=12$
Step 6. Check.
Step 7. Answer the question.		The length of the leg is 12.

Try It 3.79

Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.

Try It 3.80

Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.

Example 3.41

Kelvin is building a gazebo and wants to brace each corner by placing a $10\text{″}$ piece of wood diagonally as shown above.

If he fastens the wood so that the ends of the brace are the same distance from the corner, what is the length of the legs of the right triangle formed? Approximate to the nearest tenth of an inch.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	the distance from the corner that the bracket should be attached
Step 3. Name. Choose a variable to represent it.	Let $x=$ the distance from the corner.
Step 4. Translate Write the appropriate formula and substitute.	$\begin{array}{}\\ \hfill a^2+b^2& =\hfill & c^2\hfill \\ \hfill x^2+x^2& =\hfill & {10}^2\hfill \\ \\ \\ \end{array}$
Step 5. Solve the equation. Isolate the variable. Use the definition of square root. Simplify. Approximate to the nearest tenth.	$\begin{array}{ccc}\hfill 2x^2& =\hfill & 100\hfill \\ \hfill x^2& =\hfill & 50\hfill \\ \hfill x& =\hfill & \sqrt{50}\hfill \\ \hfill x& \approx \hfill & 7.1\hfill \end{array}$
Step 6. Check. $\begin{array}{ccc}\hfill a^2+b^2& =\hfill & c^2\hfill \\ \hfill {(7.1)}^2+{(7.1)}^2& \approx \hfill & {10}^2\text{Yes.}\hfill \end{array}$
Step 7. Answer the question.	Kelvin should fasten each piece of wood approximately 7.1" from the corner.

Table 3.10

Try It 3.81

John puts the base of a 13-foot ladder five feet from the wall of his house as shown below. How far up the wall does the ladder reach?

Try It 3.82

Randy wants to attach a 17 foot string of lights to the top of the 15 foot mast of his sailboat, as shown below. How far from the base of the mast should he attach the end of the light string?

Solve 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 3.42

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 3.43

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.	$A=LW$
Substitute.	$168=14W$
Step 5. Solve the equation.	$\frac{168}{14}=\frac{14W}{14}$ $12=W$
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 3.44

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 3.45

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.	$P=52$ 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.	$52=2L+2(L-2)$
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 3.46

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 3.47

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.	$P=150$ 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.