Which of the following expressions and equations are proportions?
1. [latex]\frac{7}{x}=\frac{3}{5}[/latex]
2. [latex]\frac{x}{2}=\frac{8}{x+2}[/latex]
3. [latex]1+\frac{x}{4}=\frac{2x}{3}[/latex]
4. [latex]\frac{6}{x}+\frac{x}{5}[/latex]
5. [latex]\frac{3}{x+1}-\frac{2x}{5}[/latex]
6. [latex]\frac{1}{x}+\frac{2}{3x}=\frac{x-2}{2}[/latex]
Solve each equation. Begin by writing an equivalent equation without fractions: multiply both sides by the LCD.
7. [latex]\frac{x}{12}=\frac{3}{x}[/latex]
8. [latex]1+\frac{x}{2}=\frac{2x}{5}[/latex]
Algebra Refresher Answers
Only 1 and 2 are proportions.
7. [latex]\pm 6[/latex]
8. [latex]-10[/latex]
Similar Triangles
Learning Objectives
Identify congruent triangles and find unknown parts.
Identify similar triangles.
Find unknown parts of similar triangles.
Solve problems using proportions and similar triangles.
Use proportions to relate sides of similar triangles.
Congruent Triangles
Two triangles are congruent if they have exactly the same size and shape. This means that their corresponding angles are equal and their corresponding sides have the same lengths, as shown below.
Example 1.17.
The two triangles below are congruent. List the corresponding parts and find the angles [latex]\theta[/latex], [latex]\phi[/latex], and [latex]\chi[/latex] and side z
Solution:
In these triangles, [latex]B = D[/latex] because they are both right angles, and [latex]BCA=DCE[/latex] because they are vertical angles, so [latex]\theta =25^{o}[/latex] The third angles, [latex]\angle CAB[/latex] and [latex]\angle CED[/latex], must also be equal, so [latex]\phi =\chi =65^{o}[/latex]. (Do you see why?) The sides opposite each pair of corresponding angles are equal, so [latex]AB=DE[/latex], [latex]BC=CD[/latex], and [latex]AC=CE[/latex]. In particular, we find that [latex]z=9[/latex].
Checkpoint 1.18.
The two triangles at right are congruent. Find the values of [latex]\alpha[/latex] [latex]\beta[/latex] and [latex]\gamma[/latex].
Recall that the altitude of a triangle is the segment from one vertex of the triangle perpendicular to the opposite side.
Example 1.19.
Show that the altitude of an equilateral triangle divides it into two congruent right triangles.
Solution.
Consider, for example, an equilateral triangle of side 8 inches, as shown above. The altitude is perpendicular to the base, so each half of the original triangle is a right triangle. Because each right triangle contains a [latex]60^{o}[/latex] angle, the remaining angle in each triangle must be [latex]90^{o} - 60^{o} = 30^{o}[/latex]. Both triangles have a side of length [latex]8[/latex] between the angles of [latex]30°[/latex] and [latex]60°[/latex], so they are congruent. (Consequently, the short sides of the congruent triangles are equal, so each is half the original base.)
The triangles in the previous example are a special type of right triangle called [latex]30^{o}-60^{o}-90^{o}[/latex] triangles. Notice that in these triangles, the leg opposite the [latex]30^{o}[/latex] angle is half the length of the hypotenuse.
Checkpoint 1.20.
The diagonal of a parallelogram divides it into two congruent triangles, as shown at right. List the corresponding parts of the two triangles and explain why each pair is equal.
Answer.
[latex]\angle BCA = \angle CAD[/latex] and [latex]\angle BAC = \angle ACD[/latex] because they are alternate interior angles. If two pairs of angles in a triangle are equal, so is the third pair, so [latex]\angle B = \angle D[/latex]. [latex]BC = AD[/latex]and [latex]AB = CD[/latex] because they are opposite sides of a parallelogram, and [latex]AC = CA[/latex].
Similar Triangles
Two triangles are similar if they have the same shape but not necessarily the same size. The corresponding angles are equal, and the corresponding sides are proportional. We can think of one similar triangle as an enlargement or a reduction of the other. (See the figures below.)
[latex]\angle A = \angle A'[/latex]
[latex]\angle B = \angle B'[/latex]
[latex]\angle C = \angle C'[/latex]
[latex]\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}[/latex]
To decide whether two triangles are similar, it turns out that we need to verify only one of the two conditions for similarity, and the other condition will be true automatically.
Similar Triangles
Two triangles are similar if either
their corresponding angles are equal, or
their corresponding sides are proportional.
Example 1.21.
Which of the pairs of triangles shown below are similar?
Solution.
a. We will check whether the corresponding sides are proportional. We compute the ratios of the corresponding sides, making sure to write each ratio in the same order, [latex]\frac{\text{larger triangle}}{\text{smaller triangle}}[/latex]. (The other order, [latex]\frac{\text{smaller triangle}}{\text{larger triangle}}[/latex], would also work, as long as we use the same order for all the ratios.) Shorter legs: [latex]\frac{9}{6}[/latex]. Longer legs: [latex]\frac{12}{8}[/latex]. Hypotenuses: [latex]\frac{15}{10}[/latex]. Because all of these ratios are equal to [latex]1.5[/latex], the triangles are similar.
b. The ratios of corresponding sides are not equal: the ratio of the longest sides is [latex]\frac{6}{5}[/latex], but the ratio of the smallest sides is [latex]\frac{4}{3}[/latex]. The triangles are not similar.
c. The missing angle of the first right triangle is [latex]48^{o}[/latex], and the missing angle in the second right triangle is [latex]42^{o}[/latex], so three pairs of angles match. The triangles are similar.
Checkpoint 1.22.
Are the triangles below similar? Explain why or why not in each case.
Answer.
a. The triangles are similar because [latex]\frac{4}{6} = \frac{6}{9} = \frac{8}{12}[/latex], so the sides are proportional.
b. The third angle in both triangles is [latex]80^{o}[/latex], so the triangles are similar because their corresponding angles are equal.
Note 1.23.
In part (b) of the previous exercise, note that the third angle in each triangle must be [latex]80^{o}[/latex] because the sum of the angles is [latex]180^{o}[/latex]. Thus, we need only show that two pairs of angles are equal to show that two triangles are similar.
Using Proportions with Similar Triangles
The figure in the next example shows a parallelogram [latex]ABCD[/latex] and two triangles, [latex]\triangle ABE[/latex] and [latex]\triangle FCE[/latex]. Can we find the unknown lengths [latex]x[/latex] and [latex]y[/latex] in the larger triangle? First note that two pairs of corresponding angles in the triangles are equal: [latex]\angle BEA[/latex] and [latex]\angle FEC[/latex] are vertical angles, and [latex]\angle EFC[/latex] and [latex]\angle BAE[/latex] are alternate interior angles. But if two pairs of corresponding angles are equal, then the third pair must be equal also. This means that the two triangles are similar, and we can use the fact that their corresponding sides are proportional to find [latex]x[/latex] and [latex]y[/latex].
Examples 1.24.
Find the value of [latex]x[/latex] in the figure at right.
Solution.
We see that [latex]x[/latex] is the length of the shortest side in triangle ABE. We know the short side in triangle FCE and the lengths of the medium sides in each triangle. If we form the ratios of the short sides and the medium sides, we obtain the following proportion.
[latex]\begin{align*} 6x = 4(15) = & 60 & \text{Divide both sides by 6} \\ x = \frac{60}{6} = 10 \end{align*}[/latex]
Caution 1.25.
Remember that “cross-multiplying” is a shortcut technique for solving proportions. It does not work for other operations involving fractions.
Checkpoint 1.26.
Find the value of [latex]y[/latex] in the previous example.
Answer.
[latex]y = 20[/latex]
Similar Right Triangles
If two right triangles have one pair of corresponding acute angles with the same measure, then the triangles are similar. We can use this fact about right triangles to make indirect measurements.
Example 1.27.
Delbert would like to know the height of a certain building. He gets Francine to hold up a [latex]5[/latex]-foot pole near the building and measures the length of its shadow. The shadow of the pole is [latex]3[/latex] feet long, and the shadow of the building is [latex]12[/latex] feet long.
Use similar triangles to write a proportion involving the height of the building.
Solve the proportion to find the height of the building.
Solution.
a. In the figure above, we see two right triangles: One triangle is formed by the building and its shadow, and the other by the pole and its shadow. Because the light rays from the sun are parallel, the two angles at the tips of the shadows are equal. Thus, the two right triangles are similar, and their corresponding sides are proportional. The ratios of heights and bases in the two triangles yield the proportion
b. To solve the proportion, we cross-multiply to get
[latex]\begin{align*} 3h = 5(12) = & 60 & \text{Divide both sides by 3}.\\ h =\frac{60}{3} = 20 \end{align*}[/latex]
The building is [latex]20[/latex] feet tall.
Checkpoint 1.28.
Earlier we created a 30°-60°-90° triangle in which the shorter leg was 4 inches and the hypotenuse was 8 inches. The hypotenuse of another 30°-60°-90° triangle is 5 feet. What is the length of the side opposite the 30° angle?
Answer.
[latex]2.5[/latex] feet
Overlapping Triangles
Example 1.29.
Example 1.29.
Identify two similar triangles in the figure at right, and write a proportion to find [latex]H[/latex]
Solution.
The two triangles overlap, sharing the marked angle, as shown below. Because each triangle also has a right angle, they are similar.
Note that the base of the larger triangle is [latex]24 + 12 = 36[/latex]. The ratio of the heights and the ratio of the bases must be equal, so we write the following proportion.
[latex]\begin{align*} \frac{H}{10} = & \frac{36}{24} & \text{Cross-multiply}.\\ 24H = & 360 & \text{Divide both sides by 24.}\\ H = \frac{360}{24} = 15 \end{align*}[/latex]
Checkpoint 1.30.
Heather wants to know the height of a street lamp. She discovers that when she is 12 feet from the lamp, her shadow is 6 feet long. Find the height of the street lamp.
Answer.[latex]15[/latex] feet
Section 1.2 Summary
Vocabulary
Look up the definitions of new terms in the Glossary.
Two triangles are congruent if they have exactly the same size and shape.
The altitude of an equilateral triangle divides it into two congruent right triangles.
In a 30°-60°-90° right triangle, the leg opposite the 30° angle is half the length of the hypotenuse.
Two triangles are similar if they have the same shape but not necessarily the same size. The corresponding angles are equal, and the corresponding sides are proportional.
Two triangles are similar if either their corresponding angles are equal or their corresponding sides are proportional.
If two right triangles have one pair of corresponding acute angles with the same measure, then the triangles are similar.
Study Questions
What is the difference between congruent triangles and similar triangles?
What is the name of the shortcut method for solving proportions? Why does the method work?
In two triangles, if two corresponding pairs of angles are equal, are the triangles similar? How do you know?
For the triangles shown, which of the following equations is true? Explain why.
[latex]\frac{4}{x} = \frac{6}{8}[/latex]
[latex]\frac{x}{4} = \frac{6}{8}[/latex]
[latex]\frac{x}{x+4} = \frac{6}{8}[/latex]
[latex]\frac{x}{x+4} = \frac{6}{14}[/latex]
definition
Two triangles are congruent if they have exactly the same size and shape.
The altitude of an equilateral triangle divides it into two congruent right triangles.
In a 30°-60°-90° right triangle, the leg opposite the 30° angle is half the length of the hypotenuse.
Two triangles are similar if they have the same shape but not necessarily the same size.
Two triangles are similar if they have the same shape but not necessarily the same size. The corresponding angles are equal, and the corresponding sides are proportional.
A right triangle is a type of triangle that has one interior angle measuring exactly 90 degrees, or a “right angle.”
In a 30°-60°-90° right triangle, the leg opposite the 30° angle is half the length of the hypotenuse.
A parallelogram is a quadrilateral with two pairs of parallel sides, where opposite sides are congruent (have the same length) and opposite angles are also congruent (have the same measure).
