Which of the following expressions and equations are proportions?
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]
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 hal