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Holt Geometry
7-1 Ratio and Proportion7-1 Ratio and Proportion
Holt Geometry
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
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Holt Geometry
7-1 Ratio and Proportion
Write and simplify ratios.
Use proportions to solve problems.
Objectives
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Holt Geometry
7-1 Ratio and Proportion
A ratio compares two numbers by division. The ratio
of two numbers a and b can be written as a to b, a:b,
or , where b ≠ 0. For example, the ratios 1 to 2,
1:2, and all represent the same comparison.
In a ratio, the denominator of the fraction cannot be zero because division by zero is undefined.
Remember!
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Holt Geometry
7-1 Ratio and Proportion
A ratio can involve more than two numbers. For the rectangle, the ratio of the side lengths may be written as 3:7:3:7.
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Holt Geometry
7-1 Ratio and Proportion
Example 2: Using Ratios
The ratio of the side lengths of a triangle is 4:7:5, and its perimeter is 96 cm. What is the length of the shortest side?
Let the side lengths be 4x, 7x, and 5x. Then 4x + 7x + 5x = 96 . After like terms are combined, 16x = 96. So x = 6. The length of the shortest side is 4x = 4(6) = 24 cm.
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Holt Geometry
7-1 Ratio and Proportion
Check It Out! Example 2
The ratio of the angle measures in a triangle is 1:6:13. What is the measure of each angle?
x + y + z = 180°
x + 6x + 13x = 180°
20x = 180°
x = 9°
y = 6x
y = 6(9°)
y = 54°
z = 13x
z = 13(9°)
z = 117°
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Holt Geometry
7-1 Ratio and Proportion
A proportion is an equation stating that two ratios
are equal. In the proportion , the values
a and d are the extremes. The values b and c
are the means. When the proportion is written as
a:b = c:d, the extremes are in the first and last
positions. The means are in the two middle positions.
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Holt Geometry
7-1 Ratio and Proportion
In Algebra 1 you learned the Cross Products Property. The product of the extremes ad and the product of the means bc are called the cross products.
The Cross Products Property can also be stated as, “In a proportion, the product of the extremes is equal to the product of the means.”
Reading Math
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Holt Geometry
7-1 Ratio and Proportion
The following table shows equivalent forms of the Cross Products Property.
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Holt Geometry
7-1 Ratio and ProportionPage 6
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Holt Geometry
7-2 Ratios in Similar Polygons7-2 Ratios in Similar Polygons
Holt Geometry
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
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Holt Geometry
7-2 Ratios in Similar Polygons
Identify similar polygons.
Apply properties of similar polygons to solve problems.
Objectives
similarsimilar polygonssimilarity ratio
Vocabulary
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Holt Geometry
7-2 Ratios in Similar Polygons
Figures that are similar (~) have the same shape but not necessarily the same size.
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Holt Geometry
7-2 Ratios in Similar Polygons
Two polygons are similar polygons if and only if their corresponding angles are congruent and their corresponding side lengths are proportional.
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Holt Geometry
7-2 Ratios in Similar Polygons
Example 1: Describing Similar Polygons
Identify the pairs of congruent angles and corresponding sides.
N Q and P R. By the Third Angles Theorem, M T.
0.5
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Holt Geometry
7-2 Ratios in Similar Polygons
Check It Out! Example 1
Identify the pairs of congruent angles and corresponding sides.
B G and C H. By the Third Angles Theorem, A J.
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Holt Geometry
7-2 Ratios in Similar Polygons
A similarity ratio is the ratio of the lengths of
the corresponding sides of two similar polygons.
The similarity ratio of ∆ABC to ∆DEF is , or .
The similarity ratio of ∆DEF to ∆ABC is , or 2.
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Holt Geometry
7-2 Ratios in Similar Polygons
Writing a similarity statement is like writing a congruence statement—be sure to list corresponding vertices in the same order.
Writing Math
When you work with proportions, be sure the ratios compare corresponding measures.
Helpful Hint
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Holt Geometry
7-2 Ratios in Similar Polygons
Example 2A: Identifying Similar Polygons
Determine whether the polygons are similar. If so, write the similarity ratio and a similarity statement.
rectangles ABCD and EFGH
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Holt Geometry
7-2 Ratios in Similar Polygons
Example 2A Continued
Step 1 Identify pairs of congruent angles.
All s of a rect. are rt. s and are .
Step 2 Compare corresponding sides.
A E, B F, C G, and D H.
Thus the similarity ratio is , and rect. ABCD ~ rect. EFGH.
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Holt Geometry
7-2 Ratios in Similar Polygons
Example 2B: Identifying Similar Polygons
Determine whether the polygons are similar. If so, write the similarity ratio and a similarity statement.
∆ABCD and ∆EFGH
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Holt Geometry
7-2 Ratios in Similar Polygons
Example 2B Continued
Step 1 Identify pairs of congruent angles.
P R and S W isos. ∆
Step 2 Compare corresponding angles.
Since no pairs of angles are congruent, the triangles are not similar.
mW = mS = 62°
mT = 180° – 2(62°) = 56°
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Holt Geometry
7-2 Ratios in Similar Polygons
Check It Out! Example 2
Step 1 Identify pairs of congruent angles.
Determine if ∆JLM ~ ∆NPS. If so, write the similarity ratio and a similarity statement.
N M, L P, S J
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Holt Geometry
7-2 Ratios in Similar Polygons
Check It Out! Example 2 Continued
Step 2 Compare corresponding sides.
Thus the similarity ratio is , and ∆LMJ ~ ∆PNS.
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Holt Geometry
7-2 Ratios in Similar Polygons
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Holt Geometry
7-2 Ratios in Similar Polygons
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Holt Geometry
7-2 Ratios in Similar Polygons
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Holt Geometry
7-2 Ratios in Similar Polygons
Lesson Quiz: Part I
1. Determine whether the polygons are similar. If so, write the similarity ratio and a similarity statement.
2. The ratio of a model sailboat’s dimensions to the
actual boat’s dimensions is . If the length of the
model is 10 inches, what is the length of the
actual sailboat in feet?
25 ft
no
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Holt Geometry
7-2 Ratios in Similar Polygons
Lesson Quiz: Part II
3. Tell whether the following statement is
sometimes, always, or never true. Two equilateral
triangles are similar. Always
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS7-3 Triangle Similarity: AA, SSS, and SAS
Holt Geometry
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Prove certain triangles are similar by using AA, SSS, and SAS.
Use triangle similarity to solve problems.
Objectives
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
There are several ways to prove certain triangles are similar. The following postulate, as well as the SSS and SAS Similarity Theorems, will be used in proofs just as SSS, SAS, ASA, HL, and AAS were used to prove triangles congruent.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Example 1: Using the AA Similarity Postulate
Explain why the triangles are similar and write a similarity statement.
Since , B E by the Alternate Interior Angles Theorem. Also, A D by the Right Angle Congruence Theorem. Therefore ∆ABC ~ ∆DEC by AA~.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Check It Out! Example 1
Explain why the trianglesare similar and write asimilarity statement.
By the Triangle Sum Theorem, mC = 47°, so C F. B E by the Right Angle Congruence Theorem. Therefore, ∆ABC ~ ∆DEF by AA ~.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Example 2A: Verifying Triangle Similarity
Verify that the triangles are similar.
∆PQR and ∆STU
Therefore ∆PQR ~ ∆STU by SSS ~.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Example 2B: Verifying Triangle Similarity
∆DEF and ∆HJK
Verify that the triangles are similar.
D H by the Definition of Congruent Angles.
Therefore ∆DEF ~ ∆HJK by SAS ~.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Check It Out! Example 2
Verify that ∆TXU ~ ∆VXW.
TXU VXW by the Vertical Angles Theorem.
Therefore ∆TXU ~ ∆VXW by SAS ~.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
A A by Reflexive Property of , and B C since they are both right angles.
Example 3: Finding Lengths in Similar Triangles
Explain why ∆ABE ~ ∆ACD, and then find CD.
Step 1 Prove triangles are similar.
Therefore ∆ABE ~ ∆ACD by AA ~.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Example 3 Continued
Step 2 Find CD.
Corr. sides are proportional. Seg. Add. Postulate.
Substitute x for CD, 5 for BE, 3 for CB, and 9 for BA.
Cross Products Prop. x(9) = 5(3 + 9)
Simplify. 9x = 60
Divide both sides by 9.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Check It Out! Example 3
Explain why ∆RSV ~ ∆RTU and then find RT.
Step 1 Prove triangles are similar.
It is given that S T. R R by Reflexive Property of .
Therefore ∆RSV ~ ∆RTU by AA ~.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Check It Out! Example 3 Continued
Step 2 Find RT.
Corr. sides are proportional.
Substitute RS for 10, 12 for TU, 8 for SV.
Cross Products Prop.
Simplify.
Divide both sides by 8.
RT(8) = 10(12)
8RT = 120
RT = 15
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Example 4: Writing Proofs with Similar Triangles
Given: 3UT = 5RT and 3VT = 5ST
Prove: ∆UVT ~ ∆RST
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Statements Reasons
1. Given1. 3UT = 5RT
2. Divide both sides by 3RT.2.
3. Given.3. 3VT = 5ST
4. Divide both sides by3ST.4.
5. Vert. s Thm.5. RTS VTU
6. SAS ~ Steps 2, 4, 56. ∆UVT ~ ∆RST
Example 4 Continued
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Check It Out! Example 4
Given: M is the midpoint of JK. N is the midpoint of KL, and P is the midpoint of JL.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Statements Reasons
Check It Out! Example 4 Continued
1. Given1. M is the mdpt. of JK, N is the mdpt. of KL,
and P is the mdpt. of JL.
2. ∆ Midsegs. Thm2.
3. Div. Prop. of =.3.
4. SSS ~ Step 34. ∆JKL ~ ∆NPM
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
You learned in Chapter 2 that the Reflexive, Symmetric, and Transitive Properties of Equality have corresponding properties of congruence. These properties also hold true for similarity of triangles.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Lesson Quiz
1. Explain why the triangles are
similar and write a similarity
statement.
2. Explain why the triangles are
similar, then find BE and CD.
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Holt Geometry
7-3 Triangle Similarity: AA, SSS, and SAS
Lesson Quiz
1. By the Isosc. ∆ Thm., A C, so by the def. of , mC = mA. Thus mC = 70° by subst. By the ∆ Sum Thm., mB = 40°. Apply the Isosc. ∆ Thm. and the ∆ Sum Thm. to ∆PQR. mR = mP = 70°. So by the def. of , A P, and C R. Therefore ∆ABC ~ ∆PQR by AA ~.
2. A A by the Reflex. Prop. of . Since BE || CD, ABE ACD by the Corr. s Post. Therefore ∆ABE ~ ∆ACD by AA ~. BE = 4 and CD = 10.