7-1 Ratio and Proportion Warm Up Lesson Presentation Lesson Quiz Holt Geometry

Objectives Write and simplify ratios. Use proportions to solve problems.

A ratio compares two numbers by division 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!

A ratio can involve more than two numbers 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.

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.

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 z = 13x y = 6(9°) z = 13(9°) y = 54° z = 117°

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.

In Algebra 1 you learned the Cross Products Property 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

The following table shows equivalent forms of the Cross Products Property.

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Ratios in Similar Polygons 7-2 Ratios in Similar Polygons Warm Up Lesson Presentation Lesson Quiz Holt Geometry

Objectives Vocabulary Identify similar polygons. Apply properties of similar polygons to solve problems. Vocabulary similar similar polygons similarity ratio

Figures that are similar (~) have the same shape but not necessarily the same size.

Two polygons are similar polygons if and only if their corresponding angles are congruent and their corresponding side lengths are proportional.

Example 1: Describing Similar Polygons Identify the pairs of congruent angles and corresponding sides. 0.5 N  Q and P  R. By the Third Angles Theorem, M  T.

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.

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.

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

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

Example 2A Continued Step 1 Identify pairs of congruent angles. A  E, B  F, C  G, and D  H. All s of a rect. are rt. s and are . Step 2 Compare corresponding sides. Thus the similarity ratio is , and rect. ABCD ~ rect. EFGH.

Example 2B: Identifying Similar Polygons Determine whether the polygons are similar. If so, write the similarity ratio and a similarity statement. ∆ABCD and ∆EFGH

Example 2B Continued Step 1 Identify pairs of congruent angles. P  R and S  W isos. ∆ Step 2 Compare corresponding angles. mW = mS = 62° mT = 180° – 2(62°) = 56° Since no pairs of angles are congruent, the triangles are not similar.

Check It Out! Example 2 Determine if ∆JLM ~ ∆NPS. If so, write the similarity ratio and a similarity statement. Step 1 Identify pairs of congruent angles. N  M, L  P, S  J

Check It Out! Example 2 Continued Step 2 Compare corresponding sides. Thus the similarity ratio is , and ∆LMJ ~ ∆PNS.

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? no 25 ft

Lesson Quiz: Part II 3. Tell whether the following statement is sometimes, always, or never true. Two equilateral triangles are similar. Always

Triangle Similarity: 7-3 AA, SSS, and SAS Warm Up Lesson Presentation Lesson Quiz Holt Geometry

Objectives Prove certain triangles are similar by using AA, SSS, and SAS. Use triangle similarity to solve problems.

There are several ways to prove certain triangles are similar 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.

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~.

Check It Out! Example 1 Explain why the triangles are similar and write a similarity statement. By the Triangle Sum Theorem, mC = 47°, so C  F. B  E by the Right Angle Congruence Theorem. Therefore, ∆ABC ~ ∆DEF by AA ~.

Example 2A: Verifying Triangle Similarity Verify that the triangles are similar. ∆PQR and ∆STU Therefore ∆PQR ~ ∆STU by SSS ~.

Example 2B: Verifying Triangle Similarity Verify that the triangles are similar. ∆DEF and ∆HJK D  H by the Definition of Congruent Angles. Therefore ∆DEF ~ ∆HJK by SAS ~.

Check It Out! Example 2 Verify that ∆TXU ~ ∆VXW. TXU  VXW by the Vertical Angles Theorem. Therefore ∆TXU ~ ∆VXW by SAS ~.

Example 3: Finding Lengths in Similar Triangles Explain why ∆ABE ~ ∆ACD, and then find CD. Step 1 Prove triangles are similar. A  A by Reflexive Property of , and B  C since they are both right angles. Therefore ∆ABE ~ ∆ACD by AA ~.

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. x(9) = 5(3 + 9) Cross Products Prop. 9x = 60 Simplify. Divide both sides by 9.

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 ~.

Check It Out! Example 3 Continued Step 2 Find RT. Corr. sides are proportional. Substitute RS for 10, 12 for TU, 8 for SV. RT(8) = 10(12) Cross Products Prop. 8RT = 120 Simplify. RT = 15 Divide both sides by 8.

Example 4: Writing Proofs with Similar Triangles Given: 3UT = 5RT and 3VT = 5ST Prove: ∆UVT ~ ∆RST

Statements Reasons Example 4 Continued 1. 3UT = 5RT 1. Given 2. 2. Divide both sides by 3RT. 3. 3VT = 5ST 3. Given. 4. 4. Divide both sides by3ST. 5. RTS  VTU 5. Vert. s Thm. 6. ∆UVT ~ ∆RST 6. SAS ~ Steps 2, 4, 5

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.

Check It Out! Example 4 Continued Statements Reasons 1. M is the mdpt. of JK, N is the mdpt. of KL, and P is the mdpt. of JL. 1. Given 2. 2. ∆ Midsegs. Thm 3. 3. Div. Prop. of =. 4. ∆JKL ~ ∆NPM 4. SSS ~ Step 3

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.

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.

Lesson Quiz 1. By the Isosc. ∆ Thm., A  C, so by the def. of , mC = mA. Thus mC = 70° by subst. By the ∆ Sum Thm., mB = 40°. Apply the Isosc. ∆ Thm. and the ∆ Sum Thm. to ∆PQR. mR = mP = 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.