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Name ________________________________________ Date ___________________ Class __________________ LESSON 7-3 Reteach Triangle Similarity: AA, SSS, and SAS Angle-Angle (AA) Similarity Side-Side-Side (SSS) Similarity Side-Angle-Side (SAS) Similarity If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. If the three sides of one triangle are proportional to the three corresponding sides of another triangle, then the triangles are similar. If two sides of one triangle are proportional to two sides of another triangle and their included angles are congruent, then the triangles are similar. UABC ∼ UDEF UABC ∼ UDEF UABC ∼ UDEF Explain how you know the triangles are similar, and write a similarity statement. 1. 2. _________________________________________ ________________________________________ _________________________________________ ________________________________________ _________________________________________ ________________________________________ _________________________________________ ________________________________________ 3. Verify that UABC ∼ UMNP. _________________________________________ _________________________________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 7-22 Holt McDougal Geometry Name ________________________________________ Date ___________________ Class __________________ LESSON 7-3 Reteach Triangle Similarity: AA, SSS, and SAS continued You can use AA Similarity, SSS Similarity, and SAS Similarity to solve problems. First, prove that the triangles are similar. Then use the properties of similarity to find missing measures. Explain why UADE ∼ UABC and then find BC. Step 1 Prove that the triangles are similar. ∠A ≅ ∠A by the Reflexive Property of ≅. AD 3 1 = = AB 6 2 AE 2 1 = = AC 4 2 Therefore, UADE ∼ UABC by SAS ∼. Step 2 Find BC. AD DE = AB BC 3 3.5 = 6 BC Corresponding sides are proportional. Substitute 3 for AD, 6 for AB, and 3.5 for DE. 3(BC) = 6(3.5) Cross Products Property 3(BC) = 21 Simplify. BC = 7 Divide both sides by 3. Explain why the triangles are similar and then find each length. 4. GK 5. US _________________________________________ ________________________________________ _________________________________________ ________________________________________ _________________________________________ ________________________________________ _________________________________________ ________________________________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 7-23 Holt McDougal Geometry 2. Possible answer: Every equilateral triangle is also equiangular, so each angle in both triangles measures 60°. Thus, UTUV ∼ UWXY by AA ∼. ∠RQS. It is given that ∠H ≅ ∠Q. So by the Angle Addition Postulate, ∠KHJ ≅ HK HJ 3 ∠TQS. = = , so UKHJ ∼ UTQS QT QS 2 by SAS∼. Because UKHJ ∼ UTQS, JK HK 3 = = . All the corresponding ST QT 2 angles are congruent; all the corresponding sides are proportional. Thus, GHIJK ∼ PQRST by the definition of similar polygons. 3. Possible answer: It is given that ∠JMN ≅ KL JL 4 ∠L. = = . Thus, UJLK ∼ UJMN MN JM 3 by SAS ∼. PQ QR PR 3 4. Possible answer: = = = . UT TS US 5 Thus, UPQR ∼ UUTS by SSS ∼. 5. Possible answer: ∠C ≅ ∠C by the Reflexive Property. ∠CGD and ∠F are right angles, so they are congruent. Thus, UCDG ∼ UCEF by AA ∼. DE = 9.75 Reteach 1. ∠Q ≅ ∠T by the Def. of ≅ ∠s . By the U Sum Thm., m∠S = 39°and m∠U = 49°, so ∠S ≅ ∠V and ∠R ≅ ∠U. UQRS ∼ UTUV by AA ∼. 2. ∠HJG ≅ ∠LJK by the Vert. ∠s Thm. HJ GJ 2 = = . UGHJ ∼ UKLJ by SAS ∼. LJ KJ 3 AB BC CA 4 3. = = = ; UABC ∼ UMNP by MN NP PM 5 SSS ∼. Practice C 1. Possible answer: UABC and UADB share ∠A. They also each have a right angle, so UABC ∼ UADB by AA ∼. They 2 have a similarity ratio of . UABC and 1 UBDC share ∠C. They also each have a right angle, so UABC ∼ UBDC by AA ∼. 2 3 to 1. They have a similarity ratio of 3 By the Transitive Property of Similarity, UADB ∼ UBDC. They have a similarity 3 to 1. ratio of 3 4. JK || FH , so ∠J ≅ ∠H, and ∠K ≅ ∠F by the Alt. Int. ∠s Thm. UJKG ∼ UHFG by 1 AA ∼. GK = 7 3 5. It is given that ∠S ≅ ∠WVU. ∠U ≅ ∠U by the Reflex. Prop. of ≅. UUVW ∼ UUST by AA ∼. US = 39 2. 6 + 2 3 3. 3 + 3;3+3 3 4. Challenge 13 3 1. Statements 5. Possible answer: Draw diagonals HK , HJ , QS, and QT . ∠G and ∠P are right angles, so they are GK GH 3 congruent. = = , so UGHK ∼ PT PQ 2 UPQT by SAS∼. It is given that ∠I ≅ HI IJ 3 ∠R. = = , so UHIJ ∼ UQRS by QR RS 2 SAS∼. Because UGHK ∼ UPQT, HQ 3 = and ∠GHK ≅ ∠PQT. Because QT 2 HJ 3 UHIJ ∼ UQRS, = and ∠IHJ ≅ QS 2 Reasons ∠PQR is a right angle. QS is the altitude of UPQR drawn from the right angle. 1. Given QS ⊥ PR 2. Definition of altitude ∠PSQ and ∠QSR are right angles. 3. Definition of perpendicular m∠PSQ = m∠QSR = m∠PQR = 90° 4. Definition of right angle ∠PSQ ≅ ∠PQR; ∠QSR ≅ ∠PQR 5. Definition of congruent angles ∠P ≅ ∠P; ∠R ≅ ∠R 6. Reflexive Property of Congruence Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. A5 Holt McDougal Geometry