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Mrs. Rivas ππ πΏπ ππΏ π¨πͺ π©πͺ π¨π© 1. π΄π΅ 2. π΄πΆ 3. πΆπ΅ 4. ππ 5. ππ 6. ππ
Name the triangle sides that are parallel to the given side. ππ πΏπ 1. π΄π΅ π΄πΆ 3. πΆπ΅ ππ 5. ππ ππ ππΏ π¨πͺ π©πͺ π¨π©
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Mrs. Rivas Points M, N, and P are the midpoints of the sides of βQRS. QR = 30, RS = 30, and SQ = 18. 7. Find MN 8. Find MQ π ππ 9. Find MP Find PS ππ π 11. Find PN Find RN ππ ππ
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Mrs. Rivas Algebra Find the value of x. 13. π+π=π(ππ) π+π=ππ π=ππ π=π
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Mrs. Rivas π ππ =πππβππ ππ=πππβππ βππ=βππ π=π.π
Algebra Find the value of x. 14. π ππ =πππβππ ππ=πππβππ βππ=βππ π=π.π
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Mrs. Rivas ππ+ππ=π(ππ) ππ+ππ=ππ ππ=ππ ππ=π
Algebra Find the value of x. 15. ππ+ππ=π(ππ) ππ+ππ=ππ ππ=ππ ππ=π
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Mrs. Rivas Use the diagram at the right for Exercises 16 and 17. 16. Which segment is shorter for kayaking across the lake, or ? Explain. π©πͺ is shorter because π©πͺ is half of 5 mi, while π¨π© is half of 6 mi.
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Mrs. Rivas Use the diagram at the right for Exercises 16 and 17. 17. Which distance is shorter, kayaking from π΄ to π΅ to πΆ, or walking from π΄ to π to πΆ? Explain. Neither; the distance is the same because π©πͺ β
π¨πΏ and π¨π© β
πΏπͺ .
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Mrs. Rivas π³π΅ is the perpendicular bisector of π΄πΆ
Use the figure at the right for Exercises 18β21. 18. What is the relationship between πΏπ and ππ ? π³π΅ is the perpendicular bisector of π΄πΆ
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Mrs. Rivas Use the figure at the right for Exercises 18β21. 19. What is the value of x? ππ+ππ=ππ ππ=ππ ππ=π
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Mrs. Rivas ππ=π π³π΅ =ππ+ππ π³π΅ =π(ππ)+ππ π³π΅ =ππ+ππ π³π΅ =ππ
Use the figure at the right for Exercises 18β21. 20. Find LM. ππ=π π³π΅ =ππ+ππ π³π΅ =π(ππ)+ππ π³π΅ =ππ+ππ π³π΅ =ππ
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Mrs. Rivas ππ=π π³πΆ =ππ π³πΆ =π(ππ) π³πΆ =ππ
Use the figure at the right for Exercises 18β21. 21. Find LO. ππ=π π³πΆ =ππ π³πΆ =π(ππ) π³πΆ =ππ
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Mrs. Rivas From A to πͺπ« =ππ From A to πͺπ© =ππ
Use the figure at the right for Exercises 22β26. 22. According to the figure, how far is A from πΆπ· ? From πΆπ΅ ? From A to πͺπ« =ππ From A to πͺπ© =ππ
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Mrs. Rivas πͺπ¨ bisects β π«πͺπ©; Converse of Angle Bisector. Theorem.
Use the figure at the right for Exercises 22β26. 23. How is πΆπ΄ related to οDCB? Explain. πͺπ¨ bisects β π«πͺπ©; Converse of Angle Bisector. Theorem.
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Mrs. Rivas Use the figure at the right for Exercises 22β26. 24. Find the value of x. ππ=ππβππ βπ=βππ π=ππ
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Mrs. Rivas π=ππ ποπ¨πͺπ« =ππ ποπ¨πͺπ« =π(ππ) ποπ¨πͺπ« =ππ ποπ¨πͺπ© =ππβππ
Use the figure at the right for Exercises 22β26. 25. Find mοACD and mοACB. π=ππ ποπ¨πͺπ« =ππ ποπ¨πͺπ« =π(ππ) ποπ¨πͺπ« =ππ ποπ¨πͺπ© =ππβππ ποπ¨πͺπ© =π ππ βππ ποπ¨πͺπ© =ππ
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Mrs. Rivas ππ+ππ+ποπ«π¨πͺ =πππ ποπ«π¨πͺ =ππ ππ+ππ+ποπ©π¨πͺ =πππ ποπ©π¨πͺ =ππ
Use the figure at the right for Exercises 22β26. 26. Find mοDAC and mοBAC. ππ+ππ+ποπ«π¨πͺ =πππ ποπ«π¨πͺ =ππ ππ+ππ+ποπ©π¨πͺ =πππ ποπ©π¨πͺ =ππ
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Mrs. Rivas ππ=π+π π=π π³πΆ =π+π π΅πΆ =ππ π³πΆ =π+π π΅πΆ =π(π) π³πΆ =ππ π΅πΆ =ππ
Algebra Find the indicated values of the variables and measures. 27. m, LO, NO ππ=π+π π=π π³πΆ =π+π π΅πΆ =ππ π³πΆ =π+π π΅πΆ =π(π) π³πΆ =ππ π΅πΆ =ππ
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Mrs. Rivas ππ=ππβππ βππ=βππ π=ππ ποπΈπ»πΊ =ππ+ππβππ ποπΈπ»πΊ =π(ππ)+π(ππ)βππ
Algebra Find the indicated values of the variables and measures. 28. π₯, πβ πππ ππ=ππβππ βππ=βππ π=ππ ποπΈπ»πΊ =ππ+ππβππ ποπΈπ»πΊ =π(ππ)+π(ππ)βππ ποπΈπ»πΊ =ππ+ππβππ ποπΈπ»πΊ =ππ
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Mrs. Rivas ππ+π=ππ+π π+π=π π=π π°π± =ππ+π π²π± =ππ+π π°π± =π(π)+π π²π± =π(π)+π
Algebra Find the indicated values of the variables and measures. 29. p, IJ, KJ ππ+π=ππ+π π+π=π π=π π°π± =ππ+π π²π± =ππ+π π°π± =π(π)+π π²π± =π(π)+π π°π± =ππ+π π²π± =π+π π°π± =ππ π²π± =ππ
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Mrs. Rivas 30. Writing Determine whether A must be on the bisector of οLMN. Explain. a. b. Yes; π¨ is equidistant from both rays of the angle, so π¨ lies on the bisector. Yes; β π³π΄π¨ β
β π΅π΄π¨, so π΄π¨ is a bisector of β π³π΄π΅.
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