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QBA 1 Review Short Answer 1. Find m . A >> 3. Violin strings are parallel. Viewed from above, a violin bow in two different positions forms two transversals to the violin strings. Find x and y in the diagram. xº 100º C (3x - 70)º >> B (4x + y)º (8x + y)º 2. Find m . R T S (3x)º >> 4. Use the information , and the theorems you have learned to show that . U (4x – 24)º >> V 1 2 5. Write a two-column proof. Given: Prove: 3 1 l 2 m Complete the proof. Proof: Statements 1. 2. 3. 4. 60º Reasons 1. Given 2. [1] 3. Substitution (Steps 1 and 2) 4. [2] l m 6. In a swimming pool, two lanes are represented by lines l and m. If a string of flags strung across the lanes is represented by transversal t, and x = 10, show that the lanes are parallel. 7. Write and solve an inequality for x. D 2x + 4 8 A B C (3x + 4)º l (4x – 6)º m t 8. Write a two-column proof. Given: Prove: t 1 2 m l Complete the proof. Proof: Statements 1. [1] 1. Given 2. 2. [2] 3. [3] 3. 9. Apply the transformation M to the triangle with the given vertices. Identify and describe the transformation. M: (x, y) (x – 6, y + 2) E(3, 0), F(1, –2), G(5, –4) 10. Apply the transformation M to the polygon with the given vertices. Identify and describe the transformation. M: (x, y) (–x, –y) A(–3, 6), B(–3, 1), C(1, 1), D(1, 6) Reasons y 6 G –6 E F P Q 6 R 11. Determine whether triangles are congruent. and –6 x 12. Prove that the triangles with the given vertices are congruent. A(3, 1), B(4, 5), C(2, 3) D(–1, –3), E(–5, –4), F(–3, –2) 13. Daphne folded a triangular sheet of paper into the shape shown. Find , given , , and m . N F D E P M E 17. Find m m , given . , , and D C 22º B A 14. One of the acute angles in a right triangle has a measure of . What is the measure of the other acute angle? 15. Find E D C 42º 61º A B F 18. Given: Identify all pairs of congruent corresponding parts. A M . ( L B (4x + 7)º C O 19. Given that m . (6x - 9)º E M 23º N A 16. Find and m 118º K and , and N , given C , . B D = 23°, find 20.Given: , , . T is the midpoint of . R S T U Prove: Complete the proof. Proof: Statements Reasons 1. 2. and are right angles. 3. 4. 5. 6. 7. T is the midpoint of . 8. 9. 10. 1. Given 2. [1] 3. Right Angle Congruence Theorem 4. Given 5. [2] 6. Given 7. Given 8. Definition of midpoint 9. [3] 10. Definition of congruent triangles 21. Tom is wearing his favorite bow tie to the school dance. The bow tie is in the shape of two triangles. Given: , , , Prove: B C A D E Complete the proof. Proof: 1. 2. 3. 4. 5. [3] , Statements , Reasons 1. Given 2. Given 3. [1] 4. [2] 5. Definition of congruent triangles 22. Given the lengths marked on the figure and that bisects , use SSS to explain why . R 4 cm E A 3 cm 3 cm D || T U Complete the explanation. 4 cm C || S B 23. The figure shows part of the roof structure of a house. Use SAS to explain why . 24. Given: P is the midpoint of and . Prove: T It is given that [1]. Since and are right angles, [2] by the Right Angle Congruence Theorem. By the Reflexive Property of Congruence, [3]. Therefore, by SAS. R P S Q Complete the proof. Proof: Statements 1. P is the midpoint of and Reasons . 1. Given 2. [1] 2. , 3. [2] 3. Vertical Angles Theorem 4. [3] 4. 25. What additional information do you need to prove 26. Determine if you can use ASA to prove by the SAS Postulate? . Explain. C B E D A A B || C || D 27. Determine if you can use the HL Congruence Theorem to prove ACD know. P A DBA. If not, tell what else you need to B | ^ ^ | C D Q 28. For these triangles, select the triangle congruence statement and the postulate or theorem that supports it. 29. A pilot uses triangles to find the angle of elevation from the ground to her plane. How can she find m ? L D C 40° 12 km O 20 km J K B A A C Find the value of x. 3x +1 31. 5 2x 30. 12 km B 20 km 31. Given: , Prove:MLP is isosceles. , L M N P O Complete the proof. Proof: Statements , Reasons 1. 2. 3. 4. 5. 6. and 7. 8. MLN PLO 9. 10. MLP is isosceles. 1. Given 2. Given 3. Definition of congruent line segments 4. Reflexive Property of Equality 5. Subtraction Property of Equality 6. Segment Addition Postulate 7. Substitution Property of Equality 8. [1] 9. [2] 10. Definition of isosceles triangle 32. Two Seyfert galaxies, BW Tauri and M77, represented by points A and B, are equidistant from Earth, represented by point C. What is m ? 33. Find m . P C | | xº 115° A B D R (2x + 15)º Q 34. Find CA. 38. Point O is the centroid of . Find A , . C ) X Y s+ 2 O B ) ) C 2 s 10 39. In B A Z , show that midsegment and that . is parallel to y 35. Find the measure of each numbered angle. | L –4 117 2 > 3 1 | C (-4, 2) 2 –2 K > 36. Find the measures and 4 6 x B (4, -2) –2 A (-4, -4) . 2 –4 B 40. Given with , find the length of midsegment 6.4 , and , . C AC = 6 X A X 2.3 C Y A 37. Given that . Y bisects and , find 3 B X 41. Vanessa wants to measure the width of a reservoir. She measures a triangle at one side of the reservoir as shown in the diagram. What is the width of the reservoir (BC across the base)? 120 m B X Z BC = 5 120 m W A 150 m 100 m Y 100 m C 42. Identify the pairs of congruent angles and corresponding sides. A 12 8 2.5 m )) C ( 18 B 1.3 m 13.5 D 44. Maya is making a miniature dinner table for her little sister. She wants the table top to be similar to their real dinner table top. Find the width of the miniature table top to the nearest tenth of a centimeter. (( ) 9 Dining table top E 9.5 cm 6 x M iniature F 43. Determine whether the rectangles are similar. If so, write the similarity ratio and a similarity statement. M 9 45. A video game designer is modeling a tower that is 320 ft high and 260 ft wide. She creates a model so that the similarity ratio of the model to the tower is . What is the height and the width of the model in inches? N 6 P R O 15 S 46. Apply the dilation D to the polygon with the given vertices. Name the coordinates of the image points. J(1, 4), K(6, 4), L(6, 1), M(1, 1) y 10 6 5 U T 4 J K M L 3 2 1 1 2 3 4 5 6 7 47. Explain why 9 x 8 and then find BC. B 28 D A 32 ( E ( 48 C 48. Find . P N 5 M Q 8 2 R 49. An artist used perspective to draw guidelines in her picture of a row of parallel buildings. How many centimeters is it from Point B to Point C? H 3 cm F 4 cm C D B 5 cm A 50. Find . B 3x - 3 18 D x+ 5 A 15 C QBA 1 Review Answer Section SHORT ANSWER 1. ANS: m = 35° Corresponding Angles Postulate Subtract x from both sides. Add 70 to both sides. Divide both sides by 2. m m PTS: OBJ: LOC: KEY: 2. ANS: m Substitute 35 for x. Simplify. 1 DIF: 2 REF: 1a24483e-4683-11df-9c7d-001185f0d2ea 3-1.1 Using the Corresponding Angles Postulate STA: MCC9-12.G.CO.9 MTH.C.11.01.03.03.01.005 TOP: 3-1 Angles Formed by Parallel Lines and Transversals corresponding angles | parallel lines DOK: DOK 2 = Alternate Exterior Angles Theorem Subtract from both sides. Divide both sides by . Substitute 24 for . m PTS: OBJ: LOC: TOP: DOK: 3. ANS: 1 DIF: 2 REF: 1a26aa9a-4683-11df-9c7d-001185f0d2ea 3-1.2 Finding Angle Measures STA: MCC9-12.G.CO.9 MTH.C.11.01.03.03.01.007 | MTH.C.11.02.01.01.007 3-1 Angles Formed by Parallel Lines and Transversals KEY: alternate exterior angles | parallel lines DOK 2 By the Corresponding Angles Postulate, By the Alternate Interior Angle Postulate, . . Subtract the first equation from the second. Divide both sides by 4. Substitute 10 for x. Simplify. PTS: 1 DIF: 3 REF: 1a26d1aa-4683-11df-9c7d-001185f0d2ea OBJ: 3-1.3 Application STA: MCC9-12.G.CO.9 LOC: MTH.C.11.01.03.03.01.005 | MTH.C.11.01.03.03.01.006 TOP: 3-1 Angles Formed by Parallel Lines and Transversals KEY: corresponding angles | parallel lines | alternate interior angles DOK: DOK 3 4. ANS: By substitution, and . By the Substitution Property of Equality, . By the Converse of the Alternate Interior Angles Theorem, . ; Substitute 20 for x. Substitution Property of Equality Converse of the Alternate Interior Angles Theorem PTS: 1 DIF: 2 REF: 1a2b9662-4683-11df-9c7d-001185f0d2ea OBJ: 3-2.2 Determining Whether Lines are Parallel STA: MCC9-12.G.CO.9 LOC: MTH.C.11.01.03.03.01.010 | MTH.C.11.01.03.03.010 TOP: 3-2 Proving Lines Parallel KEY: parallel | alternate interior angles DOK: DOK 2 5. ANS: [1] Vertical Angle Theorem [2] Converse of the Same-Side Interior Angles Theorem Proof: Statements Reasons 1. 1. Given 2. 2. Vertical Angle Theorem 3. 3. Substitution (Steps 1 and 2) 4. Converse of the Same-Side Interior Angles 4. Theorem PTS: OBJ: LOC: KEY: 6. ANS: 1 DIF: 1 REF: 1a2dd1ae-4683-11df-9c7d-001185f0d2ea 3-2.3 Proving Lines Parallel STA: MCC9-12.G.CO.9 MTH.C.11.01.03.03.01.010 | MTH.C.11.01.03.03.010 TOP: 3-2 Proving Lines Parallel parallel lines | proof DOK: DOK 1 ; The angles are alternate interior angles, and they are congruent, so the lanes are parallel by the Converse of the Alternate Interior Angles Theorem. Substitute 10 for x in each expression: The angles are alternate interior angles, and they are congruent, so the lanes are parallel by the Converse of the Alternate Interior Angles Theorem. PTS: 1 DIF: 2 REF: 1a30340a-4683-11df-9c7d-001185f0d2ea OBJ: 3-2.4 Application STA: MCC9-12.G.CO.9 LOC: MTH.C.11.01.03.03.01.010 | MTH.C.11.01.03.03.010 TOP: 3-2 Proving Lines Parallel KEY: parallel lines | proof 7. ANS: DOK: DOK 2 is the shorter segment. Substitute for and 8 for Subtract 4 from both sides. Divide both sides by 2 and simplify. . PTS: 1 DIF: 2 REF: 1a305b1a-4683-11df-9c7d-001185f0d2ea OBJ: 3-3.1 Distance From a Point to a Line STA: MCC9-12.A.REI.3 LOC: MTH.C.11.01.03.007 TOP: 3-3 Perpendicular Lines KEY: distance from a point to a line | inequality | perpendicular DOK: DOK 2 8. ANS: [1] [2] 2 intersecting lines form linear pair of s lines . [3] 2 lines to the same line lines . Proof: Statements Reasons 1. Given 1. 2. 2. If 2 intersecting lines form linear pair of s lines . 3. 3. If 2 lines to the same line lines . PTS: OBJ: LOC: TOP: DOK: 9. ANS: 1 DIF: 1 REF: 1a329666-4683-11df-9c7d-001185f0d2ea 3-3.2 Proving Properties of Lines STA: MCC9-12.G.CO.9 MTH.P.08.02.03.01.002 | MTH.C.11.01.03.03.010 | MTH.C.11.01.03.04.007 3-3 Perpendicular Lines KEY: perpendicular | parallel | proof DOK 1 y 7 E' E –7 F' 7 G' x F G –7 This is a translation 6 units left and 2 units up. PTS: 1 DIF: 2 REF: 9141cc4b-6ab2-11e0-9c90-001185f0d2ea OBJ: 4-1.1 Drawing and Identifying Transformations NAT: NT.CCSS.MTH.10.9-12.G.CO.5 STA: MCC9-12.G.CO.5 TOP: 4-1 Congruence and Transformations KEY: transformation | coordinate geometry DOK: DOK 2 10. ANS: y 7 A D B C –7 Y X Z 7 x W –7 This is a rotation of 180° about the origin. PTS: 1 DIF: 2 REF: 9141f35b-6ab2-11e0-9c90-001185f0d2ea OBJ: 4-1.1 Drawing and Identifying Transformations NAT: NT.CCSS.MTH.10.9-12.G.CO.5 STA: MCC9-12.G.CO.5 TOP: 4-1 Congruence and Transformations KEY: transformation | coordinate geometry DOK: DOK 2 11. ANS: The triangles are congruent because can be mapped to by a reflection: . PTS: 1 DIF: 2 REF: 91442ea6-6ab2-11e0-9c90-001185f0d2ea OBJ: 4-1.2 Determining Whether Figures are Congruent NAT: NT.CCSS.MTH.10.9-12.G.CO.6 STA: MCC9-12.G.CO.6 TOP: 4-1 Congruence and Transformations KEY: reflection | rotation | transformation | translation | coordinate geometry DOK: DOK 2 12. ANS: The triangles are congruent because can be mapped onto by a rotation: , followed by a reflection: . PTS: OBJ: STA: KEY: 13. ANS: 1 DIF: 2 REF: 91469101-6ab2-11e0-9c90-001185f0d2ea 4-1.3 Applying Transformations NAT: NT.CCSS.MTH.10.9-12.G.CO.6 MCC9-12.G.CO.5 TOP: 4-1 Congruence and Transformations transformation | coordinate geometry DOK: DOK 2 = Step 1 Find m . Triangle Sum Theorem Substitute 61° for and 22° for Simplify. Subtract 83° from both sides. Step 2 Find . . Linear Pair Theorem and Angle Addition Postulate Substitute 97° for and 42° for . Simplify. Subtract 139° from both sides. PTS: OBJ: LOC: KEY: 14. ANS: 1 DIF: 2 REF: 1a696caa-4683-11df-9c7d-001185f0d2ea 4-2.1 Application STA: MCC9-12.G.MG.1 MTH.C.11.02.01.01.005 | MTH.C.11.03.02.04.002 TOP: 4-2 Angle Relationships in Triangles triangle sum theorem DOK: DOK 2 Let the acute angles be m m m m PTS: OBJ: LOC: KEY: 15. ANS: and , with m . The acute angles of a right triangle are complementary. Substitute for m . Subtract from both sides. 1 DIF: 1 REF: 1a6993ba-4683-11df-9c7d-001185f0d2ea 4-2.2 Finding Angle Measures in Right Triangles STA: MCC9-12.A.CED.1 MTH.C.11.03.02.05.001 TOP: 4-2 Angle Relationships in Triangles triangle sum theorem DOK: DOK 1 = Exterior Angle Theorem Substitute for . Simplify. Add 2 to both sides. Divide both sides by 10. PTS: OBJ: LOC: KEY: 16. ANS: , for , and 118 for 1 DIF: 2 REF: 1a6bcf06-4683-11df-9c7d-001185f0d2ea 4-2.3 Applying the Exterior Angle Theorem STA: MCC9-12.A.CED.1 MTH.C.11.03.02.04.004 TOP: 4-2 Angle Relationships in Triangles exterior angle theorem DOK: DOK 2 , Third Angles Theorem Definition of congruent angles Substitute for and Subtract from both sides. Divide both sides by –3. So Since for . . , . PTS: 1 DIF: 2 REF: 1a6e3162-4683-11df-9c7d-001185f0d2ea OBJ: 4-2.4 Applying the Third Angles Theorem NAT: NT.CCSS.MTH.10.9-12.G.SRT.5 STA: MCC9-12.A.CED.1 LOC: MTH.C.11.03.02.04.005 TOP: 4-2 Angle Relationships in Triangles KEY: third angles theorem | triangle sum theorem DOK: DOK 2 17. ANS: m The Third Angles Theorem states that if two angles of one triangle are congruent to two angles of another triangle, then the third pair of angles are congruent. It is given that PTS: NAT: LOC: KEY: 18. ANS: and . Therefore, . So, m . 1 DIF: 3 REF: 1a6e5872-4683-11df-9c7d-001185f0d2ea NT.CCSS.MTH.10.9-12.G.SRT.5 STA: MCC9-12.A.CED.1 MTH.C.11.03.02.04.005 TOP: 4-2 Angle Relationships in Triangles third angles theorem | triangle sum theorem DOK: DOK 2 , , , , , Corresponding angles and corresponding sides are parts which lie in the same position in the triangles. Corresponding angles: , , Corresponding sides: , , PTS: OBJ: STA: TOP: DOK: 19. ANS: m 1 DIF: 1 REF: 1a7093be-4683-11df-9c7d-001185f0d2ea 4-3.1 Naming Congruent Corresponding Parts NAT: NT.CCSS.MTH.10.9-12.G.SRT.5 MCC9-12.G.CO.7 LOC: MTH.C.11.01.02.03.002 | MTH.C.11.02.02.002 4-3 Congruent Triangles KEY: correspondence | corresponding parts DOK 1 = 67° Triangle Sum Theorem Substitution. Simplify. Subtract 113 from both sides. Corresponding parts of congruent triangles are congruent. Definition of congruent angles Corresponding parts of congruent triangles are congruent. PTS: 1 DIF: 2 REF: 1a72f61a-4683-11df-9c7d-001185f0d2ea OBJ: 4-3.2 Using Corresponding Parts of Congruent Triangles NAT: NT.CCSS.MTH.10.9-12.G.SRT.5 STA: MCC9-12.A.CED.1 LOC: MTH.C.11.02.02.001 | MTH.C.11.03.02.04.002 TOP: 4-3 Congruent Triangles KEY: triangle sum theorem | congruent triangles | corresponding parts DOK: DOK 2 20. ANS: [1] Definition of perpendicular lines [2] Third Angles Theorem [3] Reflexive Property of Congruence Proof: Statements 1. 2. and are right angles. 3. 4. 5. 6. 7. T is the midpoint of . 8. 9. 10. Reasons 1. Given 2. Definition of perpendicular lines 3. Right Angle Congruence Theorem 4. Given 5. Third Angles Theorem 6. Given 7. Given 8. Definition of midpoint 9. Reflexive Property of Congruence 10. Definition of congruent triangles PTS: 1 DIF: 2 OBJ: 4-3.3 Proving Triangles Congruent STA: MCC9-12.G.CO.10 TOP: 4-3 Congruent Triangles DOK: DOK 1 21. ANS: [1] Vertical Angles Theorem [2] Third Angles Theorem [3] Proof: Statements 1. , , 2. 3. 4. 5. REF: NAT: LOC: KEY: PTS: OBJ: STA: TOP: DOK: 22. ANS: REF: NAT: LOC: KEY: 1 DIF: 2 4-3.4 Application MCC9-12.G.CO.10 4-3 Congruent Triangles DOK 1 1a731d2a-4683-11df-9c7d-001185f0d2ea NT.CCSS.MTH.10.9-12.G.SRT.5 MTH.P.08.02.03.017 | MTH.C.11.08.02.02.002 proof | congruent triangles Reasons 1. Given 2. Given 3. Vertical Angles Theorem 4. Third Angles Theorem 5. Definition of congruent triangles 1a755876-4683-11df-9c7d-001185f0d2ea NT.CCSS.MTH.10.9-12.G.SRT.5 MTH.P.08.02.03.017 | MTH.C.11.08.02.02.002 proof | congruent triangles It is given that , , and bisects . By the definition of segment bisector, three pairs of corresponding sides of the triangles are congruent. Therefore, ABC DEC by SSS. PTS: OBJ: LOC: KEY: 23. ANS: [1] 1 DIF: 1 REF: 1a77e1e2-4683-11df-9c7d-001185f0d2ea 5-1.1 Using SSS to Prove Triangle Congruence STA: MCC9-12.G.SRT.5 MTH.C.11.08.02.02.02.002 TOP: 5-1 Triangle Congruence: SSS and SAS SSS | congruent triangles DOK: DOK 2 . All [2] [3] It is given that . Since and are right angles, Congruence Theorem. By the Reflexive Property of Congruence, PTS: 1 DIF: 2 OBJ: 5-1.2 Application LOC: MTH.C.11.08.02.02.02.004 KEY: proof | SAS | congruent triangles 24. ANS: [1]. Definition of midpoint [2] [3] SAS Proof: Statements 1. P is the midpoint of and . 2. 3. 4. PTS: OBJ: STA: LOC: TOP: DOK: 25. ANS: , REF: STA: TOP: DOK: by the Right Angle . Therefore, by SAS. 1a7a1d2e-4683-11df-9c7d-001185f0d2ea MCC9-12.G.SRT.5 5-1 Triangle Congruence: SSS and SAS DOK 2 Reasons 1. Given 2. Definition of midpoint 3. Vertical Angles Theorem 4. SAS 1 DIF: 2 REF: 1a7ca69a-4683-11df-9c7d-001185f0d2ea 5-1.4 Proving Triangles Congruent NAT: NT.CCSS.MTH.10.9-12.G.SRT.4 MCC9-12.G.SRT.5 MTH.C.11.01.02.02.001 | MTH.C.11.08.02.02.02.004 | MTH.C.11.08.02.02.002 5-1 Triangle Congruence: SSS and SAS KEY: proof | congruent triangles DOK 1 The SAS Postulate is used when two sides and an included angle of one triangle are congruent to the corresponding sides and included angle of a second triangle. From the given, . From the figure, by the Reflexive Property of Congruence. You have two pair of congruent sides, so you need information about the included angles. Use these pairs of sides to determine the included angles. The angle between sides is . The angle between sides is . You need to know to prove PTS: STA: TOP: DOK: 26. ANS: by the SAS Postulate. 1 DIF: 3 REF: 1a7ee1e6-4683-11df-9c7d-001185f0d2ea MCC9-12.G.SRT.5 LOC: MTH.C.11.08.02.02.02.004 5-1 Triangle Congruence: SSS and SAS KEY: proof | congruent triangles | SAS DOK 2 is given. . Therefore, because both are right angles. By the Vertical Angles Theorem, by ASA. is given. . Therefore, PTS: OBJ: LOC: KEY: 27. ANS: Yes. because both are right angles. By the Vertical Angles Theorem, by ASA. 1 DIF: 1 5-2.2 Applying ASA Congruence MTH.C.11.08.02.02.02.006 proof | congruent triangles | ASA REF: STA: TOP: DOK: 1a816b52-4683-11df-9c7d-001185f0d2ea MCC9-12.G.SRT.5 5-2 Triangle Congruence: ASA, AAS, and HL DOK 2 is given. In addition, by the Reflexive Property of Congruence, . Since and , by the Perpendicular Transversal Theorem . By the definition of right angle, is a right angle. Similarly, is a right angle. Therefore, ABD DCA by the HL Congruence Theorem. PTS: OBJ: TOP: DOK: 28. ANS: 1 DIF: 2 REF: 1a8608fa-4683-11df-9c7d-001185f0d2ea 5-2.4 Applying HL Congruence STA: MCC8.MP.3 LOC: MTH.C.11.08.02.02.02.011 5-2 Triangle Congruence: ASA, AAS, and HL KEY: proof | congruent triangles | HL DOK 2 , HL Because and are right angles, are right triangles. You are given a pair of congruent legs and a pair of congruent hypotenuses . So a hypotenuse and a leg of are congruent to the corresponding hypotenuse and leg of by HL. PTS: STA: TOP: DOK: 29. ANS: . 1 DIF: 3 REF: 1a86300a-4683-11df-9c7d-001185f0d2ea MCC8.MP.3 | MCC9-12.G.SRT.5 LOC: MTH.C.11.08.02.02.02.010 | MTH.C.11.08.02.02.02.011 5-2 Triangle Congruence: ASA, AAS, and HL KEY: proof | congruent triangles | HL DOK 3 by SAS and From the figure, by SAS and by CPCTC, so m . by CPCTC. m by substitution. by the Vertical Angles Theorem. Therefore, by substitution. PTS: 1 DIF: 2 REF: 1a886b56-4683-11df-9c7d-001185f0d2ea OBJ: 5-3.1 Application STA: MCC9-12.G.CO.7 LOC: MTH.C.11.08.02.01.003 | MTH.C.11.08.02.02.02.004 TOP: 5-3 Triangle Congruence: CPCTC KEY: congruent triangles | corresponding parts | CPCTC DOK: DOK 2 30. ANS: [1] AAS [2] CPCTC [1] Steps 1 and 7 state that two angles and a nonincluded side of MLN andPLO are congruent. By AAS, MLN PLO. [2] Since MLN PLO, by CPCTC, . PTS: 1 DIF: 2 REF: 1a8af4c2-4683-11df-9c7d-001185f0d2ea OBJ: 5-3.3 Using CPCTC in a Proof STA: MCC9-12.G.CO.10 LOC: MTH.P.08.02.03.017 | MTH.C.11.08.02.01.003 TOP: 5-3 Triangle Congruence: CPCTC KEY: congruent triangles | corresponding parts | CPCTC 31. ANS: x=6 The triangles can be proved congruent by the SAS Postulate. By CPCTC, . DOK: DOK 2 Solve the equation for x. PTS: 1 DIF: 3 REF: 1a8f926a-4683-11df-9c7d-001185f0d2ea NAT: NT.CCSS.MTH.10.9-12.G.SRT.5 STA: MCC9-12.G.SRT.5 LOC: MTH.C.11.08.02.004 | MTH.C.11.08.02.02.02.004 TOP: 5-3 Triangle Congruence: CPCTC KEY: congruent triangles | corresponding parts | CPCTC DOK: DOK 2 32. ANS: m = BW Tauri and M77 are equidistant from Earth, so . By the Isosceles Triangle Theorem, From the Angle Addition Postulate, and m = . PTS: 1 DIF: 2 REF: 1a96e08e-4683-11df-9c7d-001185f0d2ea OBJ: 5-4.1 Application STA: MCC9-12.G.MG.1 LOC: MTH.C.11.03.02.03.02.002 | MTH.C.11.03.02.06.01.003 TOP: 5-4 Isosceles and Equilateral Triangles KEY: isosceles triangle DOK: DOK 1 33. ANS: m = 75 ° Isosceles Triangle Theorem m m Triangle Sum Theorem m m m Substitute for m and substitute for m and m . Simplify and subtract 30 from both sides. Divide both sides by 5. Thus m . PTS: 1 DIF: 2 REF: 1a991bda-4683-11df-9c7d-001185f0d2ea OBJ: 5-4.2 Finding the Measure of an Angle STA: MCC9-12.A.CED.1 LOC: MTH.C.11.02.01.01.007 TOP: 5-4 Isosceles and Equilateral Triangles KEY: isosceles triangle theorem DOK: DOK 2 34. ANS: CA = 14 Equiangular triangles are equilateral. ABC is equilateral. Definition of equilateral triangle. Subtract s and add 10 to both sides of the equation. Substitute 12 for s in the equation for AB. Simplify. . Definition of equilateral triangle. Substitute 14 for AB. PTS: 1 DIF: 1 REF: 1a9b7e36-4683-11df-9c7d-001185f0d2ea OBJ: 5-4.3 Using Properties of Equilateral Triangles STA: MCC9-12.A.CED.1 LOC: MTH.C.11.03.02.03.03.001 | MTH.C.11.03.02.03.05.005 TOP: 5-4 Isosceles and Equilateral Triangles KEY: equilateral triangle | side length DOK: DOK 1 35. ANS: m = ,m = ,m = Step 1: is supplementary to the angle that is . . So . Step 2: By the Alternate Interior Angles Theorem, So . . Step 3: By the Isosceles Triangle Theorem, and the angle opposite the other side of the isosceles triangle are congruent. Let be that unknown angle. Then, and . by the Triangle Sum Theorem. . So . PTS: STA: TOP: DOK: 36. ANS: 1 DIF: 3 REF: 1a9de092-4683-11df-9c7d-001185f0d2ea MCC9-12.A.CED.1 LOC: MTH.C.11.03.02.04.002 5-4 Isosceles and Equilateral Triangles KEY: multi-step | isosceles triangle theorem DOK 2 Perpendicular Bisector Theorem Substitute 6.4 for . Given Substitute. Segment Addition Postulate Substitute. Simplify. PTS: OBJ: STA: TOP: DOK: 37. ANS: 1 DIF: 2 REF: 1acdb6d2-4683-11df-9c7d-001185f0d2ea 6-1.1 Applying the Perpendicular Bisector Theorem and Its Converse MCC9-12.A.CED.1 LOC: MTH.C.11.01.02.04.01.003 | MTH.C.11.01.02.04.01.004 6-1 Perpendicular and Angle Bisectors KEY: perpendicular bisector DOK 2 Angle Bisector Theorem Substitute. PTS: 1 DIF: 1 REF: 1acff21e-4683-11df-9c7d-001185f0d2ea OBJ: 6-1.2 Applying the Angle Bisector Theorems STA: MCC9-12.A.CED.1 LOC: MTH.C.11.02.03.005 TOP: 6-1 Perpendicular and Angle Bisectors KEY: angle bisector DOK: DOK 1 38. ANS: = 2.2 Centroid Theorem Substitute 3.3 for PTS: OBJ: LOC: KEY: 39. ANS: and simplify. 1 DIF: 2 REF: 1adbddea-4683-11df-9c7d-001185f0d2ea 6-3.1 Using the Centroid to Find Segment Lengths STA: MCC9-12.G.CO.10 MTH.C.11.03.02.02.02.006 TOP: 6-3 Medians and Altitudes of Triangles centroid DOK: DOK 2 . The slope of . . The slope of The length of . The slopes are equal so . The length of . . . Find K by finding the midpoint of Find L by finding the midpoint of . Find the slope of Find the slope of The slopes are equal so Find the length of Find the length of . . . . PTS: 1 DIF: 2 REF: 1ae304fe-4683-11df-9c7d-001185f0d2ea OBJ: 6-4.1 Examining Midsegments in the Coordinate Plane STA: MCC9-12.G.GPE.4 LOC: MTH.C.11.03.02.02.05.005 TOP: 6-4 The Triangle Midsegment Theorem KEY: midsegment | coordinate geometry DOK: DOK 2 40. ANS: XY = 1.5 The length of a midsegment is half the length of the corresponding base. PTS: OBJ: LOC: KEY: 41. ANS: 1 DIF: 2 REF: 1ae32c0e-4683-11df-9c7d-001185f0d2ea 6-4.2 Using the Triangle Midsegment Theorem STA: MCC9-12.G.CO.10 MTH.C.11.03.02.02.05.005 TOP: 6-4 The Triangle Midsegment Theorem midsegment DOK: DOK 1 300 m Midsegment Theorem Substitute 150 for XY. Multiply both sides by 2 and simplify. The width of the reservoir BC is about 300 meters. PTS: OBJ: LOC: KEY: 42. ANS: 1 DIF: 2 6-4.3 Application MTH.C.11.03.02.02.05.005 midsegment DOK: DOK 1 , , and PTS: OBJ: LOC: TOP: DOK: 43. ANS: REF: 1ae5675a-4683-11df-9c7d-001185f0d2ea STA: MCC9-12.G.MG.1 TOP: 6-4 The Triangle Midsegment Theorem , . By the Third Angles Theorem . 1 DIF: 1 REF: 1b7bc30e-4683-11df-9c7d-001185f0d2ea 8-1.1 Describing Similar Polygons STA: MCC9-12.G.SRT.2 MTH.C.11.08.03.01.001 | MTH.C.11.08.03.01.002 | MTH.C.11.08.03.005 8-1 Ratios in Similar Polygons KEY: similar polygons |corresponding sides DOK 1 The similarity ratio is and rectangle MNOP ~ rectangle RSTU. Step 1 Identify pairs of congruent angles. , , All angles of a rectangle are right angles and are and congruent. Step 2 Compare corresponding sides. Thus the similarity ratio is , and rectangle MNOP ~ rectangle RSTU. PTS: 1 DIF: 1 REF: 1b7dfe5a-4683-11df-9c7d-001185f0d2ea OBJ: 8-1.2 Identifying Similar Polygons STA: MCC9-12.G.SRT.2 LOC: MTH.C.11.08.03.002 | MTH.C.11.08.03.004 TOP: 8-1 Ratios in Similar Polygons KEY: similar polygons | similarity statement DOK: DOK 1 44. ANS: 4.9 cm x is the width of the miniature in centimeters. The rectangular miniature is similar to the rectangular table top, so the corresponding lengths are proportional. Cross Product Property Simplify. Divide both sides by 2.5. The width of the miniature table top is about 4.9 cm. PTS: 1 DIF: 2 REF: 1b8060b6-4683-11df-9c7d-001185f0d2ea OBJ: 8-1.3 Application STA: MCC9-12.G.MG.1 LOC: MTH.C.09.03.01.001 | MTH.C.11.08.03.006 TOP: 8-1 Ratios in Similar Polygons KEY: similar figures | side length DOK: DOK 2 45. ANS: height = 7.68 in.; width = 6.24 in. Step 1 Convert measurements to inches. tower’s length ft in. tower’s width ft in. Step 2 Apply the scale factor formula. new dimension = (scale factor)(original dimension) model’s length in. model’s width in. PTS: 1 DIF: 3 STA: MCC9-12.G.MG.1 TOP: 8-1 Ratios in Similar Polygons DOK: DOK 2 46. ANS: J´(3, 12), K´(18, 12), L´(18, 3), M´(3, 3) PTS: OBJ: STA: KEY: 47. ANS: REF: 1b8087c6-4683-11df-9c7d-001185f0d2ea LOC: MTH.C.11.08.03.005 | MTH.C.11.08.03.006 KEY: application | similarity ratio | scale model 1 DIF: 2 REF: 914db812-6ab2-11e0-9c90-001185f0d2ea 8-2.1 Drawing and Describing Dilations NAT: NT.CCSS.MTH.10.9-12.G.CO.2 MCC9-12.G.SRT.1 TOP: 8-2 Similarity and Transformations coordinate plane | dilation DOK: DOK 2 by the Converse of the Corresponding Angles Postulate. by the Corresponding Angles Postulate. by AA Similarity. Corresponding sides are proportional, so . Step 1 Prove triangles are similar. As shown , so by the Converse of the Corresponding Angles Postulate. by the Corresponding Angles Postulate. Therefore by AA Similarity. Step 2 Find BC. Corresponding sides are proportional. Substitute 32 for DE, 48 for AC, and 28 for BE. Cross Products Property Simplify. Divide both sides by 32. PTS: 1 DIF: 2 REF: 1b85256e-4683-11df-9c7d-001185f0d2ea OBJ: 8-3.3 Finding Lengths in Similar Triangles NAT: NT.CCSS.MTH.10.9-12.G.SRT.5 STA: MCC9-12.G.SRT.5 LOC: MTH.C.11.08.03.03.002 TOP: 8-3 Triangle Similarity: AA, SSS, and SAS KEY: similar triangles | side length | AA similarity DOK: DOK 2 48. ANS: = 1.25 It is given that , so by the Triangle Proportionality Theorem. Substitute 2 for RQ, 8 for QM, and 5 for NM. Cross Products Property Divide by 8. PTS: 1 DIF: 2 REF: 1b89ea26-4683-11df-9c7d-001185f0d2ea OBJ: 8-4.1 Finding the Length of a Segment STA: MCC9-12.G.SRT.5 LOC: MTH.C.11.01.02.01.001 | MTH.C.11.08.03.04.002 TOP: 8-4 Applying Properties of Similar Triangles KEY: similar triangles | segment length DOK: DOK 2 49. ANS: 3.75 cm Two-Transversal Proportionality Corollary Substitute the given values. Cross Products Property Simplify. PTS: 1 DIF: 2 REF: 1b8c7392-4683-11df-9c7d-001185f0d2ea OBJ: 8-4.3 Application STA: MCC9-12.G.CO.9 TOP: 8-4 Applying Properties of Similar Triangles KEY: perspective drawing DOK: DOK 2 50. ANS: = 12 Since bisects , by the Triangle Angle Bisector Theorem. Substitute the given values. Cross Products Property Distributive Property Simplify. Divide. Substitute for x. Simplify. PTS: OBJ: LOC: KEY: 1 DIF: 2 REF: 1b8eaede-4683-11df-9c7d-001185f0d2ea 8-4.4 Using the Triangle Angle Bisector Theorem STA: MCC9-12.A.CED.1 MTH.C.11.08.03.04.008 TOP: 8-4 Applying Properties of Similar Triangles triangle angle bisector theorem DOK: DOK 2