Geometry Final exam Review First Semester

Name: lass: ate: Geometry Final exam Review First Semester Multiple hoice Identify the choice that best completes the statement or answers the question. 1. Find the measure of O. Then, classify the angle as acute, right, or obtuse. a. m O = 125 ; obtuse c. m O = 90 ; right b. m O = 35 ; acute d. m O = 160 ; obtuse 2. m IJK = 57 and m IJL = 20. Find m LJK. a. m LJK = 37 c. m LJK = 37 b. m LJK = 77 d. m LJK = 40 1

Name: 3. Solve the equation 4x 6 = 34. Write a justification for each step. 4x 6 = 34 Given equation +6 +6 [1] 4x = 40 Simplify. 4x = 40 4 4 [2] x = 10 Simplify. a. [1] Substitution Property of Equality; [2] ivision Property of Equality b. [1] ddition Property of Equality; [2] ivision Property of Equality c. [1] ivision Property of Equality; [2] Subtraction Property of Equality d. [1] ddition Property of Equality; [2] Reflexive Property of Equality 4. Identify the property that justifies the statement. and EF. So EF. a. Reflexive Property of ongruence c. Symmetric Property of ongruence b. Substitution Property of Equality d. Transitive Property of ongruence 5. Write a justification for each step, given that EG = FH. EG = FH Given information EG = EF + FG [1] FH = FG + GH Segment ddition Postulate EF + FG = FG + GH [2] EF = GH Subtraction Property of Equality a. [1] ngle ddition Postulate [2] Subtraction Property of Equality b. [1] Substitution Property of Equality [2] Transitive Property of Equality c. [1] Segment ddition Postulate [2] efinition of congruent segments d. [1] Segment ddition Postulate [2] Substitution Property of Equality 6. Two lines intersect to form two pairs of vertical angles. 1 with measure (20x + 7) and 3 with measure (5x + 7y + 49) are vertical angles. 2 with measure (3x 2y + 30) and 4 are vertical angles. Find the values x and y and the measures of all four angles. a. x = 6; y = 10; 127 ; 127 ; 28 ; 28 c. x = 5; y = 5; 107 ; 107 ; 73 ; 73 b. x = 8; y = 11, 167 ; 167 ; 13 ; 13 d. x = 7; y = 9; 147 ; 147 ; 33 ; 33 2

Name: 7. Find m. a. m = 40 c. m = 35 b. m = 45 d. m = 50 8. 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. a. x = 10, y = 20 c. x = 0, y = 60 b. x = 20, y = 20 d. x = 10, y = 140 3

Name: 9. What transformation is used to produce this frieze pattern? a. translation b. rotation c. reflection d. dilation 10. Find m K. a. m K = 63 c. m K = 79 b. m K = 55 d. m K = 39 4

Name: 11. What additional information do you need to prove by the SS Postulate? a. c. b. d. 12. Using the information about John, Jason, and Julie, can you uniquely determine the distances from John to Julie and from Julie to Jason? Explain your answer. Statement 1: John and Jason are standing 12 feet apart. Statement 2: The angle from Julie to John to Jason measures 31. Statement 3: The angle from John to Jason to Julie measures 49. a. No. There is no unique configuration. b. Yes. They form a unique triangle by SS. c. Yes. They form a unique triangle by S. d. Yes. They form a unique triangle by SSS. 13. etermine if you can use S to prove E. Explain. a. is given. E because both are right angles. No other congruence relationships can be determined, so S cannot be applied. b. is given. E because both are right angles. y the djacent ngles Theorem, E. Therefore, E by S. c. is given. E because both are right angles. y the Vertical ngles Theorem, E. Therefore, E by S. d. is given. E because both are right angles. y the Vertical ngles Theorem, E. Therefore, E by SS. 5

Name: 14. Given: MLN PLO, MNL POL, MO NP Prove: MLP is isosceles. omplete the proof. Proof: Statements Reasons 1. MLN PLO, MNL POL 1. Given 2. MO NP 2. Given 3. MO = NP 3. efinition of congruent line segments 4. NO = NO 4. Reflexive Property of Equality 5. MO NO = NP NO 5. Subtraction Property of Equality 6. MO NO = MN and NP NO = OP 6. Segment ddition Postulate 7. MN = OP 7. Substitution Property of Equality 8. MLN PLO 8. [1] 9. ML PL 9. [2] 10. MLP is isosceles. 10. efinition of isosceles triangle a. [1] PT [2] S b. [1] S [2] PT c. [1] PT [2] S d. [1] S [2] PT 6

Name: 15. Two Seyfert galaxies, W Tauri and M77, represented by points and, are equidistant from Earth, represented by point. What is m? a. m = 65 c. m = 50 b. m = 115 d. m = 60 16. Find m Q. a. m Q = 30 c. m Q = 70 b. m Q = 60 d. m Q = 75 17. Find the measures and. a. = 6.4, = 4.6 c. = 6.4, = 2.3 b. = 4.6, = 6.4 d. = 2.3, = 6.4 7

Name: 18. ZO, YO, and XO are the perpendicular bisectors of. Find O. a. O = 4.2 c. O = 7.4 b. O = 3.4 d. O = 14.8 19. Find the circumcenter of with vertices ( 2,4), ( 2, 2), and (4, 2). a. (1, 1) c. Ê 1 1ˆ 2 2 Ë Á b. (0, 0) d. ( 2, 2) 20. Point O is the centroid of, Y = 3.3 and O = 3. Find O. a. O = 2.2 c. O = 3.3 b. O = 1.1 d. O = 3 8

Name: 21. MNOP is a parallelogram. Find MP. a. MP = 25 c. MP = 20 b. MP = 30 d. MP = 6 22. Verify that PQR SQT. a. Q Q by the Reflexive Property of ongruence. QS QP = QT QR = 3 5 PQR SQT by SS Similarity. b. P QST and R QTS by the orresponding ngles Postulate. PQR SQT by Similarity. c. P QTS and R QST by the lternate Interior ngles Theorem. PQR SQT by Similarity. d. Q Q by the Reflexive Property of ongruence. PS QP = QT QR = 2 5 PQR SQT by SS Similarity. 9

Name: 23. Find the value of x. Express your answer in simplest radical form. a. x = 3 5 c. x = 3 3 b. x = 9 5 d. x = 5 3 24. Find the value of x. Express your answer in simplest radical form. a. x = 5 2 2 c. x = b. x = 5 2 d. x = 5 3 2 25. n architect designs the front view of a house with a gable roof that has a 45-45 -90 triangle shape. The overhangs are 0.5 meter each from the exterior walls, and the width of the house is 16 meters. What should the side length l of the triangle be? Round your answer to the nearest meter. 5 2 2 a. 12 m c. 24 m b. 11 m d. 23 m 10

Name: 26. Write the trigonometric ratio for cos X as a fraction and as a decimal rounded to the nearest hundredth. a. cos X = 12 9 1.33 c. cos X = 12 15 = 0.80 b. cos X = 9 15 = 0.60 d. cos X = 9 12 = 0.75 27. Use your calculator to find the trigonometric ratios sin 79, cos 47, and tan 77. Round to the nearest hundredth. a. sin 79 = 0.99, cos 47 = 0.44, tan 77 = 32.27 b. sin 79 = 0.44, cos 47 = 0.99, tan 77 = 32.27 c. sin 79 = 0.68, cos 47 = 0.98, tan 77 = 4.33 d. sin 79 = 0.98, cos 47 = 0.68, tan 77 = 4.33 28. Jessie is building a ramp for loading motorcycles onto a trailer. The trailer is 2.8 feet off of the ground. To avoid making it too difficult to push a motorcycle up the ramp, Jessie decides to make the angle between the ramp and the ground 15. To the nearest hundredth of a foot, find the length of the ramp. a. 10.82 feet c. 0.72 feet b. 2.90 feet d. 10.45 feet 29. Use the trigonometric ratio sin = 0.38 to determine which angle of the triangle is. a. 2 c. 3 b. 1 d. No solution. 30. Find sin to the nearest hundredth. a. sin = 0.45 c. sin = 2.24 b. sin = 0.50 d. sin = 0.89 11

Name: 31. lassify each angle in the diagram as an angle of elevation or an angle of depression. a. ngles of elevation: 1, 3 ngles of depression: 2, 4 b. ngles of elevation: 2, 4 ngles of depression: 1, 3 c. ngles of elevation: 1, 4 ngles of depression: 2, 3 d. ngles of elevation: 2, 3 ngles of depression: 1, 4 32. n eagle 300 feet in the air spots its prey on the ground. The angle of depression to its prey is 15. What is the horizontal distance between the eagle and its prey? Round to the nearest foot. a. 1,120 ft c. 310 ft b. 1,159 ft d. 723 ft 33. Rewrite the polynomial 12x 2 + 6 7x 5 + 3x 3 + 7x 4 5x in standard form. Then, identify the leading coefficient, degree, and number of terms. Name the polynomial. a. 7x 5 + 7x 4 + 3x 3 + 12x 2 5x + 6 leading coefficient: 7; degree: 5; number of terms: 6; name: quintic polynomial b. 6 5x + 12x 2 + 7x 3 + 3x 4 7x 5 leading coefficient: 6; degree: 0; number of terms: 6; name: quintic polynomial c. 7x 5 + 7x 4 + 12x 3 + 3x 2 5x + 6 leading coefficient: 7; degree: 5; number of terms: 6; name: quintic polynomial d. 6 5x + 12x 2 + 3x 3 + 7x 4 7x 5 leading coefficient: 6; degree: 0; number of terms: 6; name: quintic polynomial 34. dd. Write your answer in standard form. (5a 5 a 4 ) + (a 5 + 7a 4 2) a. 6a 5 + 6a 4 c. 6a 5 + 6a 4 2 b. 6a 10 + 6a 8 2 d. 5a 5 + 7a 4 2 35. Find the product 2c d 4 ( 4c 6 d 5 c 3 d ). a. 2c 8 d 10 + 2c 5 d 6 c. 8c 6 d 20 2c 3 d 4 b. 8c 7 d 9 2c 4 d 5 d. 2c 7 d 9 + c 4 d 5 12

Name: 36. The right triangle shown is enlarged such that each side is multiplied by the value of the hypotenuse, 3y. Find the expression that represents the perimeter of the enlarged triangle. a. 9y 2 + 8xy c. 9y 2 + 24xy b. 6y + 24xy d. 6y 2 + 24xy 13

Geometry Final exam Review First Semester nswer Section MULTIPLE HOIE 1. NS: y the Protractor Postulate, m O = m O m O. First, measure O and O. m O = m O m O = 125 35 = 90 Thus, O is a right angle. To find the measure of angle O, subtract the measure of angle O from the measure of angle O. The sum of the measure of angle O and the measure of angle O is equal to the measure of angle O. orrect! Use the Protractor Postulate. PTS: 1 IF: 2 REF: 193ac816-4683-11df-9c7d-001185f0d2ea OJ: 1-2.2 Measuring and lassifying ngles ST: M8.MP.5 LO: MTH..11.02.01.01.001 MTH..11.02.04.02.002 TOP: 1-2 Measuring and onstructing ngles KEY: angle measure protractor degrees OK: OK 1 2. NS: m IJK = m IJL + m LJK ngle ddition Postulate 57 = 20 + m LJK Substitute 57 for m IJK and 20 for m IJL. 37 = m LJK Subtract 20 from both sides. Use the ngle ddition Postulate. Subtract the smaller angle measure from the larger angle measure. orrect! Subtract the smaller angle measure from the larger angle measure. PTS: 1 IF: 1 REF: 193d0362-4683-11df-9c7d-001185f0d2ea OJ: 1-2.3 Using the ngle ddition Postulate ST: M9-12..E.1 LO: MTH..11.02.01.01.005 TOP: 1-2 Measuring and onstructing ngles KEY: angle addition postulate OK: OK 1 1

3. NS: 4x 6 = 34 Given equation +6 +6 [1] ddition Property of Equality 4x = 40 Simplify. 4x = 40 4 4 [2] ivision Property of Equality x = 10 Simplify. heck the properties. orrect! heck the properties. heck the properties. PTS: 1 IF: 1 REF: 19dce886-4683-11df-9c7d-001185f0d2ea OJ: 2-2.1 Solving an Equation in lgebra NT: NT.SS.MTH.10.9-12..REI.1 ST: M9-12..REI.1 LO: MTH.P.08.02.002 MTH..10.06.01.01.009 TOP: 2-2 lgebraic Proof KEY: algebraic proof proof OK: OK 2 4. NS: The Transitive Property of ongruence states that if figure figure and figure figure, then figure figure. The Reflexive Property of ongruence states that figure is congruent to figure. The Substitution Property of Equality states that if a = b, then b can be substituted for a in any expression. The Symmetric Property of ongruence states that if figure is congruent to figure, then figure is congruent to figure. orrect! PTS: 1 IF: 1 REF: 19e1ad3e-4683-11df-9c7d-001185f0d2ea OJ: 2-2.4 Identifying Properties of Equality and ongruence ST: M9-12.G.O.9 LO: MTH.P.08.02.001 TOP: 2-2 lgebraic Proof KEY: congruence properties reflexive symmetric transitive OK: OK 1 2

5. NS: EG = FH EG = EF + FG FH = FG + GH EF + FG = FG + GH EF = GH Given information Segment ddition Postulate Segment ddition Postulate Substitution Property of Equality Subtraction Property of Equality The ngle ddition Postulate refers to angles, not segments. heck the properties. heck the steps. orrect! PTS: 1 IF: 2 REF: 19e3e88a-4683-11df-9c7d-001185f0d2ea OJ: 2-3.1 Writing Justifications NT: NT.SS.MTH.10.9-12..REI.1 ST: M9-12.G.O.9 LO: MTH.P.08.02.001 MTH..10.06.01.01.009 MTH..11.01.02.01.003 TOP: 2-3 Geometric Proof KEY: geometric proof proof OK: OK 1 3

6. NS: Step 1 reate a system of equations. m 1 = m 3 20x + 7 = 5x + 7y + 49 15x 7y = 42 The sum of the measures of supplementary angles equals 180. m 1 + 2 = 180 20x + 7 + 3x 2y + 30 = 180 23x 2y = 143 reate a system of equations. 15x 7y = 42 23x 2y = 143 Step 2 Solve the system of equations. 15x 7y = 42 23x 2y = 143 30x + 14y = 84 Multiply the first equation by 2. 161x 14y = 1001 Multiply the second equation by 7. 131x = 917 dd the two equations together. x = 7 ivide both sides by 131. Solve for y. Substitute x = 7 into 15x 7y = 42. 15(7) 7y = 42 y = 9 The values are x = 7 and y = 9. Step 3 Solve for the four angles. ngle 1: (20(7) + 7) = 147 ngle 2: (3(7) 2(9) + 30) = 33 ngle 3: (5(7) + 7(9) + 49) = 147 ngle 4 and angle 2 are vertical and thus have equal measures. The measurement of angle 4 is 33. The measures of all four angles are 147, 147, 33, and 33. Use the definitions of supplementary and vertical angles to create a solvable system of equations. Use the definitions of supplementary and vertical angles to create a solvable system of equations. Use the definitions of supplementary and vertical angles to create a solvable system of equations. orrect! PTS: 1 IF: 3 REF: 19effb66-4683-11df-9c7d-001185f0d2ea ST: M9-12.G.O.9 TOP: 2-4 Flowchart and Paragraph Proofs KEY: vertical angles OK: OK 3 4

7. NS: (x) = (3x 70) orresponding ngles Postulate 0 = 2x 70 Subtract x from both sides. 70 = 2x dd 70 to both sides. 35 = x ivide both sides by 2. m = 3x 70 m = 3(35) 70 = 35 Substitute 35 for x. Simplify. If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent. Use the orresponding ngles Postulate. orrect! First, set the measures of the corresponding angles equal to each other. Then, solve for x and substitute in the expression (3x 70). PTS: 1 IF: 2 REF: 1a24483e-4683-11df-9c7d-001185f0d2ea OJ: 3-1.1 Using the orresponding ngles Postulate ST: M9-12.G.O.9 LO: MTH..11.01.03.03.01.005 TOP: 3-1 ngles Formed by Parallel Lines and Transversals KEY: corresponding angles parallel lines OK: OK 2 8. NS: y the orresponding ngles Postulate, (4x + y) = 60. y the lternate Interior ngle Postulate, (8x + y) = 100. 8x + y = 100 (4x + y) = 60 Subtract the first equation from the second. 4x = 40 x = 10 ivide both sides by 4. 8(10) + y = 100 Substitute 10 for x. y = 20 Simplify. orrect! When solving a system of two equations, subtract the second equation from the first. Use postulates related to parallel lines to form two equations. When solving a system of two equations, subtract the second equation from the first. PTS: 1 IF: 3 REF: 1a26d1aa-4683-11df-9c7d-001185f0d2ea OJ: 3-1.3 pplication ST: M9-12.G.O.9 LO: MTH..11.01.03.03.01.005 MTH..11.01.03.03.01.006 TOP: 3-1 ngles Formed by Parallel Lines and Transversals KEY: corresponding angles parallel lines alternate interior angles OK: OK 3 5

9. NS: orrect! Start with the left half of the figure. Which transformation would you use to create the right half? Start with the left half of the figure. Which transformation would you use to create the right half? Start with the left half of the figure. Which transformation would you use to create the right half? PTS: 1 IF: 3 REF: 9146b811-6ab2-11e0-9c90-001185f0d2ea OJ: 4-1.4 pplication ST: M9-12.G.O.6 TOP: 4-1 ongruence and Transformations KEY: frieze pattern glide reflection symmetry OK: OK 1 10. NS: m K + m L = m LMN Exterior ngle Theorem ( 6x 9) + ( 4x + 7) = 118 Substitute 6x 9 for m K, 4x + 7 for m L, and 118 for m LMN. 10x 2 = 118 Simplify. 10x = 120 dd 2 to both sides. x = 12 ivide both sides by 10. m K = 6x 9 = 6( 12) 9 = 63 orrect! You found the measure of angle L. Find the measure of angle K instead. First, use the Exterior ngle Theorem to find the value of x. Then substitute the value for x to find the measure of angle K. First, use the Exterior ngle Theorem to find the value of x. Then substitute the value for x to find the measure of angle K. PTS: 1 IF: 2 REF: 1a6bcf06-4683-11df-9c7d-001185f0d2ea OJ: 4-2.3 pplying the Exterior ngle Theorem ST: M9-12..E.1 LO: MTH..11.03.02.04.004 TOP: 4-2 ngle Relationships in Triangles KEY: exterior angle theorem OK: OK 2 6

11. NS: The SS 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 ongruence. 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 and is. The angle between sides and is. You need to know to prove by the SS Postulate. This information is needed to use the SSS Postulate. orrect! You need the included angle between the two sides. This information is already given. Find information that you need that is not given or true in the figure. PTS: 1 IF: 3 REF: 1a7ee1e6-4683-11df-9c7d-001185f0d2ea ST: M9-12.G.SRT.5 LO: MTH..11.08.02.02.02.004 TOP: 5-1 Triangle ongruence: SSS and SS KEY: proof congruent triangles SS OK: OK 2 12. NS: Statements 2 and 3 determine the measures of two angles of the triangle. Statement 1 determines the length of the included side. y S, the triangle must be unique. raw a diagram. There is enough information to determine a unique triangle. There is not enough information for SS. raw a diagram to help you. orrect! There is not enough information for SSS. raw a diagram to help you. PTS: 1 IF: 2 REF: 1a814442-4683-11df-9c7d-001185f0d2ea OJ: 5-2.1 Problem-Solving pplication ST: M9-12.G.MG.1 LO: MTH..14.06.01.005 TOP: 5-2 Triangle ongruence: S, S, and HL KEY: application triangle congruence OK: OK 2 7

13. NS: is given. E because both are right angles. y the Vertical ngles Theorem, E. Therefore, E by S. Look for vertical angles. djacent angles are angles in a plane that have their vertex and one side in common but have no interior points in common. ngle and angle E are not adjacent angles. orrect! Use S, not SS, to prove the triangles congruent. PTS: 1 IF: 1 REF: 1a816b52-4683-11df-9c7d-001185f0d2ea OJ: 5-2.2 pplying S ongruence ST: M9-12.G.SRT.5 LO: MTH..11.08.02.02.02.006 TOP: 5-2 Triangle ongruence: S, S, and HL KEY: proof congruent triangles S OK: OK 2 14. NS: [1] Steps 1 and 7 state that two angles and a nonincluded side of MLN and PLO are congruent. y S, MLN PLO. [2] Since MLN PLO, by PT, ML PL. efore using PT, you must prove that triangle MLN and triangle PLO are congruent. Since steps 1 and 7 state that two angles and a nonincluded side are congruent, which triangle congruence theorem states that the triangles are congruent? Steps 1 and 7 state that two angles and a nonincluded side of triangle MLN and triangle PLO are congruent. Which triangle congruence theorem states that the triangles are congruent? efore using PT, you must prove that triangle MLN and triangle PLO are congruent. orrect! PTS: 1 IF: 2 REF: 1a8af4c2-4683-11df-9c7d-001185f0d2ea OJ: 5-3.3 Using PT in a Proof ST: M9-12.G.O.10 LO: MTH.P.08.02.03.017 MTH..11.08.02.01.003 TOP: 5-3 Triangle ongruence: PT KEY: congruent triangles corresponding parts PT OK: OK 2 8

15. NS: W Tauri and M77 are equidistant from Earth, so. y the Isosceles Triangle Theorem,. From the ngle ddition Postulate, m = 65 and m =65. orrect! ngles and are congruent. Use the ngle ddition Postulate to find the measure of angle. ngles and are congruent. Use the ngle ddition Postulate to find the measure of angle. ngles and are congruent. Use the ngle ddition Postulate to find the measure of angle. PTS: 1 IF: 2 REF: 1a96e08e-4683-11df-9c7d-001185f0d2ea OJ: 5-4.1 pplication ST: M9-12.G.MG.1 LO: MTH..11.03.02.03.02.002 MTH..11.03.02.06.01.003 TOP: 5-4 Isosceles and Equilateral Triangles KEY: isosceles triangle OK: OK 1 16. NS: m Q = m R = ( 2x + 15) Isosceles Triangle Theorem m P + m Q + m R = 180 Triangle Sum Theorem x + ( 2x + 15) + ( 2x + 15) = 180 Substitute x for m P and substitute 2x + 15 for m Q and m R. 5x = 150 Simplify and subtract 30 from both sides. x = 30 ivide both sides by 5. Thus m Q = ( 2x + 15) = [2( 30) + 15] = 75. This is x. The measure of angle Q is 2x + 15. y the Isosceles Triangle Theorem, the measure of angle Q equals the measure of angle R. Use the Triangle Sum Theorem and solve for x. y the Isosceles Triangle Theorem, the measure of angle Q equals the measure of angle R. Use the Triangle Sum Theorem and solve for x. orrect! PTS: 1 IF: 2 REF: 1a991bda-4683-11df-9c7d-001185f0d2ea OJ: 5-4.2 Finding the Measure of an ngle ST: M9-12..E.1 LO: MTH..11.02.01.01.007 TOP: 5-4 Isosceles and Equilateral Triangles KEY: isosceles triangle theorem OK: OK 2 9

17. NS: = = 6.4 X = X X = 2.3 = X + X = 2.3 + 2.3 = 4.6 Perpendicular isector Theorem Substitute 6.4 for. Given Substitute. Segment ddition Postulate Substitute. Simplify. orrect! Switch the answers for and. is twice X. Switch the answers for and. is twice X. PTS: 1 IF: 2 REF: 1acdb6d2-4683-11df-9c7d-001185f0d2ea OJ: 6-1.1 pplying the Perpendicular isector Theorem and Its onverse ST: M9-12..E.1 LO: MTH..11.01.02.04.01.003 MTH..11.01.02.04.01.004 TOP: 6-1 Perpendicular and ngle isectors KEY: perpendicular bisector OK: OK 2 18. NS: O is the circumcenter of. y the ircumcenter Theorem, O is equidistant from the vertices of. O = O ircumcenter Theorem O = 4.2 Substitute 4.2 for O. orrect! O is equidistant from the vertices, so O = O. O is equidistant from the vertices, so O = O. O is equidistant from the vertices, so O = O. PTS: 1 IF: 2 REF: 1ad71932-4683-11df-9c7d-001185f0d2ea OJ: 6-2.1 Using Properties of Perpendicular isectors ST: M9-12.G..3 LO: MTH..11.03.02.02.03.006 TOP: 6-2 isectors of Triangles KEY: perpendicular bisector circumcenter OK: OK 1 10

19. NS: Step 1 Find equations for two perpendicular bisectors. Since two sides of the triangle lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of is y = 1, and the perpendicular bisector of is x = 1. Step 2 Find the intersection of the two equations. x = 1 and y = 1 intersect at (1,1), the circumcenter of. orrect! The equation of the perpendicular bisectors of segment and segment are y = 1 and x = 1, respectively. Their intersection is the circumcenter. The equation of the perpendicular bisectors of segment and segment are y = 1 and x = 1, respectively. Their intersection is the circumcenter. The equation of the perpendicular bisectors of segment and segment are y = 1 and x = 1, respectively. Their intersection is the circumcenter. PTS: 1 IF: 2 REF: 1ad74042-4683-11df-9c7d-001185f0d2ea OJ: 6-2.2 Finding the ircumcenter of a Triangle ST: M9-12.G..3 LO: MTH..11.03.02.02.03.002 MTH..11.03.02.02.03.004 TOP: 6-2 isectors of Triangles KEY: circumcenter OK: OK 2 20. NS: O = 2 3 Y O = 2 (3.3) = 2.2 3 entroid Theorem Substitute 3.3 for Y and simplify. orrect! Segment O is two-thirds the length of segment Y. Segment O is two-thirds the length of segment Y. Segment O is two-thirds the length of segment Y. PTS: 1 IF: 2 REF: 1adbddea-4683-11df-9c7d-001185f0d2ea OJ: 6-3.1 Using the entroid to Find Segment Lengths ST: M9-12.G.O.10 LO: MTH..11.03.02.02.02.006 TOP: 6-3 Medians and ltitudes of Triangles KEY: centroid OK: OK 2 11

21. NS: MP NO MP = NO 5x = 3x + 12 x = 6 MP = 5x = 5(6) = 30 Opposite sides of a parallelogram are congruent. efinition of congruent segments Substitute. Simplify and solve. Substitute and solve for entire segment measure. Opposite sides of a parallelogram are congruent. orrect! Set the expressions for opposite sides of the parallelogram equal to each other. Substitute the solution for x back into the original expression. PTS: 1 IF: 2 REF: 1b31b2da-4683-11df-9c7d-001185f0d2ea OJ: 7-1.2 Using Properties of Parallelograms to Find Measures ST: M9-12..E.1 LO: MTH..11.03.03.01.005 TOP: 7-1 Properties of Parallelograms KEY: parallelogram opposite sides OK: OK 2 22. NS: Q Q by the Reflexive Property of ongruence. QS QP = 6 10 = 3 5, QT QR = 9 15 = 3 5 Therefore E by SS Similarity. orrect! Is it given that segment PR is parallel to ST? Is it given that segment PR is parallel to ST? re the angle pairs in this choice alternate interior angles? Is segment PS a side of one of the triangles? re the ratios equal? PTS: 1 IF: 2 REF: 1b82ea22-4683-11df-9c7d-001185f0d2ea OJ: 8-3.2 Verifying Triangle Similarity NT: NT.SS.MTH.10.9-12.G.SRT.5 ST: M9-12.G.SRT.5 LO: MTH..11.08.03.03.007 TOP: 8-3 Triangle Similarity:, SSS, and SS KEY: similar triangles SS similarity OK: OK 2 12

23. NS: a 2 + b 2 = c 2 Pythagorean Theorem 3 2 + 6 2 = x 2 Substitute 3 for a, 6 for b, and x for c. 45 = x 2 Simplify. 45 = x Find the positive square root. x = ( 9) ( 5) = 3 5 Simplify the radical. orrect! Simplify the square root of 45 correctly. pply the Pythagorean Theorem by substituting 3 for a, 6 for b, and x for c. You reversed the order of the radicand and the number outside the radical sign. PTS: 1 IF: 2 REF: 1af8a14a-4683-11df-9c7d-001185f0d2ea OJ: 9-1.1 Using the Pythagorean Theorem ST: M9-12.G.SRT.8 LO: MTH..10.05.10.05.01.001 MTH..11.03.02.05.02.002 TOP: 9-1 The Pythagorean Theorem KEY: Pythagorean Theorem side length OK: OK 1 24. NS: The triangle is an isosceles right triangle, which is a 45 45 90 triangle. The length of the hypotenuse is 5. 5 = x 2 Hypotenuse = leg 2 5 2 = x ivide both sides by 2. 52 2 5 2 2 = x 2 = x Rationalize the denominator. orrect! In a 45 45 90 triangle, both legs are congruent, and the length of the hypotenuse is the length of a leg times the square root of 2. In a 45 45 90 triangle, both legs are congruent, and the length of the hypotenuse is the length of a leg times the square root of 2. In a 45 45 90 triangle, both legs are congruent, and the length of the hypotenuse is the length of a leg times the square root of 2. PTS: 1 IF: 1 REF: 1b0203aa-4683-11df-9c7d-001185f0d2ea OJ: 9-2.1 Finding Side Lengths in a 45-45-90 Triangle ST: M9-12.G.SRT.6 LO: MTH..11.03.02.05.03.004 TOP: 9-2 pplying Special Right Triangles KEY: special right triangle 45-45-90 isosceles right triangle OK: OK 1 13

25. NS: The roof is a 45 45 90 triangle with a hypotenuse of 17 m. 17 = l 2 Hypotenuse = leg 2 l = 17 2 12 m ivide both sides by 2 and round. orrect! The hypotenuse is 17 m (0.5 + 16 + 0.5). In a 45 45 90 triangle, the length of the hypotenuse is the length of a leg times the square root of 2. In a 45 45 90 triangle, the length of the hypotenuse is the length of a leg times the square root of 2. The hypotenuse is 17 m (0.5 + 16 + 0.5). PTS: 1 IF: 2 REF: 1b022aba-4683-11df-9c7d-001185f0d2ea OJ: 9-2.2 pplication NT: NT.SS.MTH.10.9-12.G.SRT.8 ST: M9-12.G.SRT.8 LO: MTH..11.03.02.05.03.005 TOP: 9-2 pplying Special Right Triangles KEY: special right triangles 45-45-90 isosceles right triangle OK: OK 1 26. NS: cos X = 12 15 = 0.80 adjacent leg The cosine of an is hypotenuse. osine is the ratio of the adjacent side to the hypotenuse. Sine is the ratio of the opposite side to the hypotenuse. orrect! Tangent is the ratio of the opposite to the adjacent side. PTS: 1 IF: 1 REF: 1bc0c06a-4683-11df-9c7d-001185f0d2ea OJ: 10-1.1 Finding Trigonometric Ratios NT: NT.SS.MTH.10.9-12.G.SRT.6 ST: M9-12.G.SRT.6 LO: MTH..14.02.01.002 MTH..14.02.02.004 TOP: 10-1 Trigonometric Ratios KEY: trigonometric ratio trigonometry cosine OK: OK 2 14

27. NS: Make sure your calculator is in degree mode. sin 79 = 0.98, cos 47 = 0.68, tan 77 = 4.33 hange your calculator to degree mode. hange your calculator to degree mode. Switch the first and second answers. orrect! PTS: 1 IF: 1 REF: 1bc349d6-4683-11df-9c7d-001185f0d2ea OJ: 10-1.3 alculating Trigonometric Ratios NT: NT.SS.MTH.10.K-12.5.1 TOP: 10-1 Trigonometric Ratios KEY: trigonometric ratio trigonometry cosine sine tangent OK: OK 2 28. NS: sin = sin15 = 2.8 = 2.8 sin 15 10.82 feet Write a trigonometric ratio. Substitute the given values. Multiply both sides by and divide by sin 15. Simplify the expression. orrect! Since the hypotenuse and opposite side are involved, use the sine ratio. The ramp should be substantially larger than the opposite leg, because the opposite leg is across from the smallest possible angle in the triangle. raw a picture to help determine which trigonometric ratio to use. PTS: 1 IF: 2 REF: 1bc7e77e-4683-11df-9c7d-001185f0d2ea OJ: 10-1.5 Problem-Solving pplication NT: NT.SS.MTH.10.9-12.G.SRT.8 ST: M9-12.G.SRT.8 LO: MTH..14.02.03.001 TOP: 10-1 Trigonometric Ratios KEY: trigonometric ratio trigonometry side length OK: OK 2 15

29. NS: sin = opposite hypotenuse Sine is the ratio of the opposite leg to the hypotenuse. sin 1 = 1.2 = 0.92 1.2 is the length of the leg opposite 1. 1.3 1.3 is the length of the hypotenuse. cos 1 =.5 = 0.38 0.5 is the length of the leg adjacent 1. 1.3 1.3 is the length of the hypotenuse. Since cos = cos 2, 2 is. orrect! The sine of an angle is the length of the opposite leg divided by the length of the hypotenuse. The sine of an angle is the length of the opposite leg divided by the length of the hypotenuse. The sine of an angle is the length of the opposite leg divided by the length of the hypotenuse. PTS: 1 IF: 2 REF: 1bc80e8e-4683-11df-9c7d-001185f0d2ea OJ: 10-2.1 Identifying ngles from Trigonometric Ratios ST: M9-12.G.SRT.8 LO: MTH..14.02.03.002 MTH..14.02.001 TOP: 10-2 Solving Right Triangles KEY: trigonometric ratio trigonometry OK: OK 2 30. NS: y the Pythagorean Theorem, 2 = 2 + 2 = 4 2 + 2 2 = 20. So = 20 4.47. sin = 2 4.47 0.45 orrect! Use the Pythagorean Theorem to find. Sin = /. Sin = /. Sin = /. PTS: 1 IF: 2 REF: 1bccac36-4683-11df-9c7d-001185f0d2ea OJ: 10-2.3 Solving Right Triangles ST: M9-12.G.SRT.8 LO: MTH..14.02.02.002 TOP: 10-2 Solving Right Triangles KEY: trigonometric ratio trigonometry solve right triangles OK: OK 2 16

31. NS: 1 and 3 are formed by a horizontal line and a line of sight to a point above the line. They are angles of elevation. 2 and 4 are formed by a horizontal line and a line of sight to a point below the line. They are angles of depression. orrect! n angle of elevation is formed from a horizontal line and a point above the line. n angle of elevation is formed from a horizontal line and a point above the line. n angle of elevation is formed from a horizontal line and a point above the line. PTS: 1 IF: 1 REF: 1bd170ee-4683-11df-9c7d-001185f0d2ea OJ: 10-3.1 lassifying ngles of Elevation and epression ST: M9-12.G.SRT.8 LO: MTH..11.02.04.10.001 MTH..11.02.04.10.002 TOP: 10-3 ngles of Elevation and epression KEY: angle of elevation angle of depression trigonometry OK: OK 1 32. NS: y the lternate Interior ngles Theorem, m S = 15. From the sketch, tan15 = 300 x. So x = 300 1,120 ft. tan15 orrect! ivide 300 by tan 15. ivide 300 by tan 15. ivide 300 by tan 15. PTS: 1 IF: 2 REF: 1bd3d34a-4683-11df-9c7d-001185f0d2ea OJ: 10-3.3 Finding istance by Using ngle of epression NT: NT.SS.MTH.10.9-12.G.SRT.8 ST: M9-12.G.SRT.8 LO: MTH..14.02.03.001 TOP: 10-3 ngles of Elevation and epression KEY: angle of elevation angle of depression trigonometry OK: OK 2 17

33. NS: The standard form is written with the terms in order from highest to lowest degree. In standard form, the degree of the first term is the degree of the polynomial. The polynomial has 6 terms. It is a quintic polynomial. orrect! The standard form is written with the terms in order from highest to lowest degree. Find the correct coefficient of the x-cubed term. The standard form is written with the terms in order from highest to lowest degree. PTS: 1 IF: 2 REF: 15d92892-4683-11df-9c7d-001185f0d2ea OJ: 3-1.2 lassifying Polynomials ST: M9-12..PR.1 LO: MTH..10.05.08.004 MTH..10.05.08.006 MTH..10.05.08.007 TOP: 3-1 Polynomials OK: OK 2 34. NS: (5a 5 a 4 ) + (a 5 + 7a 4 2) = (5a 5 + 7a 4 ) + ( a 4 + a 5 ) + ( 2) Identify like terms. Rearrange terms to get like terms together. = 6a 5 + 6a 4 2 ombine like terms. heck that you have included all the terms. When adding polynomials, keep the same exponents. orrect! First, identify the like terms and rearrange these terms so they are together. Then, combine the like terms. PTS: 1 IF: 1 REF: 15db8aee-4683-11df-9c7d-001185f0d2ea OJ: 3-1.3 dding and Subtracting Polynomials NT: NT.SS.MTH.10.9-12..PR.1 ST: M9-12..PR.1 LO: MTH..10.05.08.03.001 TOP: 3-1 Polynomials OK: OK 2 35. NS: Use the istributive Property to multiply the monomial by each term inside the parentheses. Group terms to get like bases together, and then multiply. on't forget to multiply the coefficients for each term. orrect! When multiplying like bases, add the exponents. Multiply the coefficients for each term; don't add. PTS: 1 IF: 1 REF: 15e076b6-4683-11df-9c7d-001185f0d2ea OJ: 3-2.1 Multiplying a Monomial and a Polynomial NT: NT.SS.MTH.10.9-12..PR.1 ST: M9-12..PR.1 LO: MTH..10.05.08.03.02.002 TOP: 3-2 Multiplying Polynomials OK: OK 3 18

36. NS: measure of leg 1 = 3y(4x) = 12xy measure of leg 2 = 3y(4x) = 12xy measure of hypotenuse = 3y(3y) = 9y 2 P = 9y 2 + 12xy + 12xy P = 9y 2 + 24xy The perimeter is the sum of all the side lengths. Multiply both side lengths and the hypotenuse by 3y. orrect! heck for algebra mistakes. PTS: 1 IF: 3 REF: 15e9d916-4683-11df-9c7d-001185f0d2ea NT: NT.SS.MTH.10.9-12..PR.1 ST: M9-12..PR.1 TOP: 3-2 Multiplying Polynomials OK: OK 3 19

Geometry Final exam Review First Semester [nswer Strip] 3. 7. 9. 11. 1. 4. 8. 12. 5. 10. 2. 13. 6.

Geometry Final exam Review First Semester [nswer Strip] 14. 15. 18. 21. 23. 24. 22. 16. 19. 25. 17. 20.

Geometry Final exam Review First Semester [nswer Strip] 26. 31. 36. 27. 32. 33. 28. 29. 34. 35. 30.