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Geometr y Final Assessment 11-12, 1st semester Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. Name three collinear points. a. P, G, and N b. R, P, and N ____ c. R, P, and G d. R, G, and N 2. Draw and label a pair of opposite rays a. b. ____ → FH c. 3. Name a plane that contains . ←→ AC c. plane WRT d. plane RCA 4. Find the length of . BC a. b. ____ and . d. a. plane ACR b. plane WCT ____ → FG BC BC = –7 = –9 c. d. BC BC =7 =8 5. Find the best sketch, drawing, or construction of a segment congruent to . JK a. b. c. d. ____ 6. D is between C and E. a. CE = 17.5 b. CE = 78 ____ CE = , 6x CD = , and DE = 27. Find CE. 4x + 8 c. CE = 105 d. CE = 57 7. The map shows a linear section of Highway 35. Today, the Ybarras plan to drive the 360 miles from Springfield to Junction City. They will stop for lunch in Roseburg, which is at the midpoint of the trip. If they have already traveled 55 miles this morning, how much farther must they travel before they stop for lunch? a. 125 mi b. 145 mi ____ ____ c. 180 mi d. 305 mi 8. K is the midpoint of . and a. JK = 1, KL = 1, JL = 2 b. JK = 6, KL = 6, JL = 12 9. m and m . Find m . JL ∠IJK = 57° ∠IJL = 20° JK = 6x . Find JK, KL, and JL. c. JK = 12, KL = 12, JL = 6 d. JK = 18, KL = 18, JL = 36 KL = 3x + 3 ∠LJK a. m b. m c. m d. m ∠LJK = −37° ∠LJK = 77° ____ 10. bisects , m a. m = 22° b. m = 3° → ∠ABC BD , and m ∠ABD = (7x − 1)° ∠ABD c. m = 40° d. m = 20° ∠ABD ∠ABD ____ ____ ∠ABD 11. Find the measure of the complement of , where m a. c. b. d. 12. Find the measure of the supplement of , where m a. c. b. d. 13. A billiard ball bounces off the sides of a rectangular billiards table in such a way that ∠M ∠M = 31.1° 58.9° 31.1° 148.9° 121.1° ∠R 44.5° (190 − 8z)° (80 − 8z)° , find m , m , and m . ∠1 = 26.5° a. b. c. d. ∠R = (8z + 10)° (170 − 8z)° are complementary. If m ____ ∠LJK = 40° . Find m . ∠DBC = (4x + 8)° ∠ABD ____ ∠LJK = 37° m m m m ;m ;m ;m ;m ∠3 ∠4 ;m ;m ;m ;m ∠3 = 26.5° ∠4 = 63.5° ∠5 = 63.5° ∠3 = 26.5° ∠4 = 63.5° ∠5 = 53° ∠3 = 63.5° ∠4 = 26.5° ∠5 = 53° ∠3 = 26.5° ∠4 = 153.5° ∠5 = 26.5° 14. Find the perimeter and area of the figure. ∠5 , ∠1 ≅ ∠3 , and ∠4 ≅ ∠6 ∠3 and ∠4 a. perimeter = ; c. perimeter = ; area = area = b. perimeter = ; d. perimeter = ; area = area = 15. Find the circumference and area of the circle. Use 3.14 for , and round your answer to the nearest tenth. 6x 2 + 14 7x + 14 3x + 24 6x + 48 7x + 14 7x + 14 6x 2 + 14 3x + 24 ____ π a. b. ____ C = 201.0 C = 50.2 ft; ft; A = 50.2 A = 25.1 ft 2 ft 2 c. d. C = 25.1 C = 50.2 ft; ft; A = 50.2 A = 201.0 ft 2 ft 2 16. The width of a rectangular mirror is the measure of the length of the mirror. If the area is 192 in , what are 2 3 4 the length and width of the mirror? a. length = 24 in., width = 8 in. b. length = 16 in., width = 12 in. ____ 17. Find the coordinates of the midpoint of c. length = 48 in., width = 4 in. d. length = 25 in., width = 71 in. CM with endpoints C(1, –6) and M(7, 5). a. (3, -1) b. (8, –1) ____ c. (4, 0/-1/2) d. (4/1/2, 0/1/2) 18. Use the Distance Formula and the Pythagorean Theorem to find the distance, to the nearest tenth, from T(4, –2) to U(–2, 3). a. –1.0 units b. 3.4 units ____ c. 0.0 units d. 7.8 units 19. R is the midpoint of . T is the midpoint of . S is the midpoint of . Use the diagram to find the coordinates AB AC BC of T, the area of RST, and . Round your answers to the nearest tenth. ∆ AB T(3, 1); area of ∆RST = 8; T(3, 1); area of ∆RST = 32; T(3, 1); area of ∆RST = 16; T(3, 1); area of ∆RST = 8; 20. Determine if the conjecture is valid by the Law of Syllogism. a. b. c. d. ____ AB ≈ 17.9 AB ≈ 17.9 AB ≈ 8.9 AB ≈ 8.9 Given: If you are in California, then you are in the west coast. If you are in Los Angeles, then you are in California. Conjecture: If you are in Los Angeles, then you are in the west coast. a. No, the conjecture is not valid. b. Yes, the conjecture is valid. ____ 21. Solve the equation 4x − 6 . Write a justification for each step. Given equation [1] Simplify. [2] Simplify. 4x − 6 = 34 = 34 +6 +6 = 40 4x 4x 4 = 40 4 x = 10 a. [1] Substitution Property of Equality; ____ c. [1] Division Property of Equality; [2] Division Property of Equality [2] Subtraction Property of Equality [1] Addition Property of Equality; b. d. [1] Addition Property of Equality; [2] Division Property of Equality [2] Reflexive Property of Equality 22. Use the given plan to write a two-column proof. Given: m + m = 90 , m + m = 90 , m = m ∠1 ° ∠2 ∠3 ∠4 ° ∠2 ∠3 Prove: m = m Plan: Since both pairs of angle measures add to 90 , use substitution to show that the sums of both pairs are equal. Since m = m , use substitution again to show that sums of the other pairs are equal. Use the Subtraction Property of Equality to conclude that m = m . ∠1 ∠4 ° ∠2 ∠3 ∠1 ∠4 Complete the proof. Proof: Statements 1. m 2. [1] 3. m 4. m 5. m 6. m ∠1 ∠1 ∠2 ∠1 ∠1 + m = 90 ° ∠2 +m =m +m =m +m =m +m =m ∠2 ∠3 ∠4 ∠2 ∠4 ∠3 ∠2 ∠4 a. [1] m + m = 90 ∠3 ∠4 ° [2] Substitution Property [3] Subtraction Property of Equality Reasons 1. Given 2. Given 3. Substitution Property 4. Given 5. [2] 6. [3] b. [1] m + m = 90 ∠5 ____ ∠6 ° [2] Substitution Property [3] Subtraction Property of Equality c. [1] m + m = 90 [2] Subtraction Property of Equality [3] Substitution Property d. [1] m + m = 90 [2] Addition Property of Equality [3] Substitution Property 23. Use the given flowchart proof to write a two-column proof of the statement ∠3 ∠4 ° ∠5 ∠6 ° . AF ≅ FD Flowchart proof: ; AB = CD AB + BF = AF FC + CD = FD BF = FC Segment Addition Postulate Given AB + BF = FC + CD Addition Property of Equality AF = FD AF ≅ FD Substitution Definition of congruent segments Complete the proof. Two-column proof: Statements 1. ; 2. [1] 3. [2] 4. 5. AB = CD BF = FC AF = FD AF ≅ FD a. [1] Reasons 1. Given 2. Addition Property of Equality 3. Segment Addition Postulate 4. Substitution 5. Definition of congruent segments AB + BF = AF ; FC + CD = FD [2] b. [1] [2] c. [1] ; [2] d. [1] [2] 24. Identify the transversal and classify the angle pair AF = FD AF = FD AB + BF = FC + CD AB = CD BF = FC AB + BF = FC + CD AB + BF = FC + CD AB + BF = AF ;FC + CD = FD ____ ∠11 and . ∠7 a. b. c. d. ____ The transversal is line l. The angles are corresponding angles. The transversal is line l. The angles are alternate interior angles. The transversal is line n. The angles are alternate exterior angles. The transversal is line m. The angles are corresponding angles. 25. Find m . ∠RST a. m = b. m = ____ ∠RST 108° ∠RST 24° 26. Classify ∆DBC c. m = d. m = by its angle measures, given m a. obtuse triangle b. acute triangle ____ 27. ∆ABC a. b. ____ is an isosceles triangle. AB AB = 110 = 24 ,m ∠DAB = 60° ∠RST 156° ∠RST 72° , and m ∠ABD = 75° . ∠BDC = 25° c. right triangle d. equiangular triangle AB is the longest side with length . 4x + 4 c. d. AB AB BC = 8x + 3 and CA = . Find . 7x + 8 AB = 43 =5 28. One of the acute angles in a right triangle has a measure of . What is the measure of the other acute angle? 34.6° a. b. ____ c. d. 145.4° 34.6° 29. Find m , given ∠DCB , ∠A ≅ ∠F , and m ∠B ≅ ∠E . c. m d. m ∠DCB = 134° ∠DCB = 67° 30. Given: P is the midpoint of Prove: 90° ∠CDE = 46° a. m b. m ____ 55.4° ∠DCB = 44° ∠DCB = 46° and . RS TQ ∆TPR ≅ ∆QPS Complete the proof. Proof: Statements 1. P is the midpoint of and . 2. , 3. [2] 4. Reasons 1. Given 2. [1] 3. Vertical Angles Theorem 4. [3] a. [1]. Definition of midpoint c. [1] Definition of midpoint TQ TP ≅ QP RS RP ≅ SP ∆TPR ≅ ∆QPS [2] [3] SAS b. [1] Definition of midpoint d. [2] [3] SSS 31. What additional information do you need to prove ∠TPR ≅ ∠QPS RT ≅ SQ ____ [2] [3] SAS [1] Definition of midpoint [2] [3] SSS by the SAS Postulate? ∠PRT ≅ ∠PSQ ∠TPR ≅ ∠QPS ∆ABC ≅ ∆ADC a. b. ____ c. d. AB ≅ AD ∠ACB ≅ ∠ACD c. x = 2 d. x = 8 33. Which of the following is not a positioning of a right triangle with leg lengths of 4 units and 5 units? a. c. b. ____ BC ≅ DC 32. Find the value of x. a. x = 6 b. x = 4 ____ ∠ABC ≅ ∠ADC d. 34. Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints . B(5, 4) a. b. ____ y−3 = y−3 = 35. In , ∆ABC -7/2 2/7 c. d. (x − 1.5) (x − 1.5) . BY = 26.4 and CO = 24 AX BY, and CZ are medians. Find . BO -2/7 7/2 y−1 = (x − 3.5) y−1 = (x − 3.5) A(−2, 2) and a. b. ____ BO BO = 17.6 = 26.4 36. Given ∆ABC c. d. with , AB = 3 , and 37. In , ∆ABC a. b. ____ , m∠ADC > m∠BDC ____ ____ XY c. XY = 2.5 d. XY = 2 , and AC = 3x + 32 . Find the range of values for x. BC = 7x + 16 c. d. 0<x<4 − 167 < x < 4 x>4 − 78 < x < 4 38. Tell if the measures 6, 14, and 13 can be side lengths of a triangle. If so, classify the triangle as acute, right, c. Yes; right triangle d. No. 39. Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. a. regular and concave b. irregular and concave ____ = 24 = 8.8 CA = 6 or obtuse. a. Yes; acute triangle b. Yes; obtuse triangle ____ BO , find the length of midsegment . BC = 5 a. XY = 3 b. XY = 1.5 ____ BO c. regular and convex d. irregular and convex 40. Find the measure of each interior angle of a regular 45-gon. a. 176 c. 172 b. 164 d. 188 41. Find the measure of each exterior angle of a regular decagon. a. 45° c. 18° b. 22.5° d. 36° 42. and . Determine if the quadrilateral must be a parallelogram. Justify your answer. KL ≅ MN ° ° ° ° ∠KLM ≅ ∠MNK a. No. Only one set of angles and sides are given as congruent. The conditions for a parallelogram are not met. b. Yes. Opposite angles are congruent to each other. This is sufficient evidence to prove that the quadrilateral is a parallelogram. c. Yes. Opposite sides are congruent to each other. This is sufficient evidence to prove that the quadrilateral is a parallelogram. d. Yes. One set of opposite sides are congruent, and one set of opposite angles are congruent. This is sufficient evidence to prove that the quadrilateral is a parallelogram. Numer ic Response 43. Find the measure of the angle formed by the hands of a clock when it is 7:00. 44. The supplement of an angle is 26 more than five times its complement. Find the measure of the angle. 45. A satellite completely orbits Earth in 216 days. Determine the angle through which the satellite travels over a period of 12 days. 46. In the quadrilateral, . If m∠1 + m∠2 + m∠3 + m∠4 = 360° , m∠2 = 3m∠1 , and m∠3 = m∠1 + 6 , find m∠4 = m∠1 m∠3 in degrees. 47. An isosceles triangle has a perimeter of 50 in. The congruent sides measure (2x + 3) cm. The length of the third side is 4x cm. What is the value of x? 48. Find the value of x. 49. Find the value of n in the triangle. 50. In parallelogram LMNO, , and NO = 10.2 . What is the perimeter of parallelogram LMNO? LO = 14.7 51. Find the value of x in the rhombus. Matching Match each vocabulary term with its definition. a. bisect e. b. midpoint f. c. angle bisector g. d. angle h. ____ ____ ____ ____ ____ ____ 52. 53. 54. 55. 56. 57. perpendicular bisector measure degree segment bisector a unit of angle measure the point that divides a segment into two congruent segments to divide into two congruent parts a figure formed by two rays with a common endpoint the use of units to find a size or quantity a line, ray, or segment that divides a segment into two congruent segments Match each vocabulary term with its definition. a. conclusion e. hypothesis converse b. f. truth value c. inverse g. contrapositive d. negation ____ ____ ____ ____ ____ 58. 59. 60. 61. 62. for a statement, either true (T) or false (F) operations that undo each other the contradiction of a statement by using “not,” written as ∼ the statement formed by exchanging the hypothesis and conclusion of a conditional statement the statement formed by both exchanging and negating the hypothesis and conclusion Geometr y Final Assessment 11-12, 1st semester Answer Section MULTIPLE CHOICE 1. ANS: D Collinear points are points that lie on the same line. R, G, and N are three collinear points. Feedback A B C D Collinear points are points that lie on the same line. Collinear points are points that lie on the same line. Points R, P, and G are noncollinear. Correct! PTS: OBJ: STA: KEY: 2. ANS: 1 DIF: Basic REF: Page 7 1-1.1 Naming Points, Lines, and Planes NAT: 12.3.1.c GE1.0 TOP: 1-1 Understanding Points Lines and Planes points | lines | planes B In the diagram, rays → FG and → FH share a common endpoint F and form the line . ← → GH Feedback A B C D Opposite rays form a line. Correct! Opposite rays form a line. Opposite rays are two rays that have a common endpoint and form a line. PTS: OBJ: TOP: 3. ANS: 1 DIF: Basic REF: Page 7 1-1.2 Drawing Segments and Rays NAT: 12.3.1.d 1-1 Understanding Points Lines and Planes C STA: GE1.0 KEY: opposite rays A plane can be described by any three noncollinear points. Of the choices given, only points W, R, and T are noncollinear. Thus, lies in plane WRT. ←→ AC Feedback A B C D Points A, C, and R are collinear. A plane can be described by any three noncollinear points. Points W, C, and T are collinear. A plane can be described by any three noncollinear points. Correct! A plane can be described by any three noncollinear points. PTS: OBJ: STA: KEY: 4. ANS: 1 DIF: Basic REF: Page 7 1-1.3 Identifying Points and Lines in a Plane NAT: 12.3.4.b GE1.0 TOP: 1-1 Understanding Points Lines and Planes points | lines | planes C = | −8 − (−1) | BC = | −8 + 1 | = | −7 | =7 Feedback A B C D The length of a segment is always positive. Find the absolute value of the difference of the coordinates. Correct! Find the absolute value of the difference of the coordinates. PTS: OBJ: STA: KEY: 5. ANS: 1 DIF: Basic REF: Page 13 1-2.1 Finding the Length of a Segment NAT: 12.2.1.e GE1.0 TOP: 1-2 Measuring and Constructing Segments segments | length B Use a ruler to measure the lengths of , , , , and . JK LM PQ GH RS Feedback A B C D Use a ruler to measure the lengths of all the segments. Correct! Use a ruler to measure the lengths of all the segments. Use a ruler to measure the lengths of all the segments. PTS: OBJ: TOP: KEY: 6. ANS: 1 DIF: Average REF: Page 14 1-2.2 Copying a Segment STA: GE16.0 1-2 Measuring and Constructing Segments copying a segment | constructions C CE = CD + DE 6x 6x = ( 4x + 8 ) + 27 6x = 4x + 35 CE 4x + 8 CD 4x 2x = 35 2x = 35 2 2 or 17.5 x = 35 2 CE Segment Addition Postulate Substitute for and for . Simplify. Subtract from both sides. Divide both sides by 2. Simplify. = 6x = 6( 17.5 ) = 105 Feedback A You found the value of x. Find the length of the specified segment. B C D You found the length of a different segment. Correct! Check your equation. Make sure you are not subtracting instead of adding. PTS: OBJ: STA: KEY: 7. ANS: 1 DIF: Average REF: Page 15 1-2.3 Using the Segment Addition Postulate NAT: 12.3.5.a GE1.0 TOP: 1-2 Measuring and Constructing Segments segment addition postulate A If the Ybarra’s current position is represented by X, then the distance they must travel before they stop for lunch is XR. Segment Addition Postulate Solve for XR. Substitute known values. R is the midpoint of , so Simplify. SX + XR = SR XR = SR − SX XR = 1 2 SJ ( 360 ) − 55 XR = 125 SR = 1 2 . SJ Feedback A B C D Correct! Use the definition of midpoint and the Segment Addition Postulate to find the distance to Roseburg. This is the distance from Springfield to Roseburg. You must subtract the distance they have already traveled. This is the distance to Junction City. Use the definition of midpoint and the Segment Addition Postulate to find the distance to Roseburg. PTS: NAT: KEY: 8. ANS: 1 DIF: Average REF: Page 15 OBJ: 1-2.4 Application 12.2.1.e STA: GE1.0 TOP: 1-2 Measuring and Constructing Segments application | segment addition postulate B Step 1 Write an equation and solve. K is the midpoint of . Substitute for and for . Subtract from both sides. Divide both sides by 3. JL JK = KL 6x 6x = 3x + 3 JK 3x + 3 KL 3x 3x = 3 x=1 Step 2 Find JK, KL, and JL. JK = 6x = 6( 1 ) = 6 KL = 3x + 3 = 3(1) + 3 = 6 JL = JK + KL = 6 + 6 = 12 Feedback A B This is the value of x. Substitute this value for x to solve for the segment lengths. Correct! C D Reverse your answers. The first two segments are half as long as the last segment. Check your simplification methods when solving for x. Use division for the last step. PTS: OBJ: STA: KEY: 9. ANS: m m m m ∠IJK = ∠IJL + m Angle Addition Postulate Substitute for m and 20° for m . Subtract ° from both sides. ∠LJK ∠LJK 57° = 20° + 37° = 1 DIF: Average REF: Page 16 1-2.5 Using Midpoints to Find Lengths NAT: 12.2.1.e GE1.0 TOP: 1-2 Measuring and Constructing Segments midpoints | length C ∠IJK 57° ∠LJK ∠IJL 20 Feedback A B C D Use the Angle Addition Postulate. Subtract the smaller angle measure from the larger angle measure. Correct! Subtract the smaller angle measure from the larger angle measure. PTS: OBJ: STA: KEY: 10. ANS: 1 DIF: Basic REF: Page 22 1-3.3 Using the Angle Addition Postulate NAT: 12.2.1.f GE1.0 TOP: 1-3 Measuring and Constructing Angles angle addition postulate D Step 1 Solve for x. m m Definition of angle bisector. (7x − 1)° = (4x + 8)° Substitute ∠ABD = ∠DBC 7x − 1 for ∠ABD and 4x + 8 for . ∠DBC Add 1 to both sides. Subtract from both sides. Divide both sides by 3. 7x = 4x + 9 4x 3x = 9 x=3 Step 2 Find m . m ∠ABD ∠ABD = 7x − 1 = 7(3) − 1 = 20° Feedback A B C D Check your simplification technique. Substitute this value of x into the expression for the angle. This answer is the entire angle. Divide by two. Correct! PTS: OBJ: STA: KEY: 11. ANS: 1 DIF: Average REF: Page 23 1-3.4 Finding the Measure of an Angle NAT: 12.2.1.f GE1.0 TOP: 1-3 Measuring and Constructing Angles angle bisectors | angle measures A Subtract from 90º and simplify. 90° − 31.1°= 58.9 ° Feedback A B C D Correct! Find the measure of a complementary angle, not a supplementary angle. Complementary angles are angles whose measures have a sum of 90 degrees. The measures of complementary angles add to 90 degrees. PTS: OBJ: NAT: KEY: 12. ANS: 1 DIF: Basic REF: Page 29 1-4.2 Finding the Measures of Complements and Supplements 12.3.3.g STA: 6MG2.2 TOP: 1-4 Pairs of Angles complementary angles | supplementary angles A Subtract from 180º and simplify. 180° − (8z + 10)° = 180 − 8z − 10 = (170 − 8z)° Feedback A B C D Correct! The measures of supplementary angles add to 180 degrees. Supplementary angles are angles whose measures have a sum of 180 degrees. Find the measure of a supplementary angle, not a complementary angle. PTS: OBJ: NAT: KEY: 13. ANS: 1 DIF: Average REF: Page 29 1-4.2 Finding the Measures of Complements and Supplements 12.3.3.g STA: 6MG2.2 TOP: 1-4 Pairs of Angles complementary angles | supplementary angles B Since , Thus m . ∠1 ≅ ∠3 . m∠1 ≅ m∠3 ∠3 = 26.5° Since ∠3 and Since , Thus m . ∠4 ≅ ∠6 ∠4 are complementary, m . ∠4 = 90° − 26.5° = 63.5° . m∠4 ≅ m∠6 ∠6 = 63.5° By the Angle Addition Postulate, 180° = m∠4 + m∠5 + m∠6 = 63.5° + m∠5 + 63.5° Thus, m . ∠5 = 53° Feedback A B C D The measure of angle 5 is 180 degrees minus the sum of the measure of angle 4 and the measure of angle 6. Correct! Angle 1 and angle 3 are congruent. Congruent angles have the same measure. Angle 3 and angle 4 are complementary, not supplementary. PTS: OBJ: TOP: KEY: 14. ANS: 1 DIF: Average REF: Page 30 1-4.4 Problem-Solving Application NAT: 12.3.3.g STA: 6MG2.2 1-4 Pairs of Angles application | complementary angles | supplementary angles B Solve for the perimeter of the triangle. P= = = Solve for the area of the triangle. A= = = a+b+c 6 + (x + 8) + 6x 7x + 14 1 2 bh 1 2 (x + 8)(6) 3x + 24 Feedback A B C D Check your algebra when adding like terms. Correct! The triangle's area is half of its base times its height. The triangle's area is half of its base times its height. PTS: OBJ: STA: KEY: 15. ANS: C = 2πr = 2π ( 4 ) ≈ 25.1 A = πr2 = π ( 4 ) 2 1 DIF: Average REF: Page 36 1-5.1 Finding the Perimeter and Area GE8.0 TOP: 1-5 Using Formulas in Geometry perimeter | area | triangles C NAT: 12.2.1.h ft ≈ 50.2 ft 2 Feedback A B Use the radius, not the diameter, in your calculations. The circumference of a circle is 2 times pi times the radius. The area of a circle is pi times the radius squared. Correct! Use the radius, not the diameter, in your calculations. C D PTS: OBJ: STA: KEY: 16. ANS: 1 DIF: Average REF: Page 37 1-5.3 Finding the Circumference and Area of a Circle GE8.0 TOP: 1-5 Using Formulas in Geometry circles | circumference | area B NAT: 12.2.1.h The area of a rectangle is found by multiplying the length and width. Let l represent the length of the mirror. Then the width of the mirror is . 3 4 l A = lw 192 = l( 34 l) 192 = 3 2 l 4 256 = l2 16 = l The length of the mirror is 16 inches. The width of the mirror is Feedback 3 4 (16) = 12 inches. A B C D First, find the length. Then, use substitution to find the width. Correct! First, find the length. Then, use substitution to find the width. The formula for the area of a rectangle is length times width. PTS: 1 DIF: Advanced TOP: 1-5 Using Formulas in Geometry 17. ANS: C = NAT: 12.2.1.h STA: GE8.0 KEY: area | rectangles | application (4, 0/-1/2) Êx + x Á y + y ˆ˜ Ê Á 1 + 7 −6 + 5 ˜˜ˆ 1 2 Á MÁ , 1 2 ˜˜˜˜ = Á , Á ˜ Á Á Á 2 ˜¯ Á 2 ˜¯ Ë 2 Ë 2 Feedback A The x- and y-coordinates of the midpoint are the averages of the x- and y-coordinates of the endpoints. The x- and y-coordinates of the midpoint are the averages of the x- and y-coordinates of the endpoints. Correct! The x- and y-coordinates of the midpoint are the averages of the x- and y-coordinates of the endpoints. B C D PTS: OBJ: STA: KEY: 18. ANS: 1 DIF: Basic REF: Page 43 1-6.1 Finding the Coordinates of a Midpoint NAT: 12.2.1.e GE17.0 TOP: 1-6 Midpoint and Distance in the Coordinate Plane midpoint formula | coordinates D Method 1 Substitute the values for the coordinates of T and U into the Distance Formula. TU = Method 2 Use the Pythagorean Theorem. Plot the points on a coordinate plane. Then draw a right triangle. ˆ˜ 2 Ê ˆ˜ 2 Ê Á Á Ëx 2 − x 1 ¯ + Ëy 2 − y 1 ¯ = ( −2 − 4 ) + ( 3 − −2 ) = 2 ( −6 ) + ( 5 ) = 61 2 2 2 ≈ 7.8 units Count the units for sides a and b. and . Then apply the Pythagorean Theorem. a=6 c 2 = a2 + b2 = 62 + 52 = 36 + 25 = 61 c ≈ 7.8 units Feedback A B C D The distance is the square root of the quantity The distance is the square root of the quantity The distance is the square root of the quantity Correct! PTS: 1 DIF: Average . . . (x2 − x1)^2 + (y2 − y1)^2 (x2 − x1)^2 + (y2 − y1)^2 (x2 − x1)^2 + (y2 − y1)^2 REF: Page 45 b=5 OBJ: STA: KEY: 19. ANS: 1-6.4 Finding Distances in the Coordinate Plane NAT: 12.2.1.e GE15.0 TOP: 1-6 Midpoint and Distance in the Coordinate Plane congruent segments | distance formula | Pythagorean Theorem D Using the given diagram, the coordinates of T are (3, 1). The area of a triangle is given by . From the diagram, the base of the triangle is . From the diagram, the height of the triangle is . Therefore the area is . To find , use the Distance Formula with points A(1,5) and B . ≈ 8.9 A= 1 2 bh b = RT = 4 h=4 A= 1 2 (4)(4) = 8 AB AB = (x 2 − x 1) 2 + (y 2 − y 1) 2 = (−3, −3) (−3 − 1) 2 + (−3 − 5) 2 = 16 + 64 = 80 Feedback A B C D Use the distance formula to find the measurement of AB. The area of a triangle is one half the measure of its base times the measure of its height. The area of a triangle is one half times the measure of its base times the measure of its height. Correct! PTS: TOP: KEY: 20. ANS: 1 DIF: Advanced NAT: 12.2.1.e 1-6 Midpoint and Distance in the Coordinate Plane area | distance formula | triangles B STA: GE17.0 Let p, q, and r represent the following. p: You are in California. q: You are in the west coast. r: You are in Los Angeles. You are given that p→q and r→p Since p is the conclusion of the second statement and the hypothesis of the first statement, reorder the statements like this and . r→p p→q By the Law of Syllogism, if and are true, then is true. is the statement, If you are in Los Angeles, then you are in the west coast. r→p p→q r→q r→q Feedback A B The Law of Syllogism states that if (if p, then q) and (if q, then r) are true statements, then (if p, then r) is a true statement. Correct! PTS: OBJ: NAT: TOP: 21. ANS: 1 DIF: Average REF: Page 89 2-3.3 Verifying Conjectures by Using the Law of Syllogism 12.3.5.a STA: GE3.0 2-3 Using Deductive Reasoning to Verify Conjectures B 4x − 6 = 34 Given equation +6 [1] Addition Property of Equality Simplify. [2] Division Property of Equality Simplify. +6 = 40 4x 4x 4 = 40 4 x = 10 Feedback A B C D Check the properties. Correct! Check the properties. Check the properties. PTS: OBJ: STA: 22. ANS: 1 DIF: Basic REF: Page 104 2-5.1 Solving an Equation in Algebra 1A5.0 TOP: 2-5 Algebraic Proof A NAT: 12.5.2.e Proof: 1. m 2. m 3. m 4. m 5. m 6. m +m +m +m =m +m =m ∠1 ∠2 ∠3 ∠4 ∠1 ∠2 ∠2 Statements = 90 = 90 =m +m Reasons 1. Given 2. Given 3. Substitution Property 4. Given 5. Substitution Property 6. Subtraction Property of Equality ° ° ∠3 ∠4 ∠3 ∠1 ∠2 ∠1 =m +m ∠2 ∠4 ∠4 Feedback A B C D Correct! Check the given information. To get from Step 4 to Step 5, use substitution, not subtraction. To get from Step 4 to Step 5, use substitution, not addition. PTS: OBJ: STA: 23. ANS: 1 DIF: Average REF: Page 112 2-6.3 Writing a Two-Column Proof from a Plan GE2.0 TOP: 2-6 Geometric Proof D NAT: 12.3.5.a In a flowchart, reasons flow from the statement above. The statement above Reason 2 is statement above Reason 3 is . AB + BF = AF ; FC + CD = FD Feedback A B C D Reasons flow from the statement above. Reasons flow from the statement above. Reasons flow from the statement above. Correct! PTS: 1 DIF: Average REF: Page 118 OBJ: 2-7.1 Reading a Flowchart Proof NAT: 12.3.5.a TOP: 2-7 Flowchart and Paragraph Proofs STA: GE2.0 . The AB + BF = FC + CD 24. ANS: A To determine which line is the transversal for a given angle pair, locate the line that connects the vertices. Corresponding angles lie on the same side of the transversal l, on the same sides of lines n and m. Feedback A B C Correct! Alternate interior angles lie on opposite sides of the transversal, between two lines. To find which line is the transversal for a given angle pair, locate the line that connects the vertices. To find which line is the transversal for a given angle pair, locate the line that connects the vertices. D PTS: OBJ: STA: 25. ANS: 1 DIF: Average REF: Page 147 3-1.3 Identifying Angle Pairs and Transversals GE7.0 TOP: 3-1 Lines and Angles D NAT: 12.3.3.g Alternate Exterior Angles Theorem Subtract from both sides. Divide both sides by . (3x)° = (4x − 24)° 4x −x = −24 −1 x = 24 m ∠RST = 3x = 3(24) = 72° Substitute 24 for . x Feedback A B C D Find the measure of angle RST, not the supplement. Find the measure of angle RST, not the value of x. After finding x, substitute to find the angle measure. Correct! PTS: OBJ: TOP: 26. ANS: ∠ABD ∆DBC 1 DIF: Average REF: Page 156 3-2.2 Finding Angle Measures NAT: 12.3.3.g 3-2 Angles Formed by Parallel Lines and Transversals A and form a linear pair, so they are supplementary. Therefore is an obtuse triangle by definition. ∠DBC STA: GE7.0 . By substitution, m∠ABD + m∠DBC = 180° Feedback A B C D Correct! An acute triangle has three acute angles. A right triangle has one right angle. An equiangular triangle has three congruent acute angles. PTS: OBJ: STA: 27. ANS: 1 DIF: Average REF: Page 216 4-1.1 Classifying Triangles by Angle Measures GE12.0 TOP: 4-1 Classifying Triangles B Step 1 Find the value of x. Step 2 Find . AB NAT: 12.3.3.f . So 75° + m∠DBC = 180° . m∠DBC = 105° BC = CA = 4x + 4 8x + 3 = 7x + 8 = 4(5) + 4 x = 5 = 24 AB Feedback A B C D First, find the value of x. Then, use substitution to find AB. Correct! This is equals BC and CA. Now find AB. This is the value of x. Now find AB. PTS: OBJ: TOP: 28. ANS: 1 DIF: Average REF: Page 217 4-1.3 Using Triangle Classification NAT: 12.3.3.f 4-1 Classifying Triangles C Let the acute angles be m m m m ∠M 34.6° + ∠N ∠N + ∠N STA: GE12.0 and , with m . The acute angles of a right triangle are complementary. Substitute for m . Subtract from both sides. ∠M ∠N = 90° = 90° ∠M = 34.6° ∠M 34.6° = 55.4° 34.6° Feedback A B C D The two acute angles in a right triangle are complementary. This is the measure of the given angle. Find the measure of the other acute angle. Correct! The measure of the other acute angle is less than 90 degrees. PTS: OBJ: STA: 29. ANS: 1 DIF: Basic REF: Page 225 4-2.2 Finding Angle Measures in Right Triangles NAT: 12.3.3.f GE12.0 TOP: 4-2 Angle Relationships in Triangles D 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 ∠A ≅ ∠F and . Therefore, ∠B ≅ ∠E . So, m ∠CDE ≅ ∠DCB . ∠DCB = 46° Feedback A B C D This is the supplement. Use the Third Angles Theorem. 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. This is the complement. Use the Third Angles Theorem. Correct! PTS: 1 DIF: Advanced NAT: 12.3.3.f TOP: 4-2 Angle Relationships in Triangles 30. ANS: A STA: GE12.0 Proof: Statements Reasons 1. P is the midpoint of 2. , 3. 4. TQ 1. Given 2. Definition of midpoint 3. Vertical Angles Theorem 4. SAS and . RS RP ≅ SP TP ≅ QP ∠TPR ≅ ∠QPS ∆TPR ≅ ∆QPS Feedback A B C D Correct! There is not enough information to show that segment RT is congruent to segment SQ. Angle PRT and angle PSQ are not vertical angles. Use the correct postulate to prove the triangles congruent. PTS: OBJ: TOP: 31. ANS: 1 DIF: Average REF: Page 244 4-4.4 Proving Triangles Congruent NAT: 12.3.5.a 4-4 Triangle Congruence: SSS and SAS B STA: GE5.0 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. BC ≅ DC AC ≅ AC 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 by the SAS Postulate. ∠ACB ≅ ∠ACD AC and BC ∠ACB AC and DC ∠ACD ∆ABC ≅ ∆ADC Feedback A B C D This information is needed to use the SSS Postulate. Correct! 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 DIF: Advanced NAT: 12.3.5.a TOP: 4-4 Triangle Congruence: SSS and SAS 32. ANS: A The triangles can be proved congruent by the SAS Postulate. By CPCTC, . 3x − 5 = 2x + 1 Solve the equation for x. 3x − 5 = 2x + 1 3x = 2x + 6 x=6 Feedback A B Correct! When solving, you can either add 5 or subtract 1 from each side. STA: GE5.0 C D Remember to combine the like terms when solving. These two triangles have SAS congruence, so the two expressions are equal by CPCTC. PTS: 1 DIF: Advanced TOP: 4-6 Triangle Congruence: CPCTC 33. ANS: B NAT: 12.3.2.e STA: GE5.0 These graphs show right triangles with leg lengths of 4 units and 5 units with one vertex at the origin. This graph shows a right triangle with leg lengths of 4 units and 5 units with the base centered at the origin. This graph shows an isosceles triangle that is not a right triangle. Feedback A B C D This is a right triangle with a vertex at the origin. Correct! This is a right triangle with a vertex at the origin. This is a right triangle with a vertex at the origin. PTS: OBJ: STA: 34. ANS: 1 DIF: Basic REF: Page 267 4-7.1 Positioning a Figure in the Coordinate Plane NAT: 12.3.4.d GE17.0 TOP: 4-7 Introduction to Coordinate Proof A Step 1 Plot . AB The perpendicular bisector of AB is perpendicular to AB at its midpoint. Step 2 Find the midpoint of . Midpoint of AB Ê −2 + 5 2 + 4 ˆ˜ Á ˜ AB = Á Á Á 2 , 2 ˜˜ = (1.5, 3) Ë ¯ Step 3 Find the slope of the perpendicular bisector. 2/7 Slope of AB = (4) − (2) = (5) − (−2) Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is -7/2. Step 4 Use point-slope form to write the equation. y − y 1 = m(x − x 1) y−3 = -7/2 (x − 1.5) Feedback A B C Correct! The perpendicular bisector has a slope that is perpendicular to the given segment. To find the midpoint, add the x-coordinates and divide by 2, and add the y-coordinates and divide by 2. The perpendicular bisector has a slope that is perpendicular to the given segment. To find the midpoint, add the x-coordinates and divide by 2, and add the y-coordinates and divide by 2. D PTS: OBJ: NAT: 35. ANS: BO = 2 3 BY BO = 2 3 (26.4) = 17.6 1 DIF: Average REF: Page 303 5-1.4 Writing Equations of Bisectors in the Coordinate Plane 12.5.2.c STA: 1A8.0 TOP: 5-1 Perpendicular and Angle Bisectors A Centroid Theorem Substitute 3.3 for and simplify. BY Feedback A B C Correct! Segment BO is two-thirds the length of segment BY. Segment BO is two-thirds the length of segment BY. D Segment BO is two-thirds the length of segment BY. PTS: OBJ: STA: 36. ANS: 1 DIF: Average REF: Page 315 5-3.1 Using the Centroid to Find Segment Lengths NAT: 12.3.3.f GE1.0 TOP: 5-3 Medians and Altitudes of Triangles B The length of a midsegment is half the length of the corresponding base. XY = 1 2 AB = 1 2 (3) = 1.5 Feedback A B C D Find the base corresponding to segment XY. Correct! Find the base corresponding to segment XY. The length of the midsegment is half the length of the corresponding base. PTS: OBJ: NAT: 37. ANS: 1 DIF: Average REF: Page 323 5-4.2 Developing and Using the Triangle Midsegment Theorem 12.3.3.f STA: GE1.0 TOP: 5-4 The Triangle Midsegment Theorem B From the diagram, .m AD ≅ BD ∠ADC >m is given, and ∠BDC CD ≅ CD by the Reflexive Property. By the Hinge Theorem, 3x + 32 > 7x + 16 16 > 4x 4>x Since , which means . So the range of values for x is BC > 0 7x + 16 > 0 x > − 167 . − 167 < x < 4 Feedback A B C D By the Hinge Theorem, AC > BC. Set up an inequality and solve for x. Correct! By the Hinge Theorem, AC > BC. Set up an inequality and solve for x. By the Hinge Theorem, AC > BC. Set up an inequality and solve for x. PTS: 1 38. ANS: A DIF: Advanced TOP: 5-6 Inequalities in Two Triangles Step 1 Determine if the sides form a triangle. By the Triangle Inequality Theorem, 6, 14, and 13 can be the sides of a triangle. Step 2 Classify the triangle. c2 ? a2 + b2 Compare c2 to a2 + b2. Substitute the longest side length for c. Multiply. Add and compare. 142 ? 62 + 132 196 ? 36 + 169 196 < 205 Since 142 < 62 + 132 , the triangle is acute. Feedback A Correct! . AC > BC B C D Compare the square of the longest side to the squares of the other two sides to determine the type of triangle. Compare the square of the longest side to the squares of the other two sides to determine the type of triangle. The largest side length is greater than the sum of the other two sides, so this can be the side lengths of a triangle. PTS: OBJ: TOP: 39. ANS: 1 DIF: Average 5-7.4 Classifying Triangles 5-7 The Pythagorean Theorem B REF: Page 351 NAT: 12.3.3.d STA: GE6.0 If the shape were regular all the sides would be congruent and all the angles would be congruent. Since neither the sides nor the angles are congruent, the shape is irregular. Because some of the diagonals of this shape contain points in the exterior of the polygon, the shape is concave. Feedback A B C D In regular shapes, all sides and all angles are congruent. Correct! In regular shapes, all sides and all angles are congruent. In convex shapes, no diagonals contain points in the exterior. PTS: OBJ: TOP: 40. ANS: 1 DIF: Basic REF: Page 383 6-1.2 Classifying Polygons NAT: 12.3.3.f 6-1 Properties and Attributes of Polygons C STA: GE12.0 Step 1 Find the sum of the interior angle measures. (n – 2)180° Polygon Angle Sum Theorem = (45 – 2)180° A 45-gon has 45 sides, so substitute 45 for n. = 7740 Simplify. Step 2 Find the measure of one interior angle. = 172 The interior angles are , so divide by 45. ≅ 7740 45 Feedback A B C D Subtract 2, not 1, from the number of sides. According to the Polygon Angle Sum Theorem, the sum of the interior angle measures is the product of 180 and the number of sides minus 2. Correct! Subtract, not add, 2 from the number of sides. PTS: OBJ: NAT: 41. ANS: 1 DIF: Average REF: Page 384 6-1.3 Finding Interior Angle Measures and Sums in Polygons 12.3.3.f STA: GE12.0 TOP: 6-1 Properties and Attributes of Polygons D A decagon has 10 sides and 10 vertices. sum of exterior angle measures = 360° Polygon Exterior Angle Sum Theorem measure of one exterior angle = ° 360 = 36 10 A regular decagon has 10 congruent exterior angles, so divide the sum by 10. The measure of each exterior angle of a regular decagon is 36°. Feedback A B C D Divide by the number of sides the polygon has. Divide 360 by the number of sides the polygon has. Divide 360 by the number of sides. Correct! PTS: OBJ: STA: 42. ANS: 1 DIF: Average REF: Page 384 6-1.4 Finding Exterior Angle Measures in Polygons NAT: 12.3.3.f GE12.0 TOP: 6-1 Properties and Attributes of Polygons A One set of opposite sides are congruent and one set of opposite angles are congruent. This is insufficient information to prove that the quadrilateral is a parallelogram. Feedback A B C D Correct! Both sets of opposite angles must be congruent in order for the quadrilateral to be a parallelogram. Both sets of opposite sides must be congruent in order for the quadrilateral to be a parallelogram. If one set of opposite sides are congruent and parallel, then the quadrilateral is a parallelogram. PTS: 1 DIF: Average REF: Page 400 OBJ: 6-3.2 Applying Conditions for Parallelograms STA: GE7.0 TOP: 6-3 Conditions for Parallelograms NAT: 12.3.3.f NUMERIC RESPONSE 43. ANS: 150 PTS: TOP: KEY: 44. ANS: 1 DIF: Average NAT: 12.2.1.f 1-3 Measuring and Constructing Angles application | angle measures 74 PTS: 1 DIF: TOP: 1-4 Pairs of Angles 45. ANS: 20 Average NAT: 12.2.1.f STA: 6MG2.2 KEY: supplementary angles | complementary angles PTS: 1 DIF: Advanced NAT: 12.2.1.f TOP: 1-7 Transformations in the Coordinate Plane KEY: transformations | application | rotations 46. ANS: 65 PTS: 1 47. ANS: 5.5 DIF: Average PTS: 1 DIF: Advanced TOP: 4-1 Classifying Triangles 48. ANS: 21.6 NAT: 12.2.1.f TOP: 2-5 Algebraic Proof NAT: 12.3.2.e STA: GE8.0 KEY: perimeter | isosceles PTS: 1 DIF: Average NAT: 12.2.1.f TOP: 4-8 Isosceles and Equilateral Triangles 49. ANS: 11 STA: GE12.0 KEY: equilateral | equiangular PTS: 1 DIF: Average NAT: 12.3.3.f TOP: 5-4 The Triangle Midsegment Theorem 50. ANS: 49.8 PTS: 1 DIF: Average TOP: 6-2 Properties of Parallelograms 51. ANS: 0.5 NAT: 12.2.1.h STA: GE7.0 PTS: 1 DIF: Advanced NAT: 12.3.3.f TOP: 6-4 Properties of Special Parallelograms STA: GE7.0 MATCHING 52. ANS: TOP: 53. ANS: TOP: 54. ANS: TOP: 55. ANS: TOP: 56. ANS: TOP: 57. ANS: TOP: G PTS: 1 DIF: 1-3 Measuring and Constructing Angles B PTS: 1 DIF: 1-2 Measuring and Constructing Segments A PTS: 1 DIF: 1-2 Measuring and Constructing Segments D PTS: 1 DIF: 1-3 Measuring and Constructing Angles F PTS: 1 DIF: 1-3 Measuring and Constructing Angles H PTS: 1 DIF: 1-1 Understanding Points Lines and Planes Basic REF: Page 20 Basic REF: Page 15 Basic REF: Page 15 Basic REF: Page 20 Basic REF: Page 20 Basic REF: Page 16 58. ANS: TOP: 59. ANS: TOP: 60. ANS: TOP: F PTS: 1 2-2 Conditional Statements C PTS: 1 2-2 Conditional Statements D PTS: 1 2-2 Conditional Statements DIF: Basic REF: Page 82 DIF: Basic REF: Page 83 DIF: Basic REF: Page 82 61. ANS: TOP: 62. ANS: TOP: B PTS: 1 2-2 Conditional Statements G PTS: 1 2-2 Conditional Statements DIF: Basic REF: Page 83 DIF: Basic REF: Page 83