86. a: 110 b: 70 c: 48 d: a: no b: yes c: no d: yes e: no f: yes g: yes h: no


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1 Lesson a: 110 b: 70 c: 48 d: a: no b: yes c: no d: yes e: no f: yes g: yes h: no 88. b: The measure of an exterior angle of a triangle equals the sum of the measures of its remote interior angles. c: a + b + c = 180º (the sum of the interior angles of a triangle is 180 ), x + c = 180º (straight angle); therefore, a + b + c = x + c (substitution) and a + b = x (subtracting c from both sides) x = 72 and y = = a: congruent (SAS ), x = 79 b: Cannot be determined. c: congruent (AAS ), x 5.9 units d: congruent (SAS ), x a: True b: False (counterexample is a quadrilateral without parallel sides.) c: True d: True e: False (counterexample is a parallelogram that is not a rhombus) Selected Answers 1
2 Lesson a: Isosceles right triangle, because AC = BC and AC BC. b: 45 ; methods vary a = 87, b = 83, c = 96, d = 94 ; A = 40.5 square miles, P 27.7 miles ( 5, 1), ( 3, 7), and ( 6, 2) a: a n = n = (n 1) b: a n = 27( 1 3 )n = 9( 1 3 )n See answers in bold below B Statements 1. BC // EF, AB // DE, and AF = DC 1. Given 2. m BCF = m EFC and m EDF = m CAB Reasons 2. If two lines cut by a transversal are parallel, then alternate interior angles are equal. 3. FC = FC 3. Reflexive Property 4. AF + FC + CD = FC 4. Additive Property of Equality (adding the same amount to both sides of an equation keeps the equation true) 5. AC = DF 5. Segment addition 6. ΔABC ΔDEF 6. ASA 7. BC EF 7. s parts Selected Answers 2
3 Lesson a: A = 36 sq. ft, P = 28 ft b: A = 600 sq. cm, P cm QP = RS and PR = SQ (given), QR = QR (Reflexive Property), so ΔPQR ΔSRQ (SSS ) and P S ( Δs parts) a: Isosceles triangles b: The central vertex must be = 36. The other two angles must be equal since the triangle is isosceles. Therefore, ( ) 2 = 72. c: = 145 square inches a: (6.5, 5) b: 3 8 c: Using the strategy developed in Lesson 7.3.2, Δx = 14 2 = 12 and Δy = 10 2 = 8. Then the xcoordinate is (12) = 6.5 and the ycoordinate is (8) = a: x + x + 82 = 180, x = 49 b: 2(71 ) + x = 180, x = a: The region can be rearranged into a rectangle with dimensions 14 and 7 units B b: (14)(7) = 98 square units Selected Answers 3
4 Lesson The reflections are all congruent triangles with equal area. Therefore, the total area is (6)(11.42) = square inches a: 1.04 b: f (t) = (1.04) t c: $199, a: Nonconvex b: Convex c: Convex d: Nonconvex a: 64 units 2 b: 27.0 units 2 c: units a: 3 b: 15 c: 4 d: All circles are similar, or use similarity transformations to justify the similarity a: A = 192 cm 2, P = 70 cm b: The length of each side is 5 times the corresponding side in the floor plan. A = 4,800 cm 2 and P = 350 cm. c: The ratio is 5 1 = 5; the ratio of the perimeters equals the zoom factor. d: The ratio of the areas is 25 1 = 25. The ratio of the areas equals the square of the zoom factor (5 2 ). Selected Answers 4
5 Lesson a: The interior and exterior angles must be supplementary = 160. b: Use = 18 sides or solve the equation 180º(n 2) n = 160 to find n = a: The perimeter of both triangles is units. b: 32.9 and Since the diagonals of a parallelogram bisect each other, they must intersect at the midpoint of BD. Thus, they intersect at (6, 21) A = mm meters a: w = ± 17 5 ± 1.84 b: w 2.17 and 1.57 c: No real solution E a: 60 b: 82 c: 14 d: a: Equilateral triangle b: Rectangle c: Nonagon d: Rhombus or kite The xcoordinate must lie on the perpendicular bisector of segment AB. Thus, since the midpoint M of segment AB is (6, 0), the xcoordinate of point C must be 6. ΔAMC is a right triangle and the hypotenuse must have a length of 12 units for ΔABC to be equilateral. Therefore, MC = = 6 3 because of the Pythagorean Theorem. So the ycoordinate of point C could be 6 3 or a: Yes; since BC = BC (Reflexive Property), AB DC (given), and ABC DCB (given), then ΔABC ΔDCB (SAS ). Therefore, AC = DB ( Δs parts). b: No; the relationships in the figure are true, as long as the two angles remain congruent. See the diagram for problem 830 for a similar diagram a: ( 2.5, 0) and (3, 0) B D b: The graph of y = (2x 2 x 15) would be the reflection of y = 2x 2 x 15 across the xaxis because each yvalue would have its sign changed. Selected Answers 5
6 Lesson a: A = 34 units 2 ; P 25.7 units b: A = 306 units 2 ; P 77 units c: ratio of the perimeters = 3; ratio of the areas = inches or 6.67 feet The area of the hexagon 23.4 ft 2. Adding the rectangles makes the total area 41.4 ft % ; k = 0, 6, 10, 12 are factorable a: Reasoning will vary. For example, it is most likely you will earn more extra credit if the class spins the spinner with the options of 5 and 10 points. b: Reasoning will vary, but now the first spinner is definitely more attractive x 2 = 2x 2 +17x 30, x = 2.5 or 6; yes, there are two possible answers B Selected Answers 6
7 Lesson a: 3 4 b: rp c: ar a: cm 2 b: cm a 150 = a: = 6, so two sides will collapse on the third side. b: Answers vary. One solution is 2, 5, and // Given in diagram A C Given in diagram ABD CDB If lines are //, then alt. int. angles are. Shared side (or Reflexive Property) ΔADB ΔCBD AAS Δs parts a: AAS, ΔABC ΔDCB D b: ASA, ΔABC ΔEDC Selected Answers 7
8 Lesson Area of the entire pentagon square units, so the shaded area 5 3 (172.05) square units a: x = 14 3, pattern b: x 5.78, Law of Sines. c: No solution, hypotenuse must be longest side. d: 24 units, triangle area formula BC DC and A E (given) and BCA DCE (vertical angles are ). So ΔABC ΔEDC (AAS ) and AB ED ( s parts ) a: x + x + 125º + 125º + 90º = 540º, x = 100 b: 6x + 18º = 2x + 12º, x = a: (1.5, 5) b: (3, 7) c: y = 4 3 x + 3 d: = 6.25 = 2.5 units B Lesson (100 25π) square units a: 8 b: 18 c: Could have the combinations (1, 33), (3, 11), (11, 3) or (33, 1) a: A = 42 square units, P 30.5 units b: A = square units, P units a: x = 26; if lines are parallel and cut by a transversal, then alt. int. angles are equal. b: x = 33, n = 59 ; If lines are parallel and cut by a transversal, then sameside exterior angles are supplementary a: 20 b: units The area of the hexagon is 24 3 square units, so the side length of the square is units D Selected Answers 8
9 Lesson a: C = 28 π units, A = 196 π units 2 b: C = 10 π units, A = 25 π units 2 c: d = 100 units, r = 50 units, area = 2500 π units 2 d: A = C 2 /4π units See bold answers in table below. Statements 1. AB DE and DE is a diameter of C. 1. Given Reasons 2. AFC and BFC are right angles. 2. Definition of Perpendicular 3. FC = FC 3. Reflexive Property 4. AC = BC 4. Definition of a Circle (radii must be equal) 5. ΔAFC ΔBFC 5. HL 6. AF FB 6. s parts a: (3) b: (1) c: (4) d: (2) a: 3 b: 4 c: 3 and 3 d: 2 and a = 97, b = 15, c = 68, d = D a: (55)(60) + 900π square feet b: π feet, = $ or approximately $2,388 c: Area is four times as large 24,509.6 square feet; perimeter is twice as long 597 units a: CD = 22, BC = 7, and ED = 6 ; the perimeter is = 48 units b: 54(4) = 216 cm a: square units b: 36 square units; more c: sq. units; its area is greater than both the square and the equilateral Δ. d: A circle a: 1 8 b: a: 2πr = 24; r = 12 π ; A = 144 π square units b: 2πr = 18π; r = 9; A = 81π square units E Selected Answers 9
86. a: 110 b: 70 c: 48 d: a: no b: yes c: no d: yes e: no f: yes g: yes h: no
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