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Geometry – AP Book 7, Part 2: Unit 5 AP Book G7-13 8. page 105 1. Teacher to check. 2. 4. J = 66° b) Teacher to check. A = E = B = D a) VW = EF C = F WX = FG ii) ∆ABC ∆EDF VX = EG iii) Answers may vary – teacher to check. JK = ST b) JL = SU a) A = D i) All the angles are equal. ii) Teacher to check – any order is correct. iii) Teacher to check – any order is correct. B = E C = F b) M = P N = Q O = R 6. A = D = K 12. C = F = M a) B = E C = F 13. b) R = W c) COPYRIGHT©2010JUMPMATH:NOTTOBECOPIED d) Measurements may vary slightly: RS = WY BC = 2.0 cm DE = 4.6 cm None of Tom’s are correct. There is more than one correct solution – teacher to check. a) Teacher to check. Teacher to check. AP Book G7-15 page 111 INVESTIGATION 1 A. a) Only one unique triangle is possible (though it may be rotated, reflected, etc.) b) Yes, there is only one way to arrange any combination of side lengths into a triangle. a) There is only one way to complete the triangle (all the triangles drawn will be congruent). b) Yes, if you know any two side lengths and the angle between those sides, there is only one way to draw the triangle. a) They are congruent. b) Yes, if you know one side length and the angles on either side of it, only one triangle is possible. a) It is impossible to make more than one unique triangle. Sample solution: ∆ABC ∆XV W AP Book G7-14 4 squares: B, C, D, E, F, G B. B F, C D, E G, A I, H K C. Yes D. No, for instance A and H have the same areas but are not congruent. E. No, if two shapes are the same size and shape (which is the definition of congruence), they must also have the same area. EF = 2.0 cm DF = 5.0 cm AB = KL = BC = LM HI = 4.6 cm AC = KM IJ = 2.0 cm E = P = G = O HJ = 5.0 cm F = N KL = 5.5 cm EF = PN = FG = NO LM = 3.9 cm EG = PO KM = 2.4 cm AnswerKeysforAPBook7.2 Teacher to check. 1. a) No, if their corresponding side lengths are different, then they won’t have the same shape. No, they are not congruent. ∆ABC has no right angles. b) B. 5 squares: A, H, I, K AB = 4.8 cm B = L d) A. AC = 3.4 cm A = K = C = M 15. ∆DEF ∆HIJ 2 BONUS Teacher to check. M = 118° PR = LM T = X RT = WX 3. L = 23° R = M S = Y ST = YX b) K = 39° c) 2 Area of ∆ADC: 6 m I = 91° INVESTIGATION a) Area of ∆ABC: 6 m H = 23° Q = L R = M 14. 2. F = 66° Q = K BC = EF AC = DF E = 91° page 110 PR = KM B = E = L 7. QR = LM a) C = 120° a) i) No, if they’re congruent, their corresponding side lengths would be equal, which means their perimeters must also be equal. B = 38° 11. Teacher to check. b) A = 22° 10. KL = TU 5. Measurements may vary slightly: D = 23° a) b) b) The angles of a triangle always add to 180° so, if any two corresponding angles are equal, the third has to be equal as well. ∆ABC ∆RST 9. 3. No, not necessarily. Kali forgot to check that the side length of the two shapes is equal (they might just be similar, not congruent). C. D. U‐23 Geometry – AP Book 7, Part 2: Unit 5 (continued) 1. a) Yes, the third angle can always be found by subtracting the first two from 180°. Three angles, along with one side length, will always form a unique triangle. 10. b) c) c) E INVESTIGATION 2 d) A A. c) Uppermost (middle) point e) B f) D BC = DE, AC = EF and B = D ASA No 3. ∆RST ∆ZXY D. No, neither one is a congruence rule. The equal angles must be between the equal sides – otherwise more than one triangle can be drawn according to the specs. B. SSS 4. Q = Y 5. Need to know if any corresponding sides are equal. Teacher to check. OA = OB a) No b) Yes a) 2 b) 1 c) 0 page 120 5. Teacher to check. 6. Teacher to check. 8. a) 13. a) Teacher to check. b) Teacher to check. c) Teacher to check. a) Yes, they are: all 6 angles will be 60° and, since one pair of corresponding sides is equal, all 6 sides will be the same length. Sample answer: b) Yes, ∆ABC ∆GHI since corresponding sides and angles are all equal. Teacher to check sketch. 15. Teacher to check – there are 2 possible locations for point D: B C A D2 2. b) E c) D d) A e) C, F a) SSS, A = D b) SAA, BC = EF or SAA, AB = DE 3. A C (SAS) 4. a) 5. ASA b) SAS c) SSS d) SSS a) Teacher to check. b) Teacher to check. c) i) ∆ADP ∆BDP by SSS ii) ∆ADP ∆BDP by SSS iii) IMPORTANT The sketch should include pre-drawn lines for AP and BP, as shown here: A B BD = CD AP Book G7-17 page 118 Teacher to check sketch. 1. Teacher to check. 2. Teacher to check. No, not necessarily. It’s possible that their side lengths are different (making them similar but not congruent). 3. If 2 of the 3 corresponding angles are equal, the third will also be equal since the sum in both cases must be 180°. 1. D1 b) BONUS In order to form a triangle, the sum of the two shorter sides must be greater than the length of the longest side. AP Book G7-18 Congruent triangles are also similar since their corresponding angles are equal. But similar triangles aren’t always congruent: their angles are equal but their corresponding side lengths may be different. Teacher to check. No, they aren’t always congruent – it depends where the equal sides are in relation to the given angle, e.g: c) 12. 14. 9. Teacher to check. Teacher to check. Sample answer: 8. b) 7. Possible answers: As above. Teacher to check. Teacher to check. A ≠ D, C ≠ F, BA ≠ ED, CA ≠ FD a) 4. a) 11. Teacher to check. b) E C Left point C. Teacher to check. a) b) Top point SAS 3. b) 1. 5. page 115 2. Teacher to check. 7. Circle: 7 No, since neither one has the equal angle between the two equal sides. 2. a) AP Book G7-16 b) ∆MNO ∆FGH 6. Check, from left to right: 2, 3, 6, 7 a) ∆ABC ∆DEF b) a) a) scalene right b) isosceles obtuse c) isosceles obtuse d) isosceles obtuse e) equilateral acute Answers will vary, re: those identified by side lengths. Teacher to check. 4. a) AC = BC b) AC = AB c) AB = AC = BC P ∆ADB ∆APB by SSS * radii of equal circles * radii of the same circle * radii of equal circles U‐24 AnswerKeysforAPBook7.2 COPYRIGHT©2010JUMPMATH:NOTTOBECOPIED b) Geometry – AP Book 7, Part 2: Unit 5 (continued) AP Book G7-19 9. Teacher to check. page 122 10. Question 5 shows that ∆ABD ∆CBD by SSS, so we know from this that ABD = CBD. 1. Teacher to check. 2. a) Teacher to check. b) Teacher to check. c) AD is also a median in i), but not in ii). 3. a) No b) No, only the median through point A is a perpendicular bisector. c) 4. In an isosceles triangle, a median to the unequal side (or, to say it slightly differently, through the vertex between the two equal sides) is a perpendicular bisector. 1. a) QR = PR, OP = OQ b) Teacher to check. Teacher to check. ∆ABD ∆CBD (SSS) c) A = C d) 6. ADB = CDB F. DE FC 3. Teacher to check. page 128 4. Teacher to check. 5. Teacher to check. The following sides will likely be marked – teacher to check. 6. a) QT = QS a) b) TU = SU c) ∆QTU ∆QSU d) SSS a) The angles opposite the equal sides are equal (isosceles). 7. 8. ABD = CBD ADB = CDB ∆DFC = ∆EFC so all their corresponding angles are equal. We also know from this that both equal 90°, since 180° ÷ 2 = 90°. AP Book G7-22 AC a) Yes Teacher to check. 90° b) b) All the angles are equal (equilateral). a) 180° ÷ 3 = 60° b) Teacher to check. c) CAD = 30° 1. a) b) COPYRIGHT©2010JUMPMATH:NOTTOBECOPIED * BM bisects ABC 7. 8. 2. Teacher to check. 3. a) b) c) Teacher to check. 6. 4 × 90° = 360° 2 × 180° = 360° c) 360° ÷ 60° = 6 a) 45° b) Teacher to check. c) Teacher to check. a) DBC = 120° b) DBC = 30° c) DBC = 30° d) DAB = 30° a) DBC = 90° b) DBC = 45° c) DBC = 135° a) Teacher to check. b) Teacher to check. AP Book G7-24 page 131 1. 2. No, the two angles already sum to 180° – this means that KFL would be 0°, which is impossible. Teacher to check. a) Teacher to check. b) AD = BD c) 90° d) 90° e) ∆ADC ∆BDC Teacher to check explanation (likely using ITT or SAS). AP Book G7-23 page 127 1. a) Teacher to check. page 129 b) Teacher to check. 1. Teacher to check. 2. QOR = 90° * ∆ABM ∆CBM A. Teacher to check. DC is an angle bisector. B. D = E Students can check by measuring to find that BC = CA. C. Yes (SAS) AnswerKeysforAPBook7.2 90° + 90° + KFL = 180° + KFL a) b) AP Book G7-21 INVESTIGATION Teacher to check for median from A to BC, which bisects CAB. Right angles occur at (and all around) points K and L. * FC is a shared side and we know, from Step 4, that DF = EF. f) AC = BC g) AE = BE h) No, any point E along CD will work since ∆AEB will always be isosceles. POR = 180° 3. BD is an extension of the median to the isosceles triangle’s unequal side, so we can use the Isosceles Triangle Theorem. * the result will be an isosceles triangle 8. Use Yen’s method, and then bisect the 60° angle to form two 30° angle. * BM is a perpendicular bisector of AC AC = AB 5. b) * BM is a median of ∆ABC * BM AC 4. 7. CDA = 90° 9. The following statements are true (the others should be crossed out as false): * BM is a line of symmetry for ∆ABC e) 2. ADB = CDB 5. E. AP Book G7-20 page 125 180° DCF + ECF = DCE which is a straight line and therefore equal to 180° Since ABD + CBD = ABC, BD bisects ABC. AB = BC, AD = DC ∆ABD ∆CBD, SSS D. a) Teacher to check. b) SSS i) No c) DBA and BDA are both equal to DBC j) Median, line of symmetry INVESTIGATION d) ∆ABE ∆CBE AC = BC Teacher to check explanations. A. Isosceles B. Teacher to check. U‐25 Geometry – AP Book 7, Part 2: Unit 5 (continued) C. No, since ∆ABC is isosceles, the angle bisector of ACB must also be the perpendicular bisector of AB. 3. BD = 3 cm Teacher to check explanation. Sample answer: COPYRIGHT©2010JUMPMATH:NOTTOBECOPIED If ∆ABC is isosceles and A = B = 60°, it follows that ∆ABC is actually an equilateral triangle. This means that AB = 6 cm. If D is the midpoint of AB, then BD = 3 cm. U‐26 AnswerKeysforAPBook7.2