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Chapter 4 Geometry Homework Answers
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Sec 4.1 73 60 110 24 3 x 360 – 180 = 900 3 x 180 – 180 = 360 a = 69, b = 47, c = 116, d = 93, e = 86 m = 30, n = 50, p = 82, q = 28, r = 32, s = 78, t = 118, u = 50 14-16) See description. False. See diagram. True. Answers in degrees unless labeled otherwise.
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Sec 4.2 79 Parallel 54 107.5 Neither 44; 35 cm
New: (6, -3), (2, -5), (3, 0). Triangles are congruent. 76; 3.5 cm 72; 10 cm 22) New: (3, -3), (-3, -1), (-1, -5). Triangles are congruent. a = 124, b = 56, c = 56, d = 38, e = 38, f = 76, g = 66, h = 104, k = 76, n = 86, p = 38 a = 36, b = 36, c = 72, d = 108, e = 36; none NCA Answers in degrees unless labeled otherwise. IEC Perpendicular
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Using Your Algebra Skills 4: Writing Linear Equations
1-3) See graphs. y = -x + 2 y = (-6/13)x + (74/13) y = x+1 y = -3x +5 y = (2/5)x – (8/5) y = x y = -3x + 26 y = (-1/4)x – 3 y = (6/5)x y = x + 1 y = (-2/9)x + (43/9)
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Sec 4.3 Yes 135 No. See diagram. 72 See description. a, b, c
a = 52, b = 38, c = 110, d = 35 c, b, a a = 90, b = 68, c = 112, d = 112, e = 68, f = 56, g = 124, h = 124 b, a, c a, c, b ABE FNK v, z, y, w, x cannot be determined 6 < length < 102 By the Triangle Inequality Conjecture, the sum of 11 cm and 25 cm should be greater than 48 cm. Answers in degrees unless labeled otherwise. By the Triangle Sum Conjecture, the third angle must measure 36° in the small triangle, but it measures 32° in the large triangle. These are the same angle, so they can’t have different measures.
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Sec 4.4 Cannot be determined. Parts do not correspond. SAS SSS
SAO by SAS AIN by SSS or SAS RAY by SAS SSS (and the converse of the Isosceles Triangle Conjecture) The midpoint of SD and PR is (0, 0). Therefore, ΔDRO ≈ ΔSPO by SAS. b = 55, but > 180, which is impossible by the Triangle Sum Conjecture. a = 37, b = 143, c = 37, d = 58, e = 37, f = 53, g = 48, h = 84, k = 96, m = 26, p = 69, r = 111, s = 69 FLE by SSS Cannot be determined. SSA is not a congruence conjecture. Answers in degrees unless labeled otherwise.
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Sec 4.5 ASA MRA by SAS Cannot be determined. AAA does not guarantee congruence. SAA WKL by ASA Cannot be determined. Yes, three exterior triangles are congruent by SAS. FED by SSS WTA by ASA or SAA SAT by SAS PRN by ASA or SAS; SRE by ASA Cannot be determined. Parts do not correspond.
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Sec 4.6 Yes; AAS a = 112, b = 68, c = 44, d = 44, e = 136, f = 68, g = 68, h = 56, k = 68, l = 56, m = 124 Yes; ASA Cannot be determined. x = 3, y = 10 Yes; SSS Yes; SAS Answers to # 18 are in degrees. Yes. The triangles are congruent by SAS. Yes. The triangles are congruent by SAS, and the angles are congruent by CPCTC. See description. KEI by ASA UTE by SAS
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Sec 4.7 The angle bisector does not go to the midpoint of the opposite side in any triangle, only in an isosceles triangle. The triangles are congruent by SSS, so the two central angles cannot have different measures. PRN by ASA, SRE by ASA 1. Given, 2. Given, 3. Vertical Angles Conjecture, 4. Δ ESM ≈ Δ USO, 5. CPCTC Cannot be determined. Parts do not correspond. ACK by SSS 2. Given, 4. Definition of midpoint, 5. Vertical Angles Conjecture, 6. Δ CIL ≈ Δ MIB by SAS, CL ≈ MB a = 72, b = 36, c = 144, d = 36, e = 144, f = 18, g = 162, h = 144, j = 36, k = 54, m = 126 5. Given, 6. Δ ESN by SSS, 7. ‹ E by CPCTC 2. ‹ 2 is an angle bisector, 3. Given, 4. Same segment, 5. ΔWNS ≈ ΔENS by SAA, 6. CPCTC, Definition of isosceles triangle Answers to # 13 are in degrees. 1. Given, 2. Given, 4. ‹ 1 ≈ ‹ 2 by AIA Conjecture, 6. Δ ESN ≈ Δ ANS by ASA, 7. SA ≈ NE by CPCTC This proof shows that in a parallelogram, opposite sides are congruent. NE, because it is across from the smallest angle in Δ NAE. It is shorter than AE, which is across from the smallest angle in Δ LAE.
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Sec 4.8 a = 128, b = 128, c = 52, d = 76, e = 104, f = 104, g = 76, h = 52, j = 70, k = 70, l = 40, m = 110, n = 58 6 90° 45° 2. ‹ 1 ≈ ‹ 2, 5. SAS 4. CPCTC, 7. Definition of perpendicular, 8. CD is an altitude Answers to # 11 are in degrees. See diagram.
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Ch 4 Review Their rigidity gives strength
The Triangle Sum Conjecture states that the sum of the measures of the angles in every triangle is 180°. Possible answers: It applies to all triangles; many other conjectures rely on it. The angle bisector of the vertex angle is also the median and the altitude. The distance between A and B is along the segment connecting them. The distance from A to C to B can’t be shorter than the distance from A to B. Therefore AC + CB > AB. Points A, B, and C form a triangle. Therefore the sum of the lengths of any two sides is greater than the length of the third side. SSS, SAS, ASA, or SAA In some cases two different triangles can be constructed using the same two sides and non-included angle. Cannot be determined ΔTOP ≅ ΔZAP by SAA ΔMSE ≅ ΔOSU by SSS ΔTRP ≅ ΔAPR by SAS ΔCGH ≅ ΔNGI by SAS ΔABE ≅ ΔDCE by SAA or ASA ΔACN ≅ ΔRBO or OBR by SAS ΔAMD ≅ ΔUMT by SAS, AD ≅ UT by CPCTC
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Ch 4 Review cont. Cannot be determined
ΔTRI ≅ ΔALS by SAA, TR ≅ AL by CPCTC ΔSVE ≅ ΔNIK by SSS, EI ≅ KV by overlapping segments property Cannot be Determined ΔLAZ ≅ ΔIAR by ASA, ΔLRI ≅ ΔIZL by ASA, ΔLRD ≅ ΔIZD by ASA ΔPTS ≅ ΔTPO by ASA or SAA ΔANG is isosceles, so ∠𝐴 ≅ ∠𝐺. However, the sum of m∠𝐴 + m∠𝑁 + m∠𝐺 = 188°. The measures of the three angles of a triangle must sum to 180°. ΔROW ≅ ΔNOG by ASA, implying that OW ≅ OG. However, the two segments shown are not equal in measure. a = g < e = d = b = f < c. Thus, c is the longest segment and a and g are the shortest. X = 20°
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