1. Chapter 4 Geometry
Homework Answers

2. 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.

3. 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

4. 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)

5. 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.

6. 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.

7. 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.

8. 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

9. 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.

10. 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.

11. 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

12. 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°