Download Lesson 7.1.1 - Solon City Schools

Lesson 7.1.1
7-6.
a: They are congruent by ASA ≅ or AAS ≅.
b: AC ≈ 9.4 units and DF = 20 units
7-7.
Relationships used will vary, but may include alternate interior angles, Triangle Angle
Sum Theorem, etc.; a = 26°, b = 65°, c = 26°, d = 117°
7-8.
width = 60 mm, area = 660 mm2
7-9.
A quadrilateral.
7-10. a: x = 18, y = 9 3
b: x = 24 2 , y = 24
c: x =
8
3
=
7-11. a: (6, –13)
Selected Answers
8 3
3
, y=
16
3
= 163 3
b: Not possible, these curves do not intersect.
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Lesson 7.1.2
7-15. Using the Pythagorean Theorem, AB = 8 and JH = 5. Then, since
ΔABC ~ ΔHGJ because of SSS ~.
3
6
=
4
8
5 ,
= 10
7-16. 7 cm
7-17. Line L: y = ! 16 x + 6 ; line M: y =
2
3
x + 1 ; point of intersection: (6, 5)
7-18. a: 3m = 5m ! 28 , m = 14°
b: 3x + 38° + 7x ! 8° = 180° , x = 15°
c: 2(n + 4) = 3n ! 1 , n = 9 units
d: 2(3x + 12) = 11x ! 1 , x = 5 units
7-19. Rotating about the midpoint of a base forms a hexagon (one convex and one nonconvex). Rotating the trapezoid about the midpoint of either of the non-base sides forms
a parallelogram.
7-20. a: 10 units
b: (–1, 4)
c: 5 units, it must be half of AB because C is the midpoint of AB .
Lesson 7.1.3 (Day 1)
7-28. a: The 90° angle is reflected, so m∠XYZ = 90º. Then m∠YZY′ = 180º.
b: They must be congruent because rigid transformations (such as reflection) do not alter
shape or size of an object.
c: XY ! XY " , XZ ! XZ , YZ ! Y "Z , ∠Y ≅ ∠Y′, ∠YXZ ≅ ∠Y′XZ, and ∠YZX ≅ ∠Y′ZX
7-29. M(0, 7)
7-30.
c 2 and a 2 + b 2
7-31. a: The triangles are similar by SSS ~.
b: The triangles are similar by AA ~.
c: Not enough information is provided.
d: The triangles are congruent by AAS ≅ or ASA ≅.
7-32. a: It is a parallelogram; opposite sides are parallel.
c: AC :!y =
1
2
x + 12 , BD :!y = !x + 5 ; No
7-33. a: No solution, lines are parallel.
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b: 63.4 °; They are equal.
d: (3, 2)
b: (0, 3) and (4, 11)
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Lesson 7.1.3 (Day 2)
7-34. Side length =
50 !·! 2 = 100 = 10 units.
50 units; diagonal is
7-35. a: It is a rhombus. It has four sides of length 5 units.
b: HJ : y = !2x + 8 and GI : y =
1
2
x+3
c: They are perpendicular.
d: (6, –1)
e: 20 square units
7-36. a: P(scalene) =
1
4
b: P(isosceles) =
2
4
=
1
2
c: P(side of the triangle is 6 cm) = 24 =
1
2
7-37. a: 6n ! 3° = n + 17° , n = 4°
b: 7x ! 19° + 3x + 14° = 180° , so x = 18.5°. Then 5y ! 2° = 7(18.5) ! 19° , so y = 22.5°
c: 5w + 36° + 3w = 180° , w = 18°
d: k 2 = 15 2 + 25 2 ! 2(15)(25) cos120° , k = 35
7-38. The graph is a parabola with roots (–3, 0) and (1, 0), and y-intercept at (0, –3).
7-39. ≈ 35.24 m
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Lesson 7.1.4
7-43. a: 360° ÷ 36° = 10 sides
b: Regular decagon
7-44. If the diagonals intersect at E, then BE = 12 mm, since the diagonals are perpendicular
bisectors. Then ΔABE is a right triangle and AE = 15 2 ! 12 2 = 9 mm.
Thus, AC = 18 mm.
7-45. Yes, she is correct. Show that the lengths on both sides of the midpoint are equal and that
(2, 4) lies on the line that connects (–3, 5) and (7, 3) .
7-46. a: See flowchart at right.
b: Not similar because corresponding sides do
not have the same ratio.
c: See below.
ΔFED ~ ΔBUG
AA ~
SSS ~
7-47. (a) and (c) are correct because if the triangles are congruent, then corresponding parts are
congruent. Since alternate interior angles are congruent, then AB // DE .
7-48. AB =
Selected Answers
40 ! 6.32 , BC =
34 ! 5.83 , therefore C is closer to B.
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Lesson 7.2.1
7-54. a: x = 8.5°
b: x = 11
7-55. a: 360° ÷ 72° = 5 sides
c: x = 14°
b: 360° ÷ 9 = 40°
7-56. ≈ 36.4 feet from the point on the street closest to the art museum.
7-57. a: an = 20 + 20n = 40 + 20(n ! 1)
b: an = 6( 12 )n = 3( 12 )n!1
7-58. a: (0.7)(0.7) = 0.49 = 49%
b: (0.3)(0.7) = 0.21 = 21%
7-59. a: Similar (SSS ~)
b: Congruent (ASA ≅ or AAS ≅)
c: Similar, because if the Pythagorean Theorem is used to solve for each unknown side,
then 3 pairs of corresponding sides have a common ratio; thus, the triangles are similar
(SSS ~).
d: Similar (AA ~) but not congruent since the two sides of length 12 are not
corresponding.
7-60. Possible response: Rotate the second triangle 180° and then translate it to match the sides
with the first triangle.
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5
Lesson 7.2.2
7-65.
4x ! 1 = x + 8 , x = 3 ; 5y + 2 = 22 , y = 4
7-66. a: 0.8
b: 1200(0.8)3 = $614.40
c: 1200(0.8)!2 = $1875
7-67. a: !a = 36°, !r = 54°, !m = 54°, !y = 72°, !z = 108°
b: Possible response: !y and !z are supplementary interior same side angles.
7-68. a: It is a parallelogram, because MN // PQ and NP // MQ .
b: (1, –5)
E is a midpoint
7-69.
Given
∠AEB ≅ ∠CED
≅
≅
Vertical angles
are congruent.
Definition of
midpoint.
Definition of
midpoint.
ΔAEB ≅ ΔCED
SAS ≅
7-70. a: 50%; The sum must be 100%.
b: central angle for red = 0.4(360°) = 144°, white = 0.1(360°) = 36°,
blue = 0.5(360°) = 180°
c: Yes; there could be more than three sections to the spinner, but the ratio of the areas for
each color must match the ratios for the spinner in part (b).
7-71.
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Lesson 7.2.3
7-75. a: Congruent (SSS ≅)
b: Not enough information
c: Congruent (ASA ≅)
d: Congruent (HL ≅)
7-76. See answers in problem 7-75.
7-77. a: 83°
b: 92°
7-78. a: Yes; HL ≅
c: tan 18° =
4
AD
b: 18°, 4
, AD ≈ 12.3 units
d: ≈ 49.2 square units
7-79. a: Parallelogram because the opposite sides are parallel.
!##"
!##"
b: AC : y = 43 x ; BD : y = ! 23 x + 9
7-80. a:
68 ≈ 8.2, since
64 = 8, then
68 must be a little greater.
b: (1) 2.2, (2) 9.2, (3) 7.1, (4) 4.7
7-81. a: 2x + 52° = 180° , 64°
c:
sin 77°
x
=
sin 72° ,
8
b: 4x ! 3° + 3x + 1° = 180° , 26°
d: 5x + 6° = 2x + 21° , x = 5°
x ≈ 8.2
Lesson 7.2.4
7-83.
36 3 ! 62.4 square units
7-84. No. Using the Pythagorean Theorem and the Law of Cosines, the perimeter of the
triangle is ≈ 26.3 feet.
7-85. a: x ≈ 10.73 cm, tangent
c: x ≈ 15.3', Law of Cosines
b: x ≈ 7.86 mi, Law of Sines
7-86. a: Congruent (SAS ≅) and x = 2
b: Congruent (HL ≅) and x = 32
7-87.
A = 24 square units
7-88. a:
b:
4
1
20 = 5
4
5 ; Since
the sum of the probabilities of finding the ring and not finding the ring is 1,
you can subtract 1! 15 = 45 .
c: No, his probability is still
sandbox is unchanged.
7-89.
4
20
=
1
5
because the ratio of the shaded region to the whole
a50 = !130
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7
Lesson 7.2.5
7-96. a: See diagram at right.
A
b: Since corresponding parts of congruent triangles are
congruent, 2y + 7 = 21 and y = 7.
7-97. y-intercept: (0, 6), x-intercept: (4, 0)
18
14
T
21
P
7-98. m∠a = 132º, m∠b = 108º, and m∠c = 120º, m∠a + m∠b + m∠c = 360º
7-99. 15%
7-100. AB ! CD and AB // CD (given), so ∠BAC ≅ ∠DCA (alt. int. angles). AC ! CA
(Reflexive Property) so ΔABC ≅ ΔCDA (SAS ≅). ∠BCA ≅ ∠DAC ( !! s!"!!!parts ).
Thus, BC // AD (if alt. int. angles are congruent, then the lines cut by the transversal are
congruent).
!
7-101. Because alternate interior angles are congruent, the angle of depression equals the angle
formed by the line of sight and the ground. Then tan!! = 52
38 , ! " 53.85° .
7-102. a: ΔADC; AAS ≅ or ASA ≅
b: ΔSQR; HL ≅
c: No solution, only angles are congruent.
d: ΔTZY; SAS ≅ and vertical angles
e: ΔGFE; alternate interior angles equal and ASA ≅
f: ΔDEF; SSS ≅
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Lesson 7.2.5
7-108. ZY ! WX , ∠YZX ≅ ∠WXZ, ∠ZYW ≅ ∠YWX (alt. int. ! ’s), ΔXWM ≅ ΔZYM, ASA ! ,
YM ! WM and XM ! ZM , ! " # ! parts .
7-109. Typical responses: right angles, congruent diagonals, 2 pairs of opposite sides that are
congruent, all sides congruent, congruent adjacent sides, diagonals that bisect each other,
congruent angles, etc.
7-110. a: The triangles should be ≅ by SSS ≅ but 80º ≠ 50º.
b: The triangles should be ≅ by SAS ≅ but 80º ≠ 90º and 40º ≠ 50º.
c: The triangles should be ≅ by SAS ≅ but 10 ≠ 12.
d: Triangle is isosceles but the base angles are not equal.
e: The large triangle is isosceles but base angles are not equal.
f: The triangles should be ≅ by SAS ≅ but sides 13 ≠ 14.
7-111. The triangles described in (a), (b), and (d) are isosceles.
7-112. a: 12
b: 15
c: 15.5
7-113. See reasons in bold below.
Statements
Reasons
1. AD // EH and BF // CG
Given
2. a = b
If lines are parallel, alternate interior
angle measures are equal.
3. b = c
If lines are parallel, corresponding
angle measures are equal.
4. a = c
Substitution
5. c = d
Vertical angle measures are equal.
6. a = d
Substitution
7-114. This problem is similar to the Interior Design problem (7-21).
Her sink should be located 3 23 feet from the right front edge of the counter. This will
make the perimeter ≈ 25.6 feet, which will meet industry standards.
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Lesson 7.3.1
7-119. a: (4.5, 3)
b: (–3, 1.5)
c: (1.5, –2)
7-120. a: ΔSHR ~ ΔSAK because ΔSHR can be dilated by a factor of 2.
b: 2HR = AK, 2SH = SA, SH = HA
c: 6 units
7-121. a: 4
b: 1
c: –8
7-122. a: ΔCED; vertical angles are equal , ASA ≅
b: ΔEFG; SAS ≅
c: ΔHJK; HI + IJ = LK + KJ, ∠J ≅ ∠J, SAS ≅.
d: Not ≅, all corresponding pairs of angles equal is not sufficient.
7-123. See answers in problem 7-122.
7-124. No. Her conclusion in Statement #3 depends on Statement #4, and thus must follow it.
7-125. a: Must be a quadrilateral with all four sides of equal length.
b: Must be a quadrilateral with one pair of opposite sides that are parallel.
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Lesson 7.3.2
7-131. a: Yes, each side has the same length ( 29 units). See graph at right.
!##"
!##"
b: BD is y = x ; AC is y = 5 ! x
c: The slopes are 1 and –1. Therefore the diagonals are perpendicular.
7-132. Multiple answers are possible. Any order is valid as long as Statement #1 is first,
Statement #6 is last, and Statement #4 follows both Statements #2 and #3. Statements #2,
#3, and #5 are independent of each other and can be in any order as long as #2 and #3
follow Statement #1.
7-133. a: 6
b: 3
c: – 6.5
7-134. a: Yes, by SAS ~.
b: ∠FGH ≅ ∠FIJ, ∠FHG ≅ ∠FJI
c: Yes, because corresponding angles are congruent. (Triangle Midsegment Theorem)
d: 2(4x ! 3) = 3x + 14 , so x = 4 and GH = 4(4) ! 3 = 13 units.
7-135. a: It is a right triangle because the slopes of AB and AC ( 43 and !
opposite reciprocals.
4
3
, respectively) are
b: B! is at (–2, 0). It is an isosceles triangle because ∠B′AB must be a straight angle
(because it is composed of two adjacent right angles) and because BC ! B"C (because
reflections preserve length).
7-136. a: ≈ 23.83 feet
c: x ≈ 66.42
b: x ≈ 7 yds
d: ≈ 334.57 feet
7-137. a: Must be: trapezoid.
Could be: isosceles trapezoid, parallelogram, rhombus, rectangle, and square.
b: Must be: parallelogram. Could be: rhombus, rectangle, and square.
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Lesson 7.3.3
7-140. a: (8, 8)
b: (6.5, 6)
7-141. a: X and Y
b: Y and Z
7-142. a: Must be: none. Could be: right trapezoid, rectangle, square.
b: Must be: none. Could be: kite, rhombus, square.
7-143. a: If a polygon is a parallelogram, then its area equals its base times its height.
b: If the area of a polygon is the base times its height, then the polygon is a
parallelogram. This is always true.
c: “If a polygon is a triangle, then its area equals one half its base times its height.”
Arrow diagram: Polygon is a triangle  area of the polygon equals one-half base
times height.
d: If a polygon’s area is one-half its base times its height, then the polygon is a triangle.
This is always true.
7-144. a: 360° ÷ 18° = 20 sides
b: It can measure 90° (which forms a square). It cannot be 180° (because this polygon
would only have 2 sides) or 13° (because 13 does not divide evenly into 360°).
7-145. The expected value is 14 ($3) + 43 ($1) = $1.50 per spin, so each player should pay $1.50
so that there is no net gain or loss over many games.
7-146. a: They are perpendicular because their slopes are opposite reciprocals.
b: It could be a kite, rhombus, or square because the diagonals are perpendicular and at
least one diagonal is bisected.
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