Download CCG CH.7

Lesson 7.1.1
7-6.
7-7.
a: They are congruent by ASA ≅ or AAS ≅. b: AC ≈
9.4 units and DF = 20 units
Relationships used will vary, but may include alternate interior angles, Triangle
AngleSum Theorem, etc.; a = 26°, b = 65°, c = 26°, d = 117°
7-8.
width = 60 mm, area = 660 mm2
7-9.
A quadrilateral.
7-1
0.
a: x = 18, y = 9 3 b:
x = 24 2 , y = 24 c: x
=83=833, y =
16 3 = 16 3 3
7-1
1.
7-3
7.
7-3
8.
7-3
9.
7-7
5.
7-7
6.
7-7
7.
7-7
8.
a: (6, –13)
b: Not possible, these curves do not intersect.
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: k2 = 152 + 252 −
2(15)(25)cos120° , k = 35
The graph is a parabola with roots (–3, 0) and (1, 0), and y-intercept at (0, –3).
≈ 35.24 m
a: Congruent (SSS ≅) c:
Congruent (ASA ≅)
b: Not enough information d:
Congruent (HL ≅)
See answers in problem 7-75.
a: 83°
b: 92°
b: 18°, 4 d: ≈
a: Yes; HL ≅ c: tan18° = 4 AD , AD
units
≈ 12.3 units
triangle is ≈ 26.3 feet.
7-8
a: x ≈ 10.73 cm, tangent
5.
c: x ≈ 15.3', Law of Cosines
7-8
6.
7-8
7.
a: Congruent (SAS ≅) and x = 2
7-8
a
A = 24 square units
42
0
b: x ≈
49.2 square
7.86 mi, Law of Sines
b: Congruent (HL ≅) and x = 32
7-15.
3
Using the Pythagorean Theorem, AB = 8 and JH = 5. Then, since 6 = 8 = 1045,
ΔABC ~ ΔHGJ because of SSS ~.
7-16.
7 cm
2
1
7-17. Line L: y = − 6 x + 6; line M: y = 3 x +1; point of intersection: (6, 5) 7-18. a:
3m = 5m−
28 , m4)=
= 14°
b: n3x=+9 38°+
8°=+12)=
180° ,11x
x = −1,
15°x = 5 units
c: 2(n+
3n−1,
units 7x
d: −2(3x
7-19. Rotating about the midpoint of a base forms a hexagon (one convex and one non-convex).
Rotating the trapezoid about the midpoint of either of the non-base sides formsa 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∠YZYʹ′ = 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 ≅
2
2
YZ ′, ∠Y ≅∠Yʹ′, ∠YXZ ≅∠Yʹ′XZ, and ∠YZX ≅∠Yʹ′ZX 7-29. M(0,
2
7) 7-30. c and a + b 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.
b: 63.4 °; They are equal.
1
1
c: AC : y = 2 x + 2 , BD : y = −x + 5; No d: (3, 2) 7-33. a: No solution, lines
are parallel. b: (0, 3) and (4, 11)
7-34. Side length =
=
50
units;
7-35. a: It is a rhombus. It has four sides of length 5 units.
diagonal is
1
b: HJ :y = −2x + 8 and GI :y = 2 x + 3
10=units.
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) =
7-6.
7-7.
2
4
1
2
=
a: They are congruent by ASA ≅ or AAS ≅. b: AC ≈
9.4 units and DF = 20 units
Relationships used will vary, but may include alternate interior angles, Triangle
AngleSum Theorem, etc.; a = 26°, b = 65°, c = 26°, d = 117°
7-8.
width = 60 mm, area = 660 mm2
7-9.
A quadrilateral.
7-1
0.
a: x = 18, y = 9 3 b:
x = 24 2 , y = 24 c: x
=83=833, y =
16 3 = 16 3 3
7-1
1.
7-3
7.
7-3
8.
7-3
9.
7-7
5.
7-7
6.
7-7
7.
a: (6, –13)
b: Not possible, these curves do not intersect.
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: k2 = 152 + 252 −
2(15)(25)cos120° , k = 35
The graph is a parabola with roots (–3, 0) and (1, 0), and y-intercept at (0, –3).
≈ 35.24 m
a: Congruent (SSS ≅) c:
Congruent (ASA ≅)
See answers in problem 7-75.
a: 83°
b: 92°
b: Not enough information d:
Congruent (HL ≅)
7-54. a: x = 8.5° b: x = 11 c: x = 14° 7-55. a: 360° ÷ 72° = 5 sides b: 360° ÷ 9 = 40°
7-56. ≈ 36.4 feet from the point on the street closest to the art museum. 7-57. a: an =
1
n
1 n−1
20+ 20n = 40+ 20(n−1) b: an = 6( 2) = 3(2 ) 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: Congruent, because if the Pythagorean Theorem is used to solve for each unknown
side, then 3 pairs of corresponding sides are congruent; thus, the triangles arecongruent
by SSS ≅).
d: Similar (AA ~) but not congruent since the two sides of length 12 are
notcorresponding.
7-60. Possible response: Rotate the second triangle 180° and then translate it to match the sides
with the first triangle.
7-6.
7-7.
a: They are congruent by ASA ≅ or AAS ≅. b: AC ≈
9.4 units and DF = 20 units
Relationships used will vary, but may include alternate interior angles, Triangle
AngleSum Theorem, etc.; a = 26°, b = 65°, c = 26°, d = 117°
7-8.
width = 60 mm, area = 660 mm2
7-9.
A quadrilateral.
7-1 a: x = 18, y = 9 3 b:
x = 24 2 , y = 24 c: x
0.
7-79. a: Parallelogram
because the opposite sides are parallel.
= 8 3y==8 3x3;,BD:
y3 = y = − 3x + 9 7-80. a: 68 ≈ 8.2, since 64 = 8, then 68 must
AC:
4
2
be a little greater.
(1) 2.2, (2) 9.2, (3) 7.1, (4) 4.7 7-81. a: 2x + 52°= 180° , 64° b: 4x − 3°+
16 3 = 16 3 3
3x +1°= 180° , 26°
7-1
a: (6, –13)
b: Not possible, these curves do not intersect.
1.
a: 6n− 3° = n+17° , n = 4° b: 7x −19° + 3x +14° = 180° , so x = 18.5°. Then 5y− 2° =
sin 77° sin 72°, so y = 22.5° c: 5w+ 36° + 3w = 180° , w = 18° d: k2 = 152 + 252 −
7-3 c:7(18.5)−19°
x8
2(15)(25)cos120°
k =d:355x + 6°= 2x + 21° , x = 5°
7.
= , x ≈ ,8.2
Lesson 7.2.4
7-3
The graph is a parabola with roots (–3, 0) and (1, 0), and y-intercept at (0, –3).
8.
7-83.
7-3 36 3 ≈ 62.4 square units 7-84. No. Using the Pythagorean Theorem and the Law of
≈ 35.24 m
9.
Cosines,
perimeter
of the≅) by
b: Not
d: DF = 20 units
Congruent
(SSS
c: ASA ≅ or AAS
7-6. a:the
They
are congruent
≅. enough
b: AC ≈information
9.4 units and
Congruent
(HL
≅)
Congruent
(ASA
≅)
7-7
5.
7-7.
7-7 Relationships used will vary, but may include alternate interior angles, Triangle
See answers in problem 7-75.
6.
AngleSum Theorem, etc.; a = 26°, b = 65°, c = 26°, d = 117°
7-7
a: 83°
b: 92°
7.
7-8.
7-7 width = 60 mm, area = 660 mm2
b: 18°, 4 d: ≈ 49.2 square
8.
a: Yes; HL ≅ c: tan18° = 4 AD , AD
units
7-9. A quadrilateral.
≈ 12.3 units
7-1 a: x = 18, y = 9 3 b:
= 24
, y =feet.
24 c: x
0.
trianglex is
≈ 226.3
=
8
3
=
8
3
3
,
y
=
7-8
a: x ≈ 10.73 cm, tangent
b: x ≈ 7.86 mi, Law of Sines
5.
c: x ≈ 15.3', Law of Cosines
16 3 = 16 3 3
7-8
7-1 a: Congruent (SAS ≅) and x = 2
b: Congruent
(HL
x = 32
a: (6, –13)
b: Not possible,
these curves
do ≅)
notand
intersect.
6.
1.
24 3°
square
units, n = 4° b: 7x −19° + 3x +14° = 180° , so x = 18.5°. Then 5y− 2° =
7-8 A
a: =6n−
= n+17°
7.
7-3 7(18.5)−19° , so y = 22.5° c: 5w+ 36° + 3w = 180° , w = 18° d: k2 = 152 + 252 −
42
, k = 35
7.
7-8 2(15)(25)cos120°
a 0
8.
: =
15
b: 45 ; Since the sum of the probabilities of finding the ring and not finding the ring is 1,
7-108. ZY ≅ WX , ∠YZX ≅∠WXZ, ∠ZYW ≅∠YWX (alt. int. ∠ ’s), ΔXWM ≅ΔZYM, ASA ≅ , YM ≅
WM and XM ≅ ZM , ≅Δ→≅ parts.
7-109. Typical responses: right angles, congruent diagonals, 2 pairs of opposite sides that
arecongruent, 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.
7-6.
7-7.
a: They are congruent by ASA ≅ or AAS ≅. b: AC ≈
9.4 units and DF = 20 units
Relationships used will vary, but may include alternate interior angles, Triangle
AngleSum Theorem, etc.; a = 26°, b = 65°, c = 26°, d = 117°
7-8.
width = 60 mm, area = 660 mm2
7-9.
A quadrilateral.
a: x = 18, y = 9 3 b:
x = 24 2 , y = 24 c: x
=83=833, y =
7-114. This problem is similar to the Interior Design problem (7-21). Her sink should be located 3
2
feet from the right front edge of the counter. This will make the perimeter ≈ 25.6 feet, which
3
16 3 = 16 3 3
will meet industry standards.
7-1
a: (6, –13)
b: Not possible, these curves do not intersect.
1.
a: 6n− 3° = n+17° , n = 4° b: 7x −19° + 3x +14° = 180° , so x = 18.5°. Then 5y− 2° =
7-3 7(18.5)−19° , so y = 22.5° c: 5w+ 36° + 3w = 180° , w = 18° d: k2 = 152 + 252 −
2(15)(25)cos120° , k = 35
7.
7-1
0.
7-3
8.
7-3
9.
The graph is a parabola with roots (–3, 0) and (1, 0), and y-intercept at (0, –3).
≈
35.24 m
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.
7-131. a: Yes, each side has the same length (
29
units). See
b: BDis y = x ; ACis y = 5− x
graph at
c: The slopes are 1 and –1. Therefore the
diagonals are perpendicular.
right.
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.
3
4
7-135. a: It is a right triangle because the slopes of AB and AC ( 4 and − 3 , respectively) are opposite
reciprocals.
b: B′ is at (–2, 0). It is an isosceles triangle because ∠Bʹ′AB must be a straight
angle(because it is composed of two adjacent right
BCangles)
(because
andreflections
because BC
preserve
≅′
length).
7-136. a: ≈
23.83 feet b: x ≈
c: x ≈ 66.42 d: ≈
7 yds
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.
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,
equals is
itsthe
base
times
its its
height.
b: If thethen
areaitsofarea
a polygon
base
times
height, then the polygon is
aparallelogram. 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°).
1
3
7-145. The expected value is 4 ($3)+ 4 ($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
atleast one diagonal is bisected.