This document provides examples and explanations for using algebraic concepts in geometry, specifically regarding angles formed when lines are cut by a transversal. It defines key terms like parallel lines, transversal, interior angles, exterior angles, alternate interior angles, alternate exterior angles, and corresponding angles. Examples are given to classify pairs of angles and find missing angle measures. Step-by-step examples walk through using properties of parallel lines cut by a transversal to find missing angles. The document also explains how to prove conjectures using paragraph and two-column proofs.
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Basic phrases for greeting and assisting costumers
Chapter 5
1. HOW can algebraic concepts be
applied to geometry?
Geometry
Course 3, Lesson 5-1
2. To
• classify the angles formed when two
lines are cut by a transversal,
• find missing angle measures when two
parallel lines are cut by a transversal
Course 3, Lesson 5-1
Geometry
4. Course 3, Lesson 5-1
Geometry
A line that intersects two or more lines is called a
, and eight angles are formed.transversal
Interior angles
Exterior angles
Alternate interior angles
Alternate exterior angles
Corresponding angles
lie inside the lines.
Examples: 3, 4, 5, 6
lie outside the lines
Examples: 1, 2, 7, 8
are interior angles that lie on opposite sides of the
transversal. When the lines are parallel, their measures are equal.
Examples: 4 6; m 3 5m m m
are exterior angles that lie on opposite sides of
the transversal. When the lines are parallel, their measures are equal.
Examples: 1 7; 2 8m m m m
are those angles that are in the same position on the two
lines in relation to the transversal. When the lines are parallel, their measures are
equal. Examples: 1 5; 2 6; 4 8; 3 7m m m m m m m m
5. 1
Need Another Example?
Step-by-Step Example
1. Classify the pair of angles in the figure
as alternate interior, alternate exterior,
or corresponding.
∠1 and ∠7
∠1 and ∠7 are exterior angles that lie on
opposite sides of the transversal. They
are alternate exterior angles.
7. 1
Need Another Example?
Step-by-Step Example
2. Classify the pair of angles in the figure
as alternate interior, alternate exterior,
or corresponding.
∠2 and ∠6
∠2 and ∠6 are in the same position on the two
lines. They are corresponding angles.
9. 1
Need Another Example?
Step-by-Step Example
3. A furniture designer built the bookcase
shown. Line a is parallel to line b.
If m∠2 = 105°, find m∠6 and m∠3.
Justify your answer.
Since ∠2 and ∠6 are supplementary,
the sum of their measures is 180°.
m∠6 = 180° – 105° or 75°
2 Since ∠6 and ∠3 are interior angles
that lie on opposite sides of the
transversal, they are alternate interior
angles. The measures of alternate
interior angles are equal. m∠3 = 75°
10. Answer
Need Another Example?
Mr. Adams installed the gate
shown. Line c is parallel to line
d. If m∠4 = 40°, find m∠6 and
m∠7. Justify your answer.
m∠6 = 40° and m∠7 = 140°; Sample answer:
∠4 and ∠6 are alternate interior angles, so they
are congruent. ∠6 and ∠7 are supplementary.
Since m∠6 = 40°, m∠7 = 140°.
11. 1
Need Another Example?
2
3
4
Step-by-Step Example
4. In the figure, line m is parallel to
line n, and line q is perpendicular
to line p. The measure of ∠1 is
40°. What is the measure of ∠7?
Since ∠1 and ∠6 are alternate
exterior angles, m∠6 = 40°.
40 + 90 + m∠7 = 180
Since ∠6, ∠7, and ∠8 form a straight line,
the sum of their measures is 180°.
So, m∠7 is 50°.
12. Answer
Need Another Example?
In the figure, line a is parallel to line b,
and line c is perpendicular to line d.
The measure of ∠7 is 125°. What is
the measure of ∠4?
35°
13. To prove a conjecture
• using a paragraph proof,
• using a two-column proof
Course 3, Lesson 5-2
Geometry
15. Course 3, Lesson 5-2
Geometry
Step 1 List the given information, or what you know. If
possible, draw a diagram to illustrate this
information.
Step 2 State what is to be proven.
Step 3 Create a deductive argument by forming a
logical chain of statements linking the given
information to what you are trying to prove.
Step 4 Justify each statement with a reason.
Reasons include definitions, algebraic
properties, and theorems.
Step 5 State what it is you have proven
16. 1
Need Another Example?
2
3
Step-by-Step Example
1. The diamondback
rattlesnake has a
diamond pattern on its
back. An enlargement
of the skin is shown.
If m∠1 = m∠4, write a
paragraph proof to show
that m∠2 = m∠3.
Given: m∠1 = m∠4
Proof:
Prove: m∠2 = m∠3
m∠1 = m∠2 because they are vertical angles. Since m∠1 = m∠4,
m∠2 = m∠4 by substitution. m∠4 = m∠3 because they are vertical
angles. Since m∠2 = m∠4, then m∠2 = m∠3 also by substitution.
Therefore, m∠2 = m∠3.
17. Answer
Need Another Example?
Refer to the diagram. If
m∠1 = m∠5, write a paragraph
proof to show that m∠1 = m∠11.
m∠1 = m∠9 because they are corresponding
angles. m∠9 = m∠11 because they are vertical
angles. Since m∠9 = m∠11, then m∠1 = m∠11
by substitution.
18. 1
Need Another Example?
2
3
4
Step-by-Step Example
2. Write a two-column proof to show that if two
angles are vertical angles, then they have the
same measure.
Statements
lines m and n intersect;
∠1 and ∠3 are vertical angles.
Given: lines m and n intersect; ∠1 and ∠3 are vertical angles
Prove: m∠1 = m∠3
Reasons
a. Given
∠1 and ∠2 are a linear pair
and ∠3 and ∠2 are a linear pair.
b. Definition of linear pair
m∠1 + m∠2 = 180º
m∠3 + m∠2 = 180º
c. Definition of supplementary angles
m∠1 + m∠2 = m∠3 + m∠2d. Substitution
5 m∠1 = m∠3 Subtraction Property of Equalitye.
19. Answer
Need Another Example?
Write a two-column proof to show that if
PQ = QS and QS = ST, then PQ = ST.
Given: PQ = QS; QS = ST
Prove: PQ = ST
Statements
PQ = QS and QS = ST
Reasons
a. Given
PQ = STb. Substitution
20. To
• find missing angle measures in a triangle,
• find missing angle measures in a triangle
using the exterior angles
Course 3, Lesson 5-3
Geometry
22. Course 3, Lesson 5-3
Geometry
Words The sum of the measures of the interior angles of a
triangle is 180˚
Model
Symbols x + y + z = 180˚
23. 1
Need Another Example?
2
Step-by-Step Example
1. Find the value of x in the
Antigua and Barbuda flag.
x + 55 + 90 = 180
The value of x is 35.
x + 145 = 180
– 145 = – 145
x = 35
Write the equation.
Simplify.
Subtract.
Simplify.
25. 1
Need Another Example?
2
3
4
Step-by-Step Example
2. The measures of the angles of ABC are in the
ratio 1:4:5. What are the measures of the angles?
Let x represent the measure of angle A.
Since x = 18, 4x = 4(18) or 72, and 5x = 5(18) or 90.
The measures of the angles are 18°, 72°, and 90°.
x + 4x + 5x = 180
x = 18
Write the equation.
Then 4x and 5x represent angle B and angle C.
10x = 180 Collect like terms.
Division Property of Equality
26. Answer
Need Another Example?
The measures of the angles of triangle DEF
are in the ratio 1:2:3. What are the measures
of the angles?
30°, 60°, and 90°
27. Course 3, Lesson 5-3
Geometry
Words The measure of an exterior angle of a triangle is equal to
the sum of the measures of its two remote interior angles.
Model
Symbol 1m A m B
28. 1
Need Another Example?
2
3
4
5
Step-by-Step Example
3. Suppose m∠4 = 135°.
Find the measure of ∠2.
Angle 4 is an exterior angle.
Its two remote interior angles
are ∠2 and ∠LKM.
So, the m∠2 = 45°.
x + 90° = 135°
x = 45°
Write the equation.m∠2 + m∠LKM = m∠4
m∠2 = x°, m∠LKM = 90°, m∠4 = 135°
Subtraction Property of Equality
32. Course3, Lesson 5-4
Geometry
Words The sum of the measures of the interior angles of a
polygon is (n – 2)180, where n represents the number of
sides.
Symbols S = (n – 2)180
33. 1
Need Another Example?
2
3
4
Step-by-Step Example
1. Find the sum of the measures of the interior
angles of a decagon.
The sum of the measures of the interior angles of a decagon
is 1,440°.
S = (8)180 or 1,440
Write the equation.S = (n – 2) 180
A decagon has 10 sides. Replace n with 10.
Simplify.
S = (10 – 2) 180
35. 1
Need Another Example?
2
3
Step-by-Step Example
2. Each chamber of a bee honeycomb is a regular
hexagon. Find the measure of an interior angle
of a regular hexagon.
Find the sum of the measures of the angles.
Write an equation.S = (n – 2) 180
Simplify.
The sum of the measures of the interior angles is 720°.
Replace n with 6.S = (6 – 2) 180
S = (4)180 or 720
Divide 720 by 6, the number of interior angles, to find the
measure of one interior angle. So, the measure of one
interior angle of a regular hexagon is 720° ÷ 6 or 120°.
36. Answer
Need Another Example?
A designer is creating a new logo for a
bank. The logo consists of a regular
pentagon surrounded by isosceles
triangles. Find the measure of an
interior angle of a regular pentagon.
108°
37. Course3, Lesson 5-4
Geometry
Words In a polygon, the sum of the measures of the exterior
angles, one at each vertex, is 360˚
Model
Symbols 1 2 3 4 5 360m m m m m
38. 1
Need Another Example?
2
3
Step-by-Step Example
3. Find the measure of an exterior angle in a
regular hexagon.
Let x represent the measure of each exterior angle.
Write an equation. A hexagon has 6 exterior angles.6x = 360
Division Property of Equalityx = 60
So, each exterior angle of a regular hexagon measures 60°.4
40. To
• find the missing side length of a right
triangle by using the Pythagorean
Theorem,
• determine whether a triangle is a right
triangle by using the converse of the
Pythagorean Theorem
Course 3, Lesson 5-5
Geometry
42. Course 3, Lesson 5-5
Geometry
Words In a right triangle, the sum of the squares of the lengths of
the legs is equal to the square of the length of the
hypotenuse.
Model
Symbols 2 2 2
a b c
43. 1
Need Another Example?
2
3
4
5
6
Step-by-Step Example
1. Write an equation you could use to find the
length of the missing side of the right triangle.
Then find the missing length. Round to the
nearest tenth if necessary.
Check:
Pythagorean Theorema2 + b2 = c2
Replace a with 12 and b with 9.
144 + 81 = c2
The equation has two solutions, 15 and –15. However, the length of a side
must be positive. So, the hypotenuse is 15 inches long.
122 + 92 = c2
Evaluate 122 and 92.
225 = c2 Add 81 and 144.
Definition of square root
c = 15 or –15 Simplify.
a2 + b2 = c2
122 + 92 = 152
144 + 81 = 225
225 = 225
?
?
±√225 = c
44. Answer
Need Another Example?
Write an equation you could use to find the
length of the missing side of the right triangle
shown. Then find the missing length. Round to
the nearest tenth if necessary.
122 + 162 = c2; 20 in.
45. 1
Need Another Example?
2
3
4
5
6
Step-by-Step Example
2. Write an equation you could use to find the length
of the missing side of the right triangle. Then find
the missing length. Round to the nearest tenth if
necessary.
Check for
Reasonableness
Pythagorean Theorema2 + b2 = c2
Replace a with 8 and c with 24.
64 + b2 = 576
The length of side b is about 22.6 meters.
82 + b2 = 242
Evaluate 82 and 242.
64 – 64 + b2 = 576 – 64 Subtract 64 from each side.
Definition of square root
b2 = 512 Simplify.
The hypotenuse is always the longest side in
a right triangle. Since 22.6 is less than 24,
the answer is reasonable.
7
b ≈ 22.6 or –22.6 Use a calculator.
b = ±√512
46. Answer
Need Another Example?
Write an equation you
could use to find the length
of the missing side of the right
triangle shown. Then find the missing length.
Round to the nearest tenth if necessary.
a2 + 282 = 332; 17.5 in.
47. Course 3, Lesson 5-5
Geometry
If the sides of a triangle have lengths a, b, and c units such that
, then the triangle is a right triangle.2 2 2
a b c
48. 1
Need Another Example?
2
3
4
5
Step-by-Step Example
3. The measures of three sides of a triangle are
5 inches, 12 inches, and 13 inches. Determine
whether the triangle is a right triangle.
Pythagorean Theorema2 + b2 = c2
a = 5, b = 12, c = 13
25 + 144 = 169
The triangle is a right triangle.
52 + 122 = 132
Evaluate 52, 122, and 132.
169 = 169 Simplify.
?
?
49. Answer
Need Another Example?
The measures of three sides of a
triangle are 24 inches, 7 inches, and
25 inches. Determine whether the
triangle is a right triangle.
yes; 72 + 242 = 252
50. To
• apply the Pythagorean Theorem to find
unknown side lengths in right triangles
in real-world situations,
• use the Pythagorean Theorem to find
missing measures in three-dimensional
figures
Course 3, Lesson 5-6
Geometry
51. 1
Need Another Example?
2
3
4
5
6
Step-by-Step Example
1. Write an equation that can be used to find the
length of the ladder. Then solve. Round to the
nearest tenth.
Pythagorean Theorema2 + b2 = c2
Replace a with 8.75 and b with 18.
76.5625 + 324 = c2
Notice that the distance from the building, the
building itself, and the ladder form a right
triangle. Use the Pythagorean Theorem.
8.752 + 182 = c2
Evaluate 8.752 and 182.
400.5625 = c2 Add 76.5625 and 324.
Definition of square root
Use a calculator.±20.0 ≈ c
Since length cannot be negative, the ladder is about 20 feet long.
±√400.5625 = c
52. Answer
Need Another Example?
Write an equation that can be
used to find the length of the
boat ramp. Then solve. Round
to the nearest tenth.
4.22 + 252 = c2; 25.4 ft
53. 1
Need Another Example?
2
3
4
5
Step-by-Step Example
2. Write an equation that can be used to find the
height of the plane. Then solve. Round to the
nearest tenth.
Pythagorean Theorema2 + b2 = c2
Replace a with 10 and c with 12.
100 + b2 = 144
The distance between the planes is the
hypotenuse of a right triangle. Use the
Pythagorean Theorem.
102 + b2 = 122
Evaluate 102 and 122.
b2 = 44 Subtraction Property of Equality
Since length cannot be negative, the height of the plane is about 6.6 miles.
Definition of square root
Use a calculator.b ≈ ±6.6
b = ±√44
54. Answer
Need Another Example?
Write an equation that can be
used to find the length of the
backboard. Then solve. Round to
the nearest tenth.
422 + x2 = 83.42; 72.1 in.
55. 1
Need Another Example?
2
3
4
5
Step-by-Step Example
3. A 12-foot flagpole is placed in the center of a
square area. To stabilize the pole, a wire will
stretch from the top of the pole to each corner of
the square. The flagpole is 7 feet from each corner
of the square. What is the length of each wire?
Round to the nearest tenth.
Pythagorean TheoremAB2 + AC2 = BC2
Replace AB with 7 and AC with 12.
49 + 144 = BC2
Draw right triangle ABC. You want to find the length of
each wire or BC. This is the hypotenuse of a right
triangle, so use the Pythagorean Theorem.
72 + 122 = BC2
Evaluate 72 and 122.
193 = BC2 Simplify
Since length cannot be negative, the length of the wire is about 13.9 feet.
Definition of square root
Use a calculator.±13.9 ≈ BC
±√193 = BC
56. Answer
Need Another Example?
The slant height of a pyramid is the
height of each lateral face. What is
the slant height of the pyramid
shown? Round to the nearest tenth.
11.7 cm
57. To
• use the Pythagorean Theorem to find
the distance between two points on a
coordinate plane,
• use the Distance Formula to find the
distance between two points on a
coordinate plane
Course 3, Lesson 5-7
Geometry
59. 1
Need Another Example?
2
3
4
5
6
Step-by-Step Example
1. Graph the ordered pairs (3, 0) and (7, –5). Then
find the distance c between the two points.
Round to the nearest tenth.
Pythagorean Theorema2 + b2 = c2
Replace a with 4 and b with 5.42 + 52 = c2
Definition of square root
41 = c2 42 + 52 = 16 + 25 or 41
The points are about 6.4 units apart.
Use a calculator.±6.4 ≈ c
±√41 = √c2
60. Answer
Need Another Example?
Graph the ordered pairs (0, –6) and (5, –1).
Then find the distance between the points.
Round to the nearest tenth.
7.1 units
61. Course 3, Lesson 5-7
Geometry
Symbols The distance d between two points with coordinates
and is given by the formula
Model
1 1
( , y )x 2 2
( , y )x
2 2
2 1 2 1
( ) ( ) .d x x y y
62. 1
Need Another Example?
2 3
4
5 6
Step-by-Step Example
2. On the map, each unit represents 45 miles. West Point,
New York, is located at (1.5, 2) and Annapolis, Maryland,
is located at (–1.5, –1.5). What is the approximate
distance between West Point and Annapolis?
a2 + b2 = c2
32 + 3.52 = c2
21.25 = c2
Let c represent the distance between West Point
and Annapolis. Then a = 3 and b = 3.5.
±4.6 ≈ c
7
Use the Pythagorean Theorem
Let (x1, y1) = (1.5, 2) and (x2, y2) = (–1.5, –1.5).
Use the Distance Formula
c ≈ ±4.6
Since each map unit equals 45 miles, the distance
between the cities is 4.6 · 45 or about 207 miles.
±√21.25 = √c2
c = √(x2 – x1)2 + (y2 – y1)2
c = √(–1.5 – 1.5)2 + (–1.5 – 2)2
c = √(–3)2 + (–3.5)2
c = √9 + 12.25
c = √21.25
63. Answer
Need Another Example?
Reed lives in Seattle,
Washington. One unit on this
map is 0.08 mile. Find the
approximate distance he
drives between Broad Street
at Denny Way (–1, 0) and
Broad Street at Dexter Ave N.
(4, 5).
0.57 mi
64. 1
Need Another Example?
2
3
4
5
Step-by-Step Example
3. Use the Distance Formula to find the distance
between X(5, –4) and Y(–3, –2). Round to the
nearest tenth if necessary.
XY ≈ ±8.2
So, the distance between points X and Y is about 8.2 units.
Distance Formula
(x1, y1) = (5, –4),
(x2, y2) = (–3, –2)
Simplify.
Evaluate (–8)2 and 22.
Add 64 and 4.
Simplify.
XY = √(–3 – 5)2 + [–2 – (–4)]2
XY = √(–8)2 + 22
XY = √64 + 4
XY = √68
d = √(x2 – x1)2 + (y2 – y1)2
65. Answer
Need Another Example?
Use the Distance Formula to find the distance
between G(–3, –2) and H(–6, 5). Round to
the nearest tenth.
7.6 units