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CHAPTER 10
Geometry
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10.2
Triangles
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Objectives
1. Solve problems involving angle relationships in
triangles.
2. Solve problems involving similar triangles.
3. Solve problems using the Pythagorean Theorem.
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Triangle
• A closed geometric figure that has three sides, all of
which lie on a flat surface or plane.
• Closed geometric figures
– If you start at any point and trace along the sides,
you end up at the starting point.
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Euclid’s Theorem
Theorem: A conclusion that is proved to be true
through deductive reasoning.
Euclid’s assumption: Given a line and a point not on
the line, one and only one line can be drawn through the
given point parallel to the given line.
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Euclid’s Theorem (cont.)
Euclid’s Theorem: The sum of the measures of the
three angles of any triangle is 180º.
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Euclid’s Theorem (cont.)
Proof:
m1 = m2 and m3 = m4 (alternate interior angles)
Angles 2, 5, and 4 form a straight angle (180º)
Therefore, m1 + m5 + m3 = 180º .
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Two-Column Proof A
Theorem: The sum of ∠’s in a △ equals 180°.
C
Given: △ABC
Prove: ∠A + ∠ B + ∠ACB = 180°
Statements
1. Through C, draw line m || to
line AB.
2. ∠1 + ∠2 + ∠3 = 180°
3. ∠1 = ∠A; ∠3 = ∠B
4. ∠A + ∠ACB + ∠B = 180°
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2
3
A
m
B
Reasons
1. Parallel postulate
2. Definition of straight angle
3. If 2 || lines, alt. int. ∠’s are equal
4. Substitution of equals axiom
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Exterior ∠ Theorem
Theorem: An exterior angle of a triangle equals the sum of
the non-adjacent interior angles.
C
A
B
D
∠CBD = ∠A + ∠C
Since ∠CBD + ∠ABC = 180 -- def. of supplementary ∠s
∠A + ∠C + ∠ABC = 180 -- sum of △’s interior ∠s
∠A + ∠C + ∠ABC = ∠CBD + ∠ABC -- axiom
∠A + ∠C = ∠CBD -- axiom
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Base ∠s of Isosceles △
• Theorem: Base ∠s of an isosceles △ are equal
C
A
B
• Given: Isosceles△ ABC, which AC = BC
Prove: ∠A = ∠B
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Two-Column Proof B
Theorem: An ∠ formed by 2 radii subtending a chord is 2 x
A
an inscribed ∠ subtending the same chord.
Given: Circle C with A, B, D on the circle
Prove: ∠ACB = 2 • ∠ADB
Statements
Reasons
1. Circle C, with central ∠C,
inscribed ∠D
2. Draw line CD, intersecting circle
at F
3. ∠y = ∠a + ∠b = 2 ∠a
∠x = ∠c + ∠d = 2 ∠c
4. ∠x + ∠y = 2(∠a + ∠c)
5. ∠ACB = 2 ∠ADB
1. Given
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F
d
x
C
c
y
b
B
a
D
2. 2 points determine a line (postulate)
3. Exterior ∠; Isosceles △ angles
4. Equal to same quantity (axiom)
5. Substitution (axiom)
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Example 1: Using Angle Relationships in
Triangles
Find the measure of angle A
for the triangle ABC.
Solution:
mA + mB + mC =180º
mA + 120º + 17º = 180º
mA + 137º = 180º
mA = 180º − 137º = 43º
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Example 2: Using Angle Relationships in
Triangles
Find the measures of angles 1 through 5.
Solution:
m1+ m2 + 43º =180º
90º + m2 + 43º = 180º
m2 + 133º = 180º
m2 = 47º
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Example 2 continued
m3 + m4+ 60º =180º
47º + m4 + 60º = 180º
m4 = 180º − 107º = 73º
m4 + m5 = 180º
73º + m5 = 180°
m5 = 107º
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Triangles and Their Characteristics
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Similar Triangles
B
△ABC ~ △XYZ iff
• Corresponding angles
are equal
A
• Corresponding sides are
proportional
•
•
∠A = ∠X; ∠B = ∠Y; ∠C = ∠Z
AB/XY = BC/YZ = AC/XZ
•
If 2 corresponding ∠s of 2 △s
are equal, then △s are similar.
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C
Y
X
Z
1
Example 1
• Given:
A
– AB || DE
x
– AB = 25; CE = 8
• Find AC
D
• Solution
• x
25
--- = ----8
12
•
8(25)
25
50
x = ------- = 2 • ---- = ----- ≈ 16.66
12
3
3
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25
B
C
8
12
E
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Example 2
• How can you estimate the height of a building, if
you know your own height (on a sunny day)?
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Example 2
6
10
--- = -----x
400
6 • 400 = 10 • x
x
x = 240
6
10
400
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Similar Triangles
• Similar figures have the same shape, but not
necessarily the same size.
• In similar triangles, the angles are equal but the
sides may or may not be the same length.
• Corresponding angles are angles that have the same
measure in the two triangles.
• Corresponding sides are the sides opposite the
corresponding angles.
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Similar Triangles (continued)
Triangles ABC and DEF are similar:
Corresponding Angles
Corresponding Sides
Angles A and D
Sides AC and DF
Angles C and F
Sides AB and DE
Angles B and E
Sides CB and FE
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Example 3: Using Similar Triangles
Find the missing length x.
Solution: Because the triangles are similar, their
corresponding sides are proportional:
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Example 3: Using Similar Triangles
Find the missing length x.
5 7

8 x
5x  8  7
5 x 56

5
5
x  11.2
The missing length of x is 11.2 inches.
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Pythagorean Theorem
The sum of the squares
of the lengths of the legs
of a right triangle equals
the square of the length
of the hypotenuse.
If the legs have lengths a
and b and the hypotenuse
has length c, then
a² + b² = c²
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Example 5: Using the Pythagorean
Theorem
Find the length of the
hypotenuse c in this right
triangle:
Solution:
Let a = 9 and b = 12
C
c2  a 2  b2
c 2  92  122
c 2  81  144
c 2  225
c  225  15
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