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Segments. Module 1 Section 2. Image courtesy of Microsoft. In this lesson you will explore: Segments Congruent Segments Segment Addition Midpoint of a Segment Segment Bisector Perpendicular Bisector. Common Core Standards: G-CO.1, G-CO.12, G-GPE.6. The Segment. m. A. B.
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Segments Module 1 Section 2 Image courtesy of Microsoft
In this lesson you will explore: • Segments • Congruent Segments • Segment Addition • Midpoint of a Segment • Segment Bisector • Perpendicular Bisector Common Core Standards: G-CO.1, G-CO.12, G-GPE.6
The Segment m A B Properties of a Segment: • A segment is part of a line. • A segment has two definite endpoints and consists of those two endpoints and all points between those endpoints. Here the segment has endpoints A and B and is a part of line m. • A segment is named using its two endpoints and has a drawn above the letters that name the segment. We would name this segment: AB Note: There are an infinite number of segments on any given line.
Identifying and Naming Segments B A D C E F G H How many segments can you name? Pause this presentation and make a list, then continue the presentation.
Identifying and Naming Segments B A D C E F G H How many segments can you name?
Identifying and Naming Segments B A D C E F G H AB, AD, AE, BC, BF, CD, CG, DH, EF, EH, FG, GH
Identifying and Naming Segments B A D C E F G H AB, AD, AE, BC, BF, CD, CG, DH, EF, EH, FG, GH
Identifying and Naming Segments B A D C E F G H AB, AD, AE, BC, BF, CD, CG, DH, EF, EH, FG, GH
Identifying and Naming Segments B A D C E F G H AB, AD, AE, BC, BF, CD, CG, DH, EF, EH, FG, GH AH, DE, AF, BE, BG, CF, CH, DG, AG, BH, AC, BD, EG, and FH
Congruency We can call two values equal because we are only talking about a measure (or a number value). We may use “equal” when talking about length or a degree. However, when we want to discuss figures that are the same size and the same shape, we call them congruent. Congruent figures are noted using a (congruent sign is an = sign with a ~ on top).
Congruent Segments Congruent Segments: • Segments that have the same measure (same length). • They usually have different endpoints, but they may share an endpoint. • Notation: • Segment AB is 3 units long can be written as AB = 3 • Segment CD is 3 units long can be written as CD = 3
Congruent Segments Congruent Segments: • Segments that have the same measure (same length). • They usually have different endpoints, but they may share an endpoint. C A Given: AB = 3 CD = 3 D B
C Figure 1: 3 cm A D B 3 cm C A Figure 2: 3 cm B 3 cm Figure 3: A C K 3 cm 3 cm B D
Figure 1: Figure 2: A B C D Segment AB, segment AC, segment AD, segment BC, segment BD, segment CD.
Segment Addition Segment addition problems can sometimes seem confusing, but as long as you remember that the two smaller parts make up the whole, you will be in good shape! This means: one small part + other small part = whole segment
Segment Addition one small part + other small part = whole segment AC C B A AB BC AB + BC = AC This is the Segment Addition Postulate
Segment Addition AC C B A AB BC AB + BC = AC Example 1: If AB = 23 and BC = 47, find AC
Segment Addition AC C B A 23 47 AB + BC = AC Example 1: If AB = 23 and BC = 47, find AC + 47 = AC 70 = AC 23
Segment Addition 60 C B A 4x + 5 15 AB + BC = AC Example 2: If AB = 4x + 5, BC = 15, and AC = 60 find x. + 15 = 60 4x + 5 4x + 20 = 60 4x = 40 x = 10
Segment Addition 58 C B A 28 ? AB + BC = AC Example 3: If AB = 28 and AC = 58 find BC. + BC = 58 + x = 58 28 28 x = 30 Which means BC = 30
Segment Addition AB + BC = AC 5x - 12 C B A 2x - 8 -x + 24 Example 4: If AB = 2x – 8, BC = - x + 24, and AC = 5x – 12, find AC. 2x - 8 + - x + 24 = 5x - 12 x + 16 = 5x – 12 16 = 4x – 12 Now that we know x = 7 we can easily find AC by replacing 7 with x in 5x – 12. 5x – 12 = AC 5(7) – 12 = AC 35 – 12 = AC 23 units = AC 28 = 4x 7 = x
Midpoint of a Segment Midpoint: • We have learned that if A, B, and C are collinear then AB + BC = AC • If AB = BC then B is called the midpoint of AC, and we can put in the tic marks to show that AB and BC are congruent A B C A B C
Midpoint of a Segment A B C AB + BC = AC Example 1: B is the midpoint of AC. If AC = 115 and AB = 5x – 10, find x. AB + BC = AC 5x – 10 + BC = 115 5x – 10 + 5x – 10 = 115 10x – 20 = 115 10x = 135 x = 13.5
Midpoint of a Segment Example 2: B is the midpoint of AC. Find x, AB, BC, and AC. A B C AB = BC 4x + 12 = 5x – 3 4x + 15 = 5x 15 = x 4x + 12 5x - 3
Midpoint of a Segment Example 2: B is the midpoint of AC. Find x, AB, BC, and AC. A B C AB = BC 4x + 12 = 5x – 3 4x + 15 = 5x 15 = x 4x + 12 5x - 3 4x + 12 = AB 4(15) + 12 = AB 60 + 12 = AB 72 = AB 72 = BC AB + BC = AC 72 + 72 = AC 144 = AC
Segment Bisector Bicycle has TWO wheels that are the same size. Biplane has TWO wings that are the same size. So what do a biplane and a bicycle have in common with a segment bisector?
Segment Bisector A segment Bisector ensures the segment has been divided into TWO parts that are the same size. The bisector goes through the segment midpoint and we know that: Line m is the segment bisector B is the midpoint of AC m C B A The marks indicate that each part is congruent
Perpendicular Segment Bisector A perpendicular segment Bisector is a segment bisector that runs perpendicular to the segment and passes through the midpoint of the segment. • Line m is the perpendicular segment bisector • B is the midpoint of AC m C B A The marks indicate that each part is congruent
Segment Addition with Bisectors m 50 C B A 2x - 4 BC Example 4: If m is a bisector of AC, AB = 2x – 4, AC = 50, find BC. Recall: AB + BC = AC AB + BC = AC 2x – 4 + BC = 50
Segment Addition with Bisectors m 50 C B A 2x - 4 BC Example 4: If m is a bisector of AC, AB = 2x – 4, AC = 50, find BC. Recall: AB + BC = AC AB + BC = AC 2x – 4 + BC = 50 2x – 4 + 2x – 4 = 50 4x – 8 = 50 4x = 58 x = 14.5
Segment Addition with Bisectors m 50 C B A 2x - 4 BC Example 4: If m is a bisector of AC, AB = 2x – 4, AC = 50, find BC. Recall: AB + BC = AC AB + BC = AC 2x – 4 + BC = 50 2x – 4 + 2x – 4 = 50 4x – 8 = 50 4x = 58 x = 14.5 2x – 4 = BC 2(14.5) – 4 = BC 29 – 4 = BC 25 = BC
Example 5: Given , find JK. mJK = mLM -2x + 33 = 6x - 23 33 = 8x – 23 56 = 8x 7 = x JK = -2x + 33 JK = -2(7) + 33 JK = 19