The segment addition postulate can be used to find the distances
between colinear locations given the total and part of the distances.
Responses:
Monitoring Progress 1 – 4
[tex]1. \hspace{0.15 cm} MN \approx \underline{2\dfrac{3}{8} \, inches}[/tex]
[tex]2. \hspace{0.15 cm} PQ \approx \underline{1\dfrac{7}{8} \, inches}[/tex]
[tex]3. \hspace{0.15 cm} UV \approx \underline{1\dfrac{2}{8} \, inches}[/tex]
[tex]4. \hspace{0.15 cm} WX \approx \underline{1\dfrac{6}{8} \, inches}[/tex]
Constructing and Comparing Congruent Segments
- A construction is a geometric drawing that uses limited set of tools usually a compass and a straight edge.
Guided Example 2
[tex]\overline{JK}[/tex] ≅ [tex]\overline{LM}[/tex] because [tex]\overline{JK}[/tex] = [tex]\overline{LM}[/tex]
Monitoring Progress 5
[tex]\overline{AB}[/tex] [tex]\ncong[/tex] [tex]\overline{CD}[/tex] because [tex]\overline{JK}[/tex] ≠ [tex]\overline{LM}[/tex]
Using the Segment Addition Postulate
- When three points are collinear, you can say that one point is between the other two.
Guided example 3
a. DF = 48
b. GH = 15
Monitoring progress 6
XZ = 73
Monitoring progress 7
- Segment Addition Postulate cannot be used to find the distance between points W and Z
Monitoring progress 8
KL = 107
Guided example 4
Tulsa to St. Louis is 361 mi.
Monitoring Progress 9
Albuquerque to Provo is 449 mi.
How can the segment addition postulate be used to find the distance between two points?
Monitoring Progress 1 – 4
Using the ruler on Snip and Sketch, we have;
36 divisions = 6.7 cm
1 inch = 2.54 cm
Which gives;
[tex]1 \ \mathbf{division} \ on \ the \ \mathbf{ruler} = \dfrac{6.7 \, cm}{36} \times \dfrac{1 \, inch}{2.54 \, cm} = \dfrac{335}{4572} \ inch[/tex]
1. MN = 33 divisions, which gives;
- [tex]\mathbf{MN} = \dfrac{335}{4572} \ inch \times 33 \approx \underline{2\dfrac{3}{8} \, inches}[/tex]
2. PQ is approximately 27 divisions, which gives;
- [tex]PQ = \dfrac{335}{4572} \ inch \times 27 \approx \underline{1\dfrac{7}{8} \, inches}[/tex]
3. UV is approximately 18 divisions, which gives;
- [tex]UV = \mathbf{\dfrac{335}{4572} \ inch \times 18} \approx \underline{1\dfrac{2}{8} \, inches}[/tex]
4. WX is approximately 25 divisions, which gives;
- [tex]WX = \mathbf{\dfrac{335}{4572} \ inch \times 25} \approx \underline{1\dfrac{6}{8} \, inches}[/tex]
Constructing and Comparing Congruent Segments
- A construction is a geometric drawing that uses limited set of tools usually a compass and a straight edge.
Guided Example 2
Using Pythagorean theorem, we have;
[tex]\overline{JK}[/tex] = √((2 - (-3))²+ (4 - 4)²) = 5
[tex]\overline{LM}[/tex] = √((1 - 1)²+ ((-2) - 3)²) = 5
- [tex]\overline{JK}[/tex] ≅ [tex]\overline{LM}[/tex] because [tex]\overline{JK}[/tex] = [tex]\overline{LM}[/tex]
Monitoring Progress 5
[tex]\mathbf{\overline{AB}}[/tex] = √((3 - (-2))²+ (4 - 4)²) = 5
[tex]\mathbf{\overline{CD}}[/tex] = √((0 - 0)²+ (2 - (-2))²) = 4
[tex]\overline{AB}[/tex] [tex]\ncong[/tex] [tex]\overline{CD}[/tex] because [tex]\overline{JK}[/tex] ≠ [tex]\overline{LM}[/tex]
- [tex]\overline{AB}[/tex] is not congruent to [tex]\overline{CD}[/tex] because [tex]\overline{JK}[/tex] is not equal [tex]\overline{LM}[/tex]
Please see attached drawing created with MS Excel
Using the Segment Addition Postulate
- When three points are collinear, you can say that one point is between the other two.
Guided example 3
a. DF = DE + DF
Which gives;
b. FH = FG + GH
Therefore;
GH = FH - FG
Which gives;
Monitoring progress 6
XZ = XY + YZ
Which gives;
Monitoring progress 7
W and Z are non colinear points
Therefore;
- Segment Addition Postulate cannot be used to find the distance between points W and Z
Monitoring progress 8
JL = JK + KL
Therefore;
KL = JL - JK
Which gives;
Guided example 4
The distance from Lubbock to St. Louis Missouri = The distance from Lubbock to Tulsa + The distance from Tulsa to St. Louis
Therefore;
Tulsa to St. Louis = Lubbock to St. Louis - Lubbock to Tulsa
Which gives;
- Tulsa to St. Louis = 738 mi. - 377 mi. = 361 mi.
Monitoring Progress 9
Carlsbad to Provo = Carlsbad to Albuquerque + Albuquerque to Provo
Which gives;
Albuquerque to Provo = Carlsbad to Provo - Carlsbad to Albuquerque
Therefore;
- Albuquerque to Provo = 680 mi. - 231 mi. = 449 mi.
Learn more about segment addition postulate here:
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