Question

need answers help please geometry homework

Answer 1

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

$1. \hspace{0.15 cm} MN \approx \underline{2\dfrac{3}{8} \, inches}$

$2. \hspace{0.15 cm} PQ \approx \underline{1\dfrac{7}{8} \, inches}$

$3. \hspace{0.15 cm} UV \approx \underline{1\dfrac{2}{8} \, inches}$

$4. \hspace{0.15 cm} WX \approx \underline{1\dfrac{6}{8} \, inches}$

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

$\overline{JK}$ ≅ $\overline{LM}$ because $\overline{JK}$ = $\overline{LM}$

Monitoring Progress 5

$\overline{AB}$ $\ncong$ $\overline{CD}$ because $\overline{JK}$ ≠ $\overline{LM}$

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;

$1 \ \mathbf{division} \ on \ the \ \mathbf{ruler} = \dfrac{6.7 \, cm}{36} \times \dfrac{1 \, inch}{2.54 \, cm} = \dfrac{335}{4572} \ inch$

1. MN = 33 divisions, which gives;

• $\mathbf{MN} = \dfrac{335}{4572} \ inch \times 33 \approx \underline{2\dfrac{3}{8} \, inches}$

2. PQ is approximately 27 divisions, which gives;

• $PQ = \dfrac{335}{4572} \ inch \times 27 \approx \underline{1\dfrac{7}{8} \, inches}$

3. UV is approximately 18 divisions, which gives;

• $UV = \mathbf{\dfrac{335}{4572} \ inch \times 18} \approx \underline{1\dfrac{2}{8} \, inches}$

4. WX is approximately 25 divisions, which gives;

• $WX = \mathbf{\dfrac{335}{4572} \ inch \times 25} \approx \underline{1\dfrac{6}{8} \, inches}$

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;

$\overline{JK}$ = √((2 - (-3))²+ (4 - 4)²) = 5

$\overline{LM}$ = √((1 - 1)²+ ((-2) - 3)²) = 5

• $\overline{JK}$ ≅ $\overline{LM}$ because $\overline{JK}$ = $\overline{LM}$

Monitoring Progress 5

$\mathbf{\overline{AB}}$ = √((3 - (-2))²+ (4 - 4)²) = 5

$\mathbf{\overline{CD}}$ = √((0 - 0)²+ (2 - (-2))²) = 4

$\overline{AB}$ $\ncong$ $\overline{CD}$ because $\overline{JK}$ ≠ $\overline{LM}$

• $\overline{AB}$ is not congruent to $\overline{CD}$ because $\overline{JK}$ is not equal $\overline{LM}$

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;

• DF = 23 + 25 = 48

b. FH = FG + GH

Therefore;

GH = FH - FG

Which gives;

• GH = 36 - 21 = 15

Monitoring progress 6

XZ = XY + YZ

Which gives;

• XZ = 23 + 50 = 73

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;

• KL = 144 - 37 = 107

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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