GEOMETRY LESSON 1-7 Basic Constructions 1. CD2. GH3. AB 4. line m5. acute ABC 6. XY || ST 7. DE =...

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GEOMETRY LESSON 1-7 Basic Constructions CD 2. GH 3. AB line m 5. acute ABC 6. XY || ST DE = 20. Point C is the midpoint of DE. Find CE. Use a protractor to draw a 60° angle. Use a protractor to draw a 120° angle. Exercises 1-6, sketch each figure.

Transcript of GEOMETRY LESSON 1-7 Basic Constructions 1. CD2. GH3. AB 4. line m5. acute ABC 6. XY || ST 7. DE =...

Page 1: GEOMETRY LESSON 1-7 Basic Constructions 1. CD2. GH3. AB 4. line m5. acute ABC 6. XY || ST 7. DE = 20. Point C is the midpoint of DE. Find CE. 8. Use a.

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7Basic Constructions

1. CD 2. GH 3. AB

4. line m 5. acute ABC 6. XY || ST

7. DE = 20. Point C is the midpoint of DE. Find CE.

8. Use a protractor to draw a 60° angle.

9. Use a protractor to draw a 120° angle.

In Exercises 1-6, sketch each figure.

Page 2: GEOMETRY LESSON 1-7 Basic Constructions 1. CD2. GH3. AB 4. line m5. acute ABC 6. XY || ST 7. DE = 20. Point C is the midpoint of DE. Find CE. 8. Use a.

1. The figure is a segment whose endpoints are C and D.

2. The figure is a ray whose endpoint is G.

3. The figure is a line passing through points A and B.

4. 5. The figure is an angle whose

measure is between 0° and 90°.

6. The figure is two segments in a plane whose corresponding 

lines are parallel.

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7Basic Constructions

Solutions

1-6. Answers may vary. Samples given:

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Page 3: GEOMETRY LESSON 1-7 Basic Constructions 1. CD2. GH3. AB 4. line m5. acute ABC 6. XY || ST 7. DE = 20. Point C is the midpoint of DE. Find CE. 8. Use a.

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7Basic Constructions

7. Since C is a midpoint, CD = CE; also, CD + CE = 20;

substituting results in CE + CE = 20, or 2CE = 20, so CE = 10.

8. 9.

Solutions (continued)

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Construction vidoes http://teachers.henrico.k12.va.us/math/igo/01Fundamentals/1_6.html

Page 5: GEOMETRY LESSON 1-7 Basic Constructions 1. CD2. GH3. AB 4. line m5. acute ABC 6. XY || ST 7. DE = 20. Point C is the midpoint of DE. Find CE. 8. Use a.

Step 2: Open the compass to the length of KM.

Construct TW congruent to KM.

Step 1: Draw a ray with endpoint T.

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7Basic Constructions

Step 3: With the same compass setting, put the compass point on point T. Draw an arc that intersects the ray. Label the point of intersection W.

TW KM

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Construct Y so that Y G.

Step 1: Draw a ray with endpoint Y.

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7

Basic Constructions

Step 3: With the same compass setting, put the compass point on point Y. Draw an arc that intersects the ray. Label the point of intersection Z.

Step 2: With the compass point on point G, draw an arc that intersects both sides of G. Label the points of intersection E and F.

75°

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(continued)

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7

Basic Constructions

Y G

Step 5: Draw YX to complete Y.

Step 4: Open the compass to the length EF. Keeping the same compass setting, put the compass point on Z. Draw an arc that intersects the arc you drew in Step 3. Label the point of intersection X.

Quick Check

Page 8: GEOMETRY LESSON 1-7 Basic Constructions 1. CD2. GH3. AB 4. line m5. acute ABC 6. XY || ST 7. DE = 20. Point C is the midpoint of DE. Find CE. 8. Use a.

Start with AB.

Step 2: With the same compass setting, put the compass point on point B and draw a short arc.

Without two points of intersection, no line can be drawn, so the perpendicular bisector cannot be drawn.

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7

Basic ConstructionsUse a compass opening less than AB. Explain why the

construction of the perpendicular bisector of AB shown in the text is

not possible.

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Step 1: Put the compass point on

point A and draw a short arc. Make

sure that the opening is less than AB.12

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Quick Check

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–3x = –48 Subtract 4x from each side. x = 16 Divide each side by –3.

m AWR = m BWR Definition of angle bisector x = 4x – 48 Substitute x for m AWR and

4x – 48 for m BWR.

m AWB = m AWR + m BWR Angle Addition Postulatem AWB = 16 + 16 = 32 Substitute 16 for m AWR and

for m BWR.

Draw and label a figure to illustrate the problem

WR bisects AWB. m AWR = x and m BWR = 4x – 48. Find m AWB.

m AWR = 16 m BWR = 4(16) – 48 = 16 Substitute 16 for x.

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7

Basic Constructions

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Quick Check

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Step 4: Draw MX. MX is the angle bisector of M.

(continued)

Step 3: Put the compass point on point C. Using the same compass setting, draw an arc in the interior of M. Make sure that the arcs intersect. Label the point where the two arcs intersect X.

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7

Basic Constructions

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For problems 1-4, check students’ work.

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NQ bisects DNB.

1. Construct AC so that AC NB.

2. Construct the perpendicular bisector of AC.

3. Construct RST so that RST QNB.

4. Construct the bisector of RST.

5. Find x.

6. Find m DNB.

Use the figure at right.

GEOMETRY LESSON 1-7GEOMETRY LESSON 1-7

Basic Constructions