McDougal Littell Jurgensen Geometry: Student Edition Geometry
McDougal Littell Jurgensen Geometry: Student Edition Geometry
5th Edition
ISBN: 9780395977279
Author: Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Publisher: Houghton Mifflin Company College Division
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Chapter 7.5, Problem 1MRE

a.

To determine

To complete the given statement.

a.

Expert Solution
Check Mark

Answer to Problem 1MRE

  GFGC and EDEF .

Explanation of Solution

Given:

  AF is parallel to is parallel to CD .

Also, ABBC.

  McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.5, Problem 1MRE , additional homework tip  1

Given statement: GF? and ED?

Calculation:

In ΔACF ,

  ABBC (given)

   B is the midpoint of AC .

Now, AF is parallel to BE ,

   BG is parallel to AF (parts of parallel lines are also parallel)

So, by mid − point theorem,

  G is the mid − point of FC .

   GFGC .

In ΔDCF ,

  GFGC (proved above)

   G is the midpoint of CF .

Now, BE is parallel to CD ,

   GE is parallel to CD (parts of parallel lines are also parallel)

So, by Midpoint Theorem,

  E is the midpoint of DF .

   EDEF .

Conclusion:

Therefore, the complete statement is: GFGC and EDEF .

b.

To determine

To complete the given statement.

b.

Expert Solution
Check Mark

Answer to Problem 1MRE

If AB=9 , then AC=18 .

Explanation of Solution

Given:

  AF is parallel to BE is parallel to CD .

  ABBC.

  McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.5, Problem 1MRE , additional homework tip  2

  AB=9.

Given statement: if AB=9 , then AC=?

Calculation:

Since, AC=AB+BC , and

  ABBC (given),

   AC=AB+AB ,

   AC=9+9 ( Since, AB=9 ),

   AC=18 .

Conclusion:

Therefore, the complete statement is: if AB=9 , then AC=18.

c.

To determine

To complete the given statement.

c.

Expert Solution
Check Mark

Answer to Problem 1MRE

If FG=3x+2 and GC=7x10 , then x=1.2 .

Explanation of Solution

Given:

  AF is parallel to BE is parallel to CD .

  ABBC.

  McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.5, Problem 1MRE , additional homework tip  3

  FG=3x+2 , and

  GC=7x10.

Given statement: if FG=3x+2 and GC=7x10 , then x=?

Calculation:

Since FGGC (proved above).

So,

  3x+2=7x10 ,

    7x+3x=10+2, 10x=12, x=1210, x=1.2.

Conclusion:

Therefore, the complete statement is: if FG=3x+2 and GC=7x10 , then x=1.2 .

d.

To determine

To find the measure of mFEG.

d.

Expert Solution
Check Mark

Answer to Problem 1MRE

  mFEG=90° .

Explanation of Solution

Given:

  AB is parallel to BE is parallel to CD .

  ABBC.

  McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.5, Problem 1MRE , additional homework tip  4

  mEDC=90°.

Given statement: mFEG=?

Concept used:

Properties of parallel and transversal lines.

Calculation:

Since BE is parallel to DC (given), and FD is transversal.

Then, FEGEDC (corresponding angles on the same side of the transversal are congruent).

   mFEG=mEDC

   mFEG=90°. ( Since mEDC=90° )Conclusion:

Therefore, the complete statement is: mFEG=90° .

Chapter 7 Solutions

McDougal Littell Jurgensen Geometry: Student Edition Geometry

Ch. 7.1 - Prob. 11CECh. 7.1 - Prob. 12CECh. 7.1 - Prob. 13CECh. 7.1 - Prob. 14CECh. 7.1 - Prob. 15CECh. 7.1 - Prob. 16CECh. 7.1 - Prob. 17CECh. 7.1 - Prob. 18CECh. 7.1 - Prob. 1WECh. 7.1 - Prob. 2WECh. 7.1 - Prob. 3WECh. 7.1 - Prob. 4WECh. 7.1 - Prob. 5WECh. 7.1 - Prob. 6WECh. 7.1 - Prob. 7WECh. 7.1 - Prob. 8WECh. 7.1 - Prob. 9WECh. 7.1 - Prob. 10WECh. 7.1 - Prob. 11WECh. 7.1 - Prob. 12WECh. 7.1 - Prob. 13WECh. 7.1 - Prob. 14WECh. 7.1 - Prob. 15WECh. 7.1 - Prob. 16WECh. 7.1 - Prob. 17WECh. 7.1 - Prob. 18WECh. 7.1 - Prob. 19WECh. 7.1 - Prob. 20WECh. 7.1 - Prob. 21WECh. 7.1 - Prob. 22WECh. 7.1 - Prob. 23WECh. 7.1 - Prob. 24WECh. 7.1 - Prob. 25WECh. 7.1 - Prob. 26WECh. 7.1 - Prob. 27WECh. 7.1 - Prob. 28WECh. 7.1 - Prob. 29WECh. 7.1 - Prob. 30WECh. 7.1 - Prob. 31WECh. 7.1 - Prob. 32WECh. 7.1 - Prob. 33WECh. 7.1 - Prob. 34WECh. 7.1 - Prob. 35WECh. 7.1 - Prob. 36WECh. 7.2 - Prob. 1CECh. 7.2 - Prob. 2CECh. 7.2 - Prob. 3CECh. 7.2 - Prob. 4CECh. 7.2 - Prob. 5CECh. 7.2 - Prob. 6CECh. 7.2 - Prob. 7CECh. 7.2 - Prob. 8CECh. 7.2 - Prob. 9CECh. 7.2 - Prob. 10CECh. 7.2 - Prob. 11CECh. 7.2 - Prob. 12CECh. 7.2 - Prob. 13CECh. 7.2 - Prob. 14CECh. 7.2 - Prob. 15CECh. 7.2 - Prob. 16CECh. 7.2 - Prob. 1WECh. 7.2 - Prob. 2WECh. 7.2 - Prob. 3WECh. 7.2 - Prob. 4WECh. 7.2 - Prob. 5WECh. 7.2 - Prob. 6WECh. 7.2 - Prob. 7WECh. 7.2 - Prob. 8WECh. 7.2 - Prob. 9WECh. 7.2 - Prob. 10WECh. 7.2 - Prob. 11WECh. 7.2 - Prob. 12WECh. 7.2 - Prob. 13WECh. 7.2 - Prob. 14WECh. 7.2 - Prob. 15WECh. 7.2 - Prob. 16WECh. 7.2 - Prob. 17WECh. 7.2 - Prob. 18WECh. 7.2 - Prob. 19WECh. 7.2 - Prob. 20WECh. 7.2 - Prob. 21WECh. 7.2 - Prob. 22WECh. 7.2 - Prob. 23WECh. 7.2 - Prob. 24WECh. 7.2 - Prob. 25WECh. 7.2 - Prob. 26WECh. 7.2 - Prob. 27WECh. 7.2 - Prob. 28WECh. 7.2 - Prob. 29WECh. 7.2 - Prob. 30WECh. 7.2 - Prob. 31WECh. 7.2 - Prob. 32WECh. 7.2 - Prob. 33WECh. 7.2 - Prob. 34WECh. 7.2 - Prob. 35WECh. 7.2 - Prob. 36WECh. 7.2 - Prob. 37WECh. 7.2 - Prob. 38WECh. 7.2 - Prob. 39WECh. 7.2 - Prob. 40WECh. 7.2 - Prob. 41WECh. 7.2 - Prob. 42WECh. 7.2 - 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