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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Question
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Chapter 14.8, Problem 7E

a.

To determine

To describe all the rotational symmetries of the figure equilateral triangle.

a.

Expert Solution
Check Mark

Answer to Problem 7E

The group table for six symmetriesare given below,

    ° I R0,120 R0,240
    I R0,120 R0,120 R0,240
    R0,240 R0,240 I I

Explanation of Solution

Given information:

The figure of an equilateral triangle.

The figure of equilateral triangle has the three rotational symmetry. Each symmetry has center O and rotates the figure onto itself. Note that 120° rotational symmetry is another name for point symmetry.

  (1) 120° rotational symmetry: R0,120(2) 240° rotational symmetry: R0,240(3) 360° rotational symmetry: I is the identity.

The identity mapping always maps a figure onto itself, we usually include the identity when listing the symmetries of a figure.

The group table for six symmetries are given below,

    ° I R0,120 R0,240
    I R0,120 R0,120 R0,240
    R0,240 R0,240 I I

b.

To determine

The find that the group is not commutative.

b.

Expert Solution
Check Mark

Answer to Problem 7E

The group is not commutative as it is not a non-abelian group.

Explanation of Solution

Given:

The figure equilateral triangle has 3 rotational symmetry.

Concept Used:

The group is commutative if it has four property,

  1. The product of two symmetry is another symmetry.
  2. The set of symmetries contains the identity.
  3. Each symmetry has an inverse that is also a symmetry.
  4. Forming of product is an associative group,

  A°(B°C)=(A°B)°C for any three group.

Therefore, the group is not commutative as it is not abelian group.

Chapter 14 Solutions

McDougal Littell Jurgensen Geometry: Student Edition Geometry

Ch. 14.1 - Prob. 1WECh. 14.1 - Prob. 2WECh. 14.1 - Prob. 3WECh. 14.1 - Prob. 4WECh. 14.1 - Prob. 5WECh. 14.1 - Prob. 6WECh. 14.1 - Prob. 7WECh. 14.1 - Prob. 8WECh. 14.1 - Prob. 9WECh. 14.1 - Prob. 10WECh. 14.1 - Prob. 11WECh. 14.1 - Prob. 12WECh. 14.1 - Prob. 13WECh. 14.1 - Prob. 14WECh. 14.1 - Prob. 15WECh. 14.1 - Prob. 16WECh. 14.1 - Prob. 17WECh. 14.1 - Prob. 18WECh. 14.1 - Prob. 19WECh. 14.1 - Prob. 20WECh. 14.1 - Prob. 21WECh. 14.1 - Prob. 22WECh. 14.1 - Prob. 23WECh. 14.2 - Prob. 1CECh. 14.2 - Prob. 2CECh. 14.2 - Prob. 3CECh. 14.2 - Prob. 4CECh. 14.2 - Prob. 5CECh. 14.2 - Prob. 6CECh. 14.2 - Prob. 7CECh. 14.2 - Prob. 8CECh. 14.2 - Prob. 9CECh. 14.2 - Prob. 10CECh. 14.2 - Prob. 11CECh. 14.2 - Prob. 12CECh. 14.2 - Prob. 13CECh. 14.2 - Prob. 14CECh. 14.2 - Prob. 15CECh. 14.2 - Prob. 16CECh. 14.2 - Prob. 17CECh. 14.2 - Prob. 18CECh. 14.2 - Prob. 19CECh. 14.2 - Prob. 20CECh. 14.2 - Prob. 1WECh. 14.2 - Prob. 2WECh. 14.2 - Prob. 3WECh. 14.2 - Prob. 4WECh. 14.2 - Prob. 5WECh. 14.2 - Prob. 6WECh. 14.2 - 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Prob. 6ST2Ch. 14.8 - Prob. 7ST2Ch. 14.8 - Prob. 8ST2Ch. 14.8 - Prob. 9ST2Ch. 14.8 - Prob. 10ST2Ch. 14.8 - Prob. 11ST2Ch. 14.8 - Prob. 1ECh. 14.8 - Prob. 2ECh. 14.8 - Prob. 3ECh. 14.8 - Prob. 4ECh. 14.8 - Prob. 5ECh. 14.8 - Prob. 6ECh. 14.8 - Prob. 7ECh. 14.8 - Prob. 8ECh. 14.8 - Prob. 9ECh. 14.8 - Prob. 10ECh. 14 - Prob. 1CRCh. 14 - Prob. 2CRCh. 14 - Prob. 3CRCh. 14 - Prob. 4CRCh. 14 - Prob. 5CRCh. 14 - Prob. 6CRCh. 14 - Prob. 7CRCh. 14 - Prob. 8CRCh. 14 - Prob. 9CRCh. 14 - Prob. 10CRCh. 14 - Prob. 11CRCh. 14 - Prob. 12CRCh. 14 - Prob. 13CRCh. 14 - Prob. 14CRCh. 14 - Prob. 15CRCh. 14 - Prob. 16CRCh. 14 - Prob. 17CRCh. 14 - Prob. 18CRCh. 14 - Prob. 19CRCh. 14 - Prob. 20CRCh. 14 - Prob. 1CTCh. 14 - Prob. 2CTCh. 14 - Prob. 3CTCh. 14 - Prob. 4CTCh. 14 - Prob. 5CTCh. 14 - Prob. 6CTCh. 14 - Prob. 7CTCh. 14 - Prob. 8CTCh. 14 - Prob. 9CTCh. 14 - Prob. 10CTCh. 14 - Prob. 11CTCh. 14 - Prob. 12CTCh. 14 - Prob. 13CTCh. 14 - Prob. 14CTCh. 14 - Prob. 15CTCh. 14 - Prob. 16CTCh. 14 - Prob. 17CTCh. 14 - 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