(a) Answer 5.25 Explanation Given: Figure is shown below. Diagram which has to be completed is. Calculation: If a line segment parallel to one side of a triangle interests the other two sides, then it divides those sides proportionally. Proportion can be written as. ( 14 − x ) x = ( 16 − 6 ) 6 14 − x x = 10 6 ( 14 − x ) x = 5 3 3 × ( 14 − x ) = 5 × x 42 − 3 x = 5 x 8 x = 42 x = 42 8 x = 5.25 Therefore, the answer is 5.25 . (b) To determine To Find : the length of upper left segment in the diagram shown and complete as shown in another figure. Answer 5.25 Explanation Given: Figure is shown below. Diagram which has to be completed is. Calculation: If a line segment parallel to one side of a triangle interests the other two sides, then it divides those sides proportionally. Proportion can be written as. ( 14 − x ) x = 10 6 14 − x x = 10 6 ( 14 − x ) x = 5 3 3 × ( 14 − x ) = 5 × x 42 − 3 x = 5 x 8 x = 42 x = 42 8 x = 5.25 Therefore, the answer is 5.25 . (c) To determine To Find : the length of upper left segment in the diagram shown and complete as shown in another figure. Answer 5.25 Explanation Given: Figure is shown below. Diagram which has to be completed is. Calculation: Using addition. 14 = 3 y + 5 y 14 = 8 y y = 14 8 y = 1.75 Measure of upper length is. = 3 y = 3 × 1.75 = 5.25 Therefore, the answer is 5.25 .

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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Equations and inequalities describe the relationship between two mathematical expressions.

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Question

Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

Chapter 7.6, Problem 6CE

(a)

To determine

To Find: the length of upper left segment in the diagram shown and complete as shown in another figure.

(a)

Expert Solution

Answer to Problem 6CE

5.25

Explanation of Solution

Given:

Figure is shown below.

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.6, Problem 6CE , additional homework tip 1

Diagram which has to be completed is.

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.6, Problem 6CE , additional homework tip 2

Calculation:

If a line segment parallel to one side of a triangle interests the other two sides, then it divides those sides proportionally.

Proportion can be written as.

(14−x)x=(16−6)614−xx=106(14−x)x=533×(14−x)=5×x42−3x=5x8x=42x=428x=5.25

Therefore, the answer is 5.25 .

(b)

To determine

To Find: the length of upper left segment in the diagram shown and complete as shown in another figure.

(b)

Expert Solution

Answer to Problem 6CE

5.25

Explanation of Solution

Given:

Figure is shown below.

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.6, Problem 6CE , additional homework tip 3

Diagram which has to be completed is.

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.6, Problem 6CE , additional homework tip 4

Calculation:

If a line segment parallel to one side of a triangle interests the other two sides, then it divides those sides proportionally.

Proportion can be written as.

(14−x)x=10614−xx=106(14−x)x=533×(14−x)=5×x42−3x=5x8x=42x=428x=5.25

Therefore, the answer is 5.25 .

(c)

To determine

To Find: the length of upper left segment in the diagram shown and complete as shown in another figure.

(c)

Expert Solution

Answer to Problem 6CE

5.25

Explanation of Solution

Given:

Figure is shown below.

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.6, Problem 6CE , additional homework tip 5

Diagram which has to be completed is.

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.6, Problem 6CE , additional homework tip 6

Calculation:

Using addition.

14=3y+5y14=8yy=148y=1.75

Measure of upper length is.

=3y=3×1.75=5.25

Therefore, the answer is 5.25 .

Chapter 7.6, Problem 5CE

Chapter 7.6, Problem 7CE

Chapter 7 Solutions

McDougal Littell Jurgensen Geometry: Student Edition Geometry

Ch. 7.1 - Prob. 1CE Ch. 7.1 - Prob. 2CE Ch. 7.1 - Prob. 3CE Ch. 7.1 - Prob. 4CE Ch. 7.1 - Prob. 5CE Ch. 7.1 - Prob. 6CE Ch. 7.1 - Prob. 7CE Ch. 7.1 - Prob. 8CE Ch. 7.1 - Prob. 9CE Ch. 7.1 - Prob. 10CE

Ch. 7.1 - Prob. 11CE Ch. 7.1 - Prob. 12CE Ch. 7.1 - Prob. 13CE Ch. 7.1 - Prob. 14CE Ch. 7.1 - Prob. 15CE Ch. 7.1 - Prob. 16CE Ch. 7.1 - Prob. 17CE Ch. 7.1 - Prob. 18CE Ch. 7.1 - Prob. 1WE Ch. 7.1 - Prob. 2WE Ch. 7.1 - Prob. 3WE Ch. 7.1 - Prob. 4WE Ch. 7.1 - Prob. 5WE Ch. 7.1 - Prob. 6WE Ch. 7.1 - Prob. 7WE Ch. 7.1 - Prob. 8WE Ch. 7.1 - Prob. 9WE Ch. 7.1 - Prob. 10WE Ch. 7.1 - Prob. 11WE Ch. 7.1 - Prob. 12WE Ch. 7.1 - Prob. 13WE Ch. 7.1 - Prob. 14WE Ch. 7.1 - Prob. 15WE Ch. 7.1 - Prob. 16WE Ch. 7.1 - Prob. 17WE Ch. 7.1 - Prob. 18WE Ch. 7.1 - Prob. 19WE Ch. 7.1 - Prob. 20WE Ch. 7.1 - Prob. 21WE Ch. 7.1 - Prob. 22WE Ch. 7.1 - Prob. 23WE Ch. 7.1 - Prob. 24WE Ch. 7.1 - Prob. 25WE Ch. 7.1 - Prob. 26WE Ch. 7.1 - Prob. 27WE Ch. 7.1 - Prob. 28WE Ch. 7.1 - Prob. 29WE Ch. 7.1 - Prob. 30WE Ch. 7.1 - Prob. 31WE Ch. 7.1 - Prob. 32WE Ch. 7.1 - Prob. 33WE Ch. 7.1 - Prob. 34WE Ch. 7.1 - Prob. 35WE Ch. 7.1 - Prob. 36WE Ch. 7.2 - Prob. 1CE Ch. 7.2 - Prob. 2CE Ch. 7.2 - Prob. 3CE Ch. 7.2 - Prob. 4CE Ch. 7.2 - Prob. 5CE Ch. 7.2 - Prob. 6CE Ch. 7.2 - Prob. 7CE Ch. 7.2 - Prob. 8CE Ch. 7.2 - Prob. 9CE Ch. 7.2 - Prob. 10CE Ch. 7.2 - Prob. 11CE Ch. 7.2 - Prob. 12CE Ch. 7.2 - Prob. 13CE Ch. 7.2 - Prob. 14CE Ch. 7.2 - Prob. 15CE Ch. 7.2 - Prob. 16CE Ch. 7.2 - Prob. 1WE Ch. 7.2 - Prob. 2WE Ch. 7.2 - Prob. 3WE Ch. 7.2 - Prob. 4WE Ch. 7.2 - Prob. 5WE Ch. 7.2 - Prob. 6WE Ch. 7.2 - Prob. 7WE Ch. 7.2 - Prob. 8WE Ch. 7.2 - Prob. 9WE Ch. 7.2 - Prob. 10WE Ch. 7.2 - Prob. 11WE Ch. 7.2 - Prob. 12WE Ch. 7.2 - Prob. 13WE Ch. 7.2 - Prob. 14WE Ch. 7.2 - Prob. 15WE Ch. 7.2 - Prob. 16WE Ch. 7.2 - Prob. 17WE Ch. 7.2 - Prob. 18WE Ch. 7.2 - Prob. 19WE Ch. 7.2 - Prob. 20WE Ch. 7.2 - Prob. 21WE Ch. 7.2 - Prob. 22WE Ch. 7.2 - Prob. 23WE Ch. 7.2 - Prob. 24WE Ch. 7.2 - Prob. 25WE Ch. 7.2 - Prob. 26WE Ch. 7.2 - Prob. 27WE Ch. 7.2 - Prob. 28WE Ch. 7.2 - Prob. 29WE Ch. 7.2 - Prob. 30WE Ch. 7.2 - Prob. 31WE Ch. 7.2 - Prob. 32WE Ch. 7.2 - Prob. 33WE Ch. 7.2 - Prob. 34WE Ch. 7.2 - Prob. 35WE Ch. 7.2 - Prob. 36WE Ch. 7.2 - Prob. 37WE Ch. 7.2 - 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Prob. 34WE Ch. 7.3 - Prob. 35WE Ch. 7.3 - Prob. 36WE Ch. 7.3 - Prob. 37WE Ch. 7.3 - Prob. 1ST1 Ch. 7.3 - Prob. 2ST1 Ch. 7.3 - Prob. 3ST1 Ch. 7.3 - Prob. 4ST1 Ch. 7.3 - Prob. 5ST1 Ch. 7.3 - Prob. 6ST1 Ch. 7.3 - Prob. 7ST1 Ch. 7.3 - Prob. 8ST1 Ch. 7.3 - Prob. 9ST1 Ch. 7.3 - Prob. 10ST1 Ch. 7.3 - Prob. 11ST1 Ch. 7.3 - Prob. 12ST1 Ch. 7.3 - Prob. 13ST1 Ch. 7.3 - Prob. 14ST1 Ch. 7.3 - Prob. 15ST1 Ch. 7.3 - Prob. 1E Ch. 7.3 - Prob. 2E Ch. 7.4 - Prob. 1CE Ch. 7.4 - Prob. 2CE Ch. 7.4 - Prob. 3CE Ch. 7.4 - Prob. 4CE Ch. 7.4 - Prob. 5CE Ch. 7.4 - Prob. 6CE Ch. 7.4 - Prob. 7CE Ch. 7.4 - Prob. 8CE Ch. 7.4 - Prob. 9CE Ch. 7.4 - Prob. 10CE Ch. 7.4 - Prob. 11CE Ch. 7.4 - Prob. 12CE Ch. 7.4 - Prob. 13CE Ch. 7.4 - Prob. 1WE Ch. 7.4 - Prob. 2WE Ch. 7.4 - Prob. 3WE Ch. 7.4 - Prob. 4WE Ch. 7.4 - Prob. 5WE Ch. 7.4 - Prob. 6WE Ch. 7.4 - Prob. 7WE Ch. 7.4 - Prob. 8WE Ch. 7.4 - Prob. 9WE Ch. 7.4 - Prob. 10WE Ch. 7.4 - Prob. 11WE Ch. 7.4 - Prob. 12WE Ch. 7.4 - Prob. 13WE Ch. 7.4 - Prob. 14WE Ch. 7.4 - Prob. 15WE Ch. 7.4 - 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