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

See similar textbooks

Concept explainers

Percentage

A percentage is a number indicated as a fraction of 100. It is a dimensionless number often expressed using the symbol %.

Algebraic Expressions

In mathematics, an algebraic expression consists of constant(s), variable(s), and mathematical operators. It is made up of terms.

Numbers

Numbers are some measures used for counting. They can be compared one with another to know its position in the number line and determine which one is greater or lesser than the other.

Subtraction

Before we begin to understand the subtraction of algebraic expressions, we need to list out a few things that form the basis of algebra.

Addition

Before we begin to understand the addition of algebraic expressions, we need to list out a few things that form the basis of algebra.

Videos

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 5, Problem 16CUR

To determine

To Find: The numerical length of each side of triangle SUN such that ∠S≅∠N,SU≅2x+7 , UN=4x−1 and SN=3x+4 .

Expert Solution & Answer

Answer to Problem 16CUR

SN=16,UN=15 and SU=15

Explanation of Solution

Given:

In a triangle SUN, ∠S≅∠N,SU≅2x+7 , UN=4x−1 and SN=3x+4 .

Calculation:

In triangle if two angles are equal the sides opposite to these angles are equal.

Since, ∠S≅∠N .

Therefore, SU=UN .

SU=UN2x+7=4x−12x−4x=−1−7−2x=−8x=4

Substituting x=4 in SN=3x+4 .

SN=3x+4SN=3×4+4SN=16

Substituting x=4 in UN=4x−1 .

UN=4x−1UN=4×4−1UN=15

Substituting x=4 in SU≅2x+7 .

SU=2x+7SU=2×4+7SU=15

Therefore, the answer is SN=16,UN=15 and SU=15 .

Chapter 5, Problem 15CUR

Chapter 5, Problem 17CUR

Chapter 5 Solutions

McDougal Littell Jurgensen Geometry: Student Edition Geometry

Ch. 5.1 - Prob. 1CE Ch. 5.1 - Prob. 2CE Ch. 5.1 - Prob. 3CE Ch. 5.1 - Prob. 4CE Ch. 5.1 - Prob. 5CE Ch. 5.1 - Prob. 6CE Ch. 5.1 - Prob. 7CE Ch. 5.1 - Prob. 8CE Ch. 5.1 - Prob. 9CE Ch. 5.1 - Prob. 10CE

Ch. 5.1 - Prob. 11CE Ch. 5.1 - Prob. 12CE Ch. 5.1 - Prob. 13CE Ch. 5.1 - Prob. 14CE Ch. 5.1 - Prob. 15CE Ch. 5.1 - Prob. 16CE Ch. 5.1 - Prob. 17CE Ch. 5.1 - Prob. 18CE Ch. 5.1 - Prob. 1WE Ch. 5.1 - Prob. 2WE Ch. 5.1 - Prob. 3WE Ch. 5.1 - Prob. 4WE Ch. 5.1 - Prob. 5WE Ch. 5.1 - Prob. 6WE Ch. 5.1 - Prob. 7WE Ch. 5.1 - Prob. 8WE Ch. 5.1 - Prob. 9WE Ch. 5.1 - Prob. 10WE Ch. 5.1 - Prob. 11WE Ch. 5.1 - Prob. 12WE Ch. 5.1 - Prob. 13WE Ch. 5.1 - Prob. 14WE Ch. 5.1 - Prob. 15WE Ch. 5.1 - Prob. 16WE Ch. 5.1 - Prob. 17WE Ch. 5.1 - Prob. 18WE Ch. 5.1 - Prob. 19WE Ch. 5.1 - Prob. 20WE Ch. 5.1 - Prob. 21WE Ch. 5.1 - Prob. 22WE Ch. 5.1 - Prob. 23WE Ch. 5.1 - Prob. 24WE Ch. 5.1 - Prob. 25WE Ch. 5.1 - Prob. 26WE Ch. 5.1 - Prob. 27WE Ch. 5.1 - Prob. 28WE Ch. 5.1 - Prob. 29WE Ch. 5.1 - Prob. 30WE Ch. 5.1 - Prob. 31WE Ch. 5.1 - Prob. 32WE Ch. 5.1 - Prob. 33WE Ch. 5.1 - Prob. 34WE Ch. 5.1 - Prob. 35WE Ch. 5.1 - Prob. 36WE Ch. 5.1 - Prob. 37WE Ch. 5.1 - Prob. 38WE Ch. 5.1 - Prob. 39WE Ch. 5.2 - Prob. 1CE Ch. 5.2 - Prob. 2CE Ch. 5.2 - Prob. 3CE Ch. 5.2 - Prob. 4CE Ch. 5.2 - Prob. 5CE Ch. 5.2 - Prob. 6CE Ch. 5.2 - Prob. 7CE Ch. 5.2 - Prob. 8CE Ch. 5.2 - Prob. 9CE Ch. 5.2 - Prob. 10CE Ch. 5.2 - Prob. 11CE Ch. 5.2 - Prob. 12CE Ch. 5.2 - Prob. 13CE Ch. 5.2 - Prob. 1WE Ch. 5.2 - Prob. 2WE Ch. 5.2 - Prob. 3WE Ch. 5.2 - Prob. 4WE Ch. 5.2 - Prob. 5WE Ch. 5.2 - Prob. 6WE Ch. 5.2 - Prob. 7WE Ch. 5.2 - Prob. 8WE Ch. 5.2 - Prob. 9WE Ch. 5.2 - Prob. 10WE Ch. 5.2 - Prob. 11WE Ch. 5.2 - Prob. 12WE Ch. 5.2 - Prob. 13WE Ch. 5.2 - Prob. 14WE Ch. 5.2 - Prob. 15WE Ch. 5.2 - Prob. 16WE Ch. 5.2 - Prob. 17WE Ch. 5.2 - Prob. 18WE Ch. 5.2 - Prob. 19WE Ch. 5.2 - Prob. 20WE Ch. 5.2 - Prob. 21WE Ch. 5.2 - Prob. 22WE Ch. 5.2 - Prob. 23WE Ch. 5.2 - Prob. 24WE Ch. 5.2 - Prob. 25WE Ch. 5.3 - Prob. 1CE Ch. 5.3 - Prob. 2CE Ch. 5.3 - Prob. 3CE Ch. 5.3 - Prob. 4CE Ch. 5.3 - Prob. 5CE Ch. 5.3 - Prob. 6CE Ch. 5.3 - Prob. 7CE Ch. 5.3 - Prob. 8CE Ch. 5.3 - Prob. 9CE Ch. 5.3 - Prob. 1WE Ch. 5.3 - Prob. 2WE Ch. 5.3 - Prob. 3WE Ch. 5.3 - Prob. 4WE Ch. 5.3 - Prob. 5WE Ch. 5.3 - Prob. 6WE Ch. 5.3 - Prob. 7WE Ch. 5.3 - Prob. 8WE Ch. 5.3 - Prob. 9WE Ch. 5.3 - Prob. 10WE Ch. 5.3 - Prob. 11WE Ch. 5.3 - Prob. 12WE Ch. 5.3 - Prob. 13WE Ch. 5.3 - Prob. 14WE Ch. 5.3 - Prob. 15WE Ch. 5.3 - Prob. 16WE Ch. 5.3 - Prob. 17WE Ch. 5.3 - Prob. 18WE Ch. 5.3 - Prob. 19WE Ch. 5.3 - Prob. 20WE Ch. 5.3 - Prob. 21WE Ch. 5.3 - Prob. 22WE Ch. 5.3 - Prob. 23WE Ch. 5.3 - Prob. 24WE Ch. 5.3 - Prob. 25WE Ch. 5.3 - Prob. 1ST1 Ch. 5.3 - Prob. 2ST1 Ch. 5.3 - Prob. 3ST1 Ch. 5.3 - Prob. 4ST1 Ch. 5.3 - Prob. 5ST1 Ch. 5.3 - Prob. 6ST1 Ch. 5.3 - Prob. 7ST1 Ch. 5.3 - Prob. 8ST1 Ch. 5.3 - Prob. 1E Ch. 5.3 - Prob. 2E Ch. 5.3 - Prob. 3E Ch. 5.3 - Prob. 4E Ch. 5.3 - Prob. 5E Ch. 5.3 - Prob. 6E Ch. 5.4 - Prob. 1CE Ch. 5.4 - Prob. 2CE Ch. 5.4 - Prob. 3CE Ch. 5.4 - Prob. 4CE Ch. 5.4 - Prob. 5CE Ch. 5.4 - Prob. 6CE Ch. 5.4 - Prob. 7CE Ch. 5.4 - Prob. 8CE Ch. 5.4 - Prob. 9CE Ch. 5.4 - Prob. 10CE Ch. 5.4 - Prob. 11CE Ch. 5.4 - Prob. 1WE Ch. 5.4 - Prob. 2WE Ch. 5.4 - Prob. 3WE Ch. 5.4 - Prob. 4WE Ch. 5.4 - Prob. 5WE Ch. 5.4 - Prob. 6WE Ch. 5.4 - Prob. 7WE Ch. 5.4 - Prob. 8WE Ch. 5.4 - Prob. 9WE Ch. 5.4 - Prob. 10WE Ch. 5.4 - Prob. 11WE Ch. 5.4 - Prob. 12WE Ch. 5.4 - Prob. 13WE Ch. 5.4 - Prob. 14WE Ch. 5.4 - Prob. 15WE Ch. 5.4 - Prob. 16WE Ch. 5.4 - Prob. 17WE Ch. 5.4 - Prob. 18WE Ch. 5.4 - Prob. 19WE Ch. 5.4 - Prob. 20WE Ch. 5.4 - Prob. 21WE Ch. 5.4 - Prob. 22WE Ch. 5.4 - Prob. 23WE Ch. 5.4 - Prob. 24WE Ch. 5.4 - Prob. 25WE Ch. 5.4 - Prob. 26WE Ch. 5.4 - Prob. 27WE Ch. 5.4 - Prob. 28WE Ch. 5.4 - Prob. 29WE Ch. 5.4 - Prob. 30WE Ch. 5.4 - Prob. 31WE Ch. 5.4 - Prob. 32WE Ch. 5.4 - Prob. 33WE Ch. 5.4 - Prob. 34WE Ch. 5.4 - Prob. 35WE Ch. 5.4 - Prob. 36WE Ch. 5.4 - Prob. 37WE Ch. 5.4 - Prob. 38WE Ch. 5.4 - Prob. 39WE Ch. 5.4 - Prob. 40WE Ch. 5.4 - Prob. 41WE Ch. 5.4 - Prob. 42WE Ch. 5.4 - Prob. 1MRE Ch. 5.4 - Prob. 2MRE Ch. 5.4 - Prob. 3MRE Ch. 5.4 - Prob. 4MRE Ch. 5.4 - Prob. 5MRE Ch. 5.4 - Prob. 6MRE Ch. 5.4 - Prob. 7MRE Ch. 5.4 - Prob. 8MRE Ch. 5.4 - Prob. 9MRE Ch. 5.5 - Prob. 1CE Ch. 5.5 - Prob. 2CE Ch. 5.5 - Prob. 3CE Ch. 5.5 - Prob. 4CE Ch. 5.5 - Prob. 5CE Ch. 5.5 - Prob. 6CE Ch. 5.5 - Prob. 7CE Ch. 5.5 - Prob. 8CE Ch. 5.5 - Prob. 9CE Ch. 5.5 - Prob. 10CE Ch. 5.5 - Prob. 11CE Ch. 5.5 - Prob. 1WE Ch. 5.5 - Prob. 2WE Ch. 5.5 - Prob. 3WE Ch. 5.5 - Prob. 4WE Ch. 5.5 - Prob. 5WE Ch. 5.5 - Prob. 6WE Ch. 5.5 - Prob. 7WE Ch. 5.5 - Prob. 8WE Ch. 5.5 - Prob. 9WE Ch. 5.5 - Prob. 10WE Ch. 5.5 - Prob. 11WE Ch. 5.5 - Prob. 12WE Ch. 5.5 - Prob. 13WE Ch. 5.5 - Prob. 14WE Ch. 5.5 - Prob. 15WE Ch. 5.5 - Prob. 16WE Ch. 5.5 - Prob. 17WE Ch. 5.5 - Prob. 18WE Ch. 5.5 - Prob. 19WE Ch. 5.5 - Prob. 20WE Ch. 5.5 - Prob. 21WE Ch. 5.5 - Prob. 22WE Ch. 5.5 - Prob. 23WE Ch. 5.5 - Prob. 24WE Ch. 5.5 - Prob. 25WE Ch. 5.5 - Prob. 26WE Ch. 5.5 - Prob. 27WE Ch. 5.5 - Prob. 28WE Ch. 5.5 - Prob. 29WE Ch. 5.5 - Prob. 30WE Ch. 5.5 - Prob. 31WE Ch. 5.5 - Prob. 32WE Ch. 5.5 - Prob. 33WE Ch. 5.5 - Prob. 34WE Ch. 5.5 - Prob. 35WE Ch. 5.5 - Prob. 36WE Ch. 5.5 - Prob. 1ST2 Ch. 5.5 - Prob. 2ST2 Ch. 5.5 - Prob. 3ST2 Ch. 5.5 - Prob. 4ST2 Ch. 5.5 - Prob. 5ST2 Ch. 5.5 - Prob. 6ST2 Ch. 5.5 - Prob. 7ST2 Ch. 5.5 - Prob. 8ST2 Ch. 5 - Prob. 1CR Ch. 5 - Prob. 2CR Ch. 5 - Prob. 3CR Ch. 5 - Prob. 4CR Ch. 5 - Prob. 5CR Ch. 5 - Prob. 6CR Ch. 5 - Prob. 7CR Ch. 5 - Prob. 8CR Ch. 5 - Prob. 9CR Ch. 5 - Prob. 10CR Ch. 5 - Prob. 11CR Ch. 5 - Prob. 12CR Ch. 5 - Prob. 13CR Ch. 5 - Prob. 14CR Ch. 5 - Prob. 15CR Ch. 5 - Prob. 16CR Ch. 5 - Prob. 17CR Ch. 5 - Prob. 18CR Ch. 5 - Prob. 19CR Ch. 5 - Prob. 20CR Ch. 5 - Prob. 21CR Ch. 5 - Prob. 22CR Ch. 5 - Prob. 1CT Ch. 5 - Prob. 2CT Ch. 5 - Prob. 3CT Ch. 5 - Prob. 4CT Ch. 5 - Prob. 5CT Ch. 5 - Prob. 6CT Ch. 5 - Prob. 7CT Ch. 5 - Prob. 8CT Ch. 5 - Prob. 9CT Ch. 5 - Prob. 10CT Ch. 5 - Prob. 11CT Ch. 5 - Prob. 12CT Ch. 5 - Prob. 13CT Ch. 5 - Prob. 14CT Ch. 5 - Prob. 15CT Ch. 5 - Prob. 16CT Ch. 5 - Prob. 17CT Ch. 5 - Prob. 18CT Ch. 5 - Prob. 19CT Ch. 5 - Prob. 1CUR Ch. 5 - Prob. 2CUR Ch. 5 - Prob. 3CUR Ch. 5 - Prob. 4CUR Ch. 5 - Prob. 5CUR Ch. 5 - Prob. 6CUR Ch. 5 - Prob. 7CUR Ch. 5 - Prob. 8CUR Ch. 5 - Prob. 9CUR Ch. 5 - Prob. 10CUR Ch. 5 - Prob. 11CUR Ch. 5 - Prob. 12CUR Ch. 5 - Prob. 13CUR Ch. 5 - Prob. 14CUR Ch. 5 - Prob. 15CUR Ch. 5 - Prob. 16CUR Ch. 5 - Prob. 17CUR Ch. 5 - Prob. 18CUR Ch. 5 - Prob. 19CUR Ch. 5 - Prob. 20CUR Ch. 5 - Prob. 21CUR Ch. 5 - Prob. 22CUR Ch. 5 - Prob. 23CUR Ch. 5 - Prob. 24CUR Ch. 5 - Prob. 25CUR Ch. 5 - Prob. 26CUR Ch. 5 - Prob. 27CUR Ch. 5 - Prob. 28CUR Ch. 5 - Prob. 29CUR Ch. 5 - Prob. 30CUR Ch. 5 - Prob. 31CUR

Additional Math Textbook Solutions

Find more solutions based on key concepts

A categorical variable has three categories, with the following frequencies of occurrence: a. Compute the perce...

Basic Business Statistics, Student Value Edition

In Problems 19-30, state the domain and range for each relation. Then determine whether each relation represent...

Precalculus

Find the areas of the parallelograms whose vertices are given in Exercises 35–40. 35. A(1, 0), B(0, 1), C(−1, 0...

University Calculus: Early Transcendentals (4th Edition)

The equivalent expression of x(y+z) by using the commutative property.

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

Maximum product What two nonnegative real numbers with a sum of 23 have the largest possible product?

Calculus: Early Transcendentals (2nd Edition)

Interpreting a P-Value In Exercises 3–8, the P-value for a hypothesis test is shown. Use the P-value to decide ...

Elementary Statistics: Picturing the World (7th Edition)

Knowledge Booster

$Geometry$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, geometry and related others by exploring similar questions and additional content below.

The numerical length of each side of triangle S U N such that ∠ S ≅ ∠ N , S U ≅ 2 x + 7 , U N = 4 x − 1 and S N = 3 x + 4 .