a. Answer Δ A B C ∼ Δ E D C Explanation Δ A B C ∼ Δ E D C because these two are similar triangles by their corresponding angles. b. To determine To find: A B ? = A C ? = B C ? . Answer A B E D = A C E C = B C D C Explanation All these triangles are similar and these are their corresponding sides. c. To determine To find: The value of 15 ? = 21 ? and x . Answer 15 10 = 21 x and x = 14 Explanation Given: 15 ? = 21 ? From the figure, the ratio is as follows: 15 10 = 21 x 15 x = 210 x = 210 / 15 x = 15 d. To determine To find: The value of 15 ? = ? 12 and y . Answer 15 y = 10 12 and y =18. Explanation Given ratio: 15 ? = ? 12 From the figure, the ratio is as follows: In Δ D C E and Δ B C A D E / B A = C E / C A 12 y = 10 15 y 12 = 15 10 y =18

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

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 4ST2

To determine

To find: The value of ΔABC∼ _____.

Expert Solution

Answer to Problem 4ST2

ΔABC∼ΔEDC

Explanation of Solution

ΔABC∼ΔEDC because these two are similar triangles by their corresponding angles.

To determine

To find: AB?=AC?=BC?.

Expert Solution

Answer to Problem 4ST2

ABED=ACEC=BCDC

Explanation of Solution

All these triangles are similar and these are their corresponding sides.

To determine

To find:The value of 15?=21? and x .

Expert Solution

Answer to Problem 4ST2

1510=21x and x=14

Explanation of Solution

Given:

15?=21?

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.6, Problem 4ST2 , additional homework tip 1

From the figure, the ratio is as follows:

1510=21x

15x=210x=210/15x=15

To determine

To find:The value of 15?=?12 and y .

Expert Solution

Answer to Problem 4ST2

15y=1012 and y =18.

Explanation of Solution

Given ratio:

15?=?12

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.6, Problem 4ST2 , additional homework tip 2

From the figure, the ratio is as follows:

In ΔDCE and ΔBCADE/BA=CE/CA

12y=1015y12=1510

y =18

Chapter 7.6, Problem 3ST2

Chapter 7.6, Problem 5ST2

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 - 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Prob. 27WE Ch. 7.6 - Prob. 28WE Ch. 7.6 - Prob. 29WE Ch. 7.6 - Prob. 30WE Ch. 7.6 - Prob. 31WE Ch. 7.6 - Prob. 32WE Ch. 7.6 - Prob. 33WE Ch. 7.6 - Prob. 1ST2 Ch. 7.6 - Prob. 2ST2 Ch. 7.6 - Prob. 3ST2 Ch. 7.6 - Prob. 4ST2 Ch. 7.6 - Prob. 5ST2 Ch. 7.6 - Prob. 6ST2 Ch. 7.6 - Prob. 7ST2 Ch. 7.6 - Prob. 8ST2 Ch. 7.6 - Prob. 9ST2 Ch. 7.6 - Prob. 10ST2 Ch. 7.6 - Prob. 11ST2 Ch. 7.6 - Prob. 1E Ch. 7.6 - Prob. 2E Ch. 7.6 - Prob. 3E Ch. 7.6 - Prob. 4E Ch. 7.6 - Prob. 5E Ch. 7.6 - Prob. 6E Ch. 7.6 - Prob. 7E Ch. 7 - Prob. 1CR Ch. 7 - Prob. 2CR Ch. 7 - Prob. 3CR Ch. 7 - Prob. 4CR Ch. 7 - Prob. 5CR Ch. 7 - Prob. 6CR Ch. 7 - Prob. 7CR Ch. 7 - Prob. 8CR Ch. 7 - Prob. 9CR Ch. 7 - Prob. 10CR Ch. 7 - Prob. 11CR Ch. 7 - Prob. 12CR Ch. 7 - Prob. 13CR Ch. 7 - Prob. 14CR Ch. 7 - Prob. 15CR Ch. 7 - Prob. 16CR Ch. 7 - Prob. 17CR Ch. 7 - Prob. 18CR Ch. 7 - Prob. 19CR Ch. 7 - Prob. 20CR Ch. 7 - Prob. 21CR Ch. 7 - Prob. 22CR Ch. 7 - Prob. 23CR Ch. 7 - Prob. 24CR Ch. 7 - Prob. 1CT Ch. 7 - Prob. 2CT Ch. 7 - Prob. 3CT Ch. 7 - 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