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

Chapter 7, Problem 12AE

To determine

To find: the lengths of the sides of a parallelogram similar to a given parallelogram and with perimeter 90 cm

Expert Solution & Answer

Answer to Problem 12AE

The lengths of sides of the parallelogram is 20 cm, 25 cm

Explanation of Solution

Given Information: The sides of a parallelogram are 12 cm and 15 cm. A similar parallelogram has a perimeter of 90 cm.

Formula used:

Two similar parallelograms satisfy the following conditions.

a) The ratio of lengths of corresponding sides of the two parallelograms is equal and is referred to as scale factor.

b) The perimeter of a parallelogram is the sum of its sides. Since the opposite sides of a parallelogram are equal in measure, it can also be expressed as twice the sum of adjacent sides.

Calculation: Consider the two parallelograms ABCD and A’B’C’D’.

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7, Problem 12AE , additional homework tip 1

The scale factor of the given parallelograms as per the definition in a) can be expressed as,

ABA'B'=ACA'C'

Substituting the measure of the sides of the parallelogram ABCD in the above expression,

15A'B'=12A'C'A'BA'C'=1512=54

Thus, one of the sides A’B’ can be expressed in terms of the other side A’C’ as ,

A'B'=54A'C' ............ (1)

Now, the perimeter of the first parallelogram ABCD is obtained as,

P1=2(AB+AC)=2(15+12)=54 cm

Now, the perimeter of the second parallelogram A’B’C’D’ is given as 90 cm. This can be

expressed in terms of its sides as,

P2=2(A'B'+A'C')=90

Substituting the relation in (1) , the perimeter can be rewritten as,

90=2(54A'C'+A'C')90=92A'C'A'C'=20 cm

The value of the other side A’B’ can be calculated by substituting the value of A’C’ in (1) as,

A'B'=54×20=25 cm

Chapter 7, Problem 11AE

Chapter 7, Problem 13AE

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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