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
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’.
The scale factor of the given parallelograms as per the definition in a) can be expressed as,
Substituting the measure of the sides of the parallelogram ABCD in the above expression,
Thus, one of the sides A’B’ can be expressed in terms of the other side A’C’ as ,
Now, the perimeter of the first parallelogram ABCD is obtained as,
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,
Substituting the relation in (1) , the perimeter can be rewritten as,
The value of the other side A’B’ can be calculated by substituting the value of A’C’ in (1) as,
Chapter 7 Solutions
McDougal Littell Jurgensen Geometry: Student Edition Geometry
Ch. 7.1 - Prob. 1CECh. 7.1 - Prob. 2CECh. 7.1 - Prob. 3CECh. 7.1 - Prob. 4CECh. 7.1 - Prob. 5CECh. 7.1 - Prob. 6CECh. 7.1 - Prob. 7CECh. 7.1 - Prob. 8CECh. 7.1 - Prob. 9CECh. 7.1 - Prob. 10CE
Ch. 7.1 - Prob. 11CECh. 7.1 - Prob. 12CECh. 7.1 - Prob. 13CECh. 7.1 - Prob. 14CECh. 7.1 - Prob. 15CECh. 7.1 - Prob. 16CECh. 7.1 - Prob. 17CECh. 7.1 - Prob. 18CECh. 7.1 - Prob. 1WECh. 7.1 - Prob. 2WECh. 7.1 - Prob. 3WECh. 7.1 - Prob. 4WECh. 7.1 - Prob. 5WECh. 7.1 - Prob. 6WECh. 7.1 - Prob. 7WECh. 7.1 - Prob. 8WECh. 7.1 - Prob. 9WECh. 7.1 - Prob. 10WECh. 7.1 - Prob. 11WECh. 7.1 - Prob. 12WECh. 7.1 - Prob. 13WECh. 7.1 - Prob. 14WECh. 7.1 - Prob. 15WECh. 7.1 - Prob. 16WECh. 7.1 - Prob. 17WECh. 7.1 - Prob. 18WECh. 7.1 - Prob. 19WECh. 7.1 - Prob. 20WECh. 7.1 - Prob. 21WECh. 7.1 - Prob. 22WECh. 7.1 - Prob. 23WECh. 7.1 - Prob. 24WECh. 7.1 - Prob. 25WECh. 7.1 - Prob. 26WECh. 7.1 - Prob. 27WECh. 7.1 - Prob. 28WECh. 7.1 - Prob. 29WECh. 7.1 - Prob. 30WECh. 7.1 - Prob. 31WECh. 7.1 - Prob. 32WECh. 7.1 - Prob. 33WECh. 7.1 - Prob. 34WECh. 7.1 - Prob. 35WECh. 7.1 - Prob. 36WECh. 7.2 - Prob. 1CECh. 7.2 - Prob. 2CECh. 7.2 - Prob. 3CECh. 7.2 - Prob. 4CECh. 7.2 - Prob. 5CECh. 7.2 - Prob. 6CECh. 7.2 - Prob. 7CECh. 7.2 - 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Prob. 23WECh. 7.5 - Prob. 1MRECh. 7.5 - Prob. 2MRECh. 7.6 - Prob. 1CECh. 7.6 - Prob. 2CECh. 7.6 - Prob. 3CECh. 7.6 - Prob. 4CECh. 7.6 - Prob. 5CECh. 7.6 - Prob. 6CECh. 7.6 - Prob. 7CECh. 7.6 - Prob. 8CECh. 7.6 - Prob. 9CECh. 7.6 - Prob. 1WECh. 7.6 - Prob. 2WECh. 7.6 - Prob. 3WECh. 7.6 - Prob. 4WECh. 7.6 - Prob. 5WECh. 7.6 - Prob. 6WECh. 7.6 - Prob. 7WECh. 7.6 - Prob. 8WECh. 7.6 - Prob. 9WECh. 7.6 - Prob. 10WECh. 7.6 - Prob. 11WECh. 7.6 - Prob. 12WECh. 7.6 - Prob. 13WECh. 7.6 - Prob. 14WECh. 7.6 - Prob. 15WECh. 7.6 - Prob. 16WECh. 7.6 - Prob. 17WECh. 7.6 - Prob. 18WECh. 7.6 - Prob. 19WECh. 7.6 - Prob. 20WECh. 7.6 - Prob. 21WECh. 7.6 - Prob. 22WECh. 7.6 - Prob. 23WECh. 7.6 - Prob. 24WECh. 7.6 - Prob. 25WECh. 7.6 - Prob. 26WECh. 7.6 - Prob. 27WECh. 7.6 - Prob. 28WECh. 7.6 - Prob. 29WECh. 7.6 - Prob. 30WECh. 7.6 - Prob. 31WECh. 7.6 - Prob. 32WECh. 7.6 - Prob. 33WECh. 7.6 - Prob. 1ST2Ch. 7.6 - Prob. 2ST2Ch. 7.6 - Prob. 3ST2Ch. 7.6 - Prob. 4ST2Ch. 7.6 - Prob. 5ST2Ch. 7.6 - Prob. 6ST2Ch. 7.6 - 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