Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00 Edition
ISBN: 9780395977224
Author: Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher: McDougal Littell
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Question
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Chapter 3.8, Problem 1ST
To determine

To find the dimensions of each rectangle and make a sketch of them.

Expert Solution & Answer
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Answer to Problem 1ST

First rectangle:

Length: 59cm.

Width: 51cm.

Second rectangle:

Length: 65cm.

Width: 56cm.

Explanation of Solution

Given information:

The length of first rectangle is 8 cm more than its width.

The second rectangle is 6 cm longer and 5 cm wider than the first rectangle.

The perimeter of second rectangle is 242 cm.

Formula Used : Perimeter of a rectangle = 2(length + width).

Calculation:

Let the width of the first rectangle = x cm

Length of first rectangle is 8cm more than width = x+8 cm

  Algebra: Structure And Method, Book 1, Chapter 3.8, Problem 1ST , additional homework tip 1

Width of second rectangle (5cm more than width of first rectangle) =

  x+5 cm

Length of second rectangle(5cm more than length of first rectangle) = x+8+6=x+14 cm

  Algebra: Structure And Method, Book 1, Chapter 3.8, Problem 1ST , additional homework tip 2

Perimeter of second rectangle = 2(l1+w1) , where l1 and w1 are length and width of second rectangle.

Substituting l1=x+5, w1=x+14 and Perimerter = 242  in above formula we get

  242=2(x+5+(x+14))

Switching sides

  2(x+5+(x+14))=242

Simplify

  2(x+5+x+14)=242

  ⇒2(2x+19)=242

Distribute 2 over (2x+19)

  4x+38=242

Subtract 38 on both sides:

  4x+38−38=242−38

  ⇒4x=204

Divide both sides by 4

  4x4=2044

  ⇒x=51

So, the width of first rectangle is 51cm.

The length of first rectangle = x+8=51+8=59cm.

Therefore, the width of second rectangle = x+5=51+5=56cm.

The length of the second rectangle = x+14=51+14=65cm.

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Chapter 3.8, Problem 12MRE
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Chapter 3.8, Problem 2ST

Chapter 3 Solutions

Algebra: Structure And Method, Book 1

Ch. 3.1 - Prob. 1OECh. 3.1 - Prob. 2OECh. 3.1 - Prob. 3OECh. 3.1 - Prob. 4OECh. 3.1 - Prob. 5OECh. 3.1 - Prob. 6OECh. 3.1 - Prob. 7OECh. 3.1 - Prob. 8OECh. 3.1 - Prob. 9OECh. 3.1 - Prob. 10OE
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