1. Chapter 3 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-1 Applications of Algebra

2. 2 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-2 3.1 – Changing Application Problems into Equations 3.2 – Solving Application Problems 3.3 – Geometric Problems 3.4 – Motion, Money and Mixture Problems Chapter Sections

3. 3 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-3 Geometric Problems

4. 4 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-4 Solving Geometric Problems Two angles are complementary angles if the sum of their measures is 90º. Two angles are supplementary angles if the sum of their measures is 180º. When two lines cross, the opposite angles are called vertical angles.

5. 5 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-5 Solving Geometric Problems Example: Christine O’Conner is planning to build a sandbox for her daughter. She has 30 feet of lumber with which to build the perimeter. What should be the dimensions of the rectangular sandbox if the length is to be 3 feet longer than the width. Let w = the width of the sandbox Then w + 3 = length of the sandbox P = 2l + 2w 30 = 2(w + 3 ) + 2w Solve the equation. Continued.

6. 6 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-6 Solving Geometric Problems 30 = 2(w + 3) + 2w 30 = 2w + 6 + 2w 30 = 6 + 4w 24 = 4w 6 = w The width is 6 feet. The length = 3 + w = 6 + 3 = 9 feet Check: 2(9) + 2(6) = 30 Example continued:

7. 7 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-7 Solving Geometric Problems Example: Mr. and Mrs. Harmon Katz have a corner lot that is in the shape of an isosceles triangle. Two angles of their triangular lot have the same measure and the measure of the third angle is 30 degrees greater than the measure of the other two. Find the measure of all three angles. Let x = the measure of each smaller angle. Then x + 30 = the measure of the larger angle. Sum of the 3 angles = 180 Continued.

8. 8 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-8 Solving Geometric Problems x + x + (x + 30) = 180 3x + 30 = 180 3x = 150 x = 150/3 = 50 Check: 50° + 50° + 80° = 180° The two smaller angles each measure 50º. The measure of the larger angle is x + 30° or 50° + 30° = 80°. Example continued: Carry Out