1. Chapter 2 Equations and Inequalities in One Variable Section 7 Problem Solving: Geometry and Uniform Motion

2. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 2 © 2010 Pearson Education, Inc. Section 2.7 Objectives 1Set Up and Solve Complementary and Supplementary Angle Problems 2Set Up and Solve Angles of Triangle Problems 3Use Geometry Formulas to Solve Problems 4Set Up and Solve Uniform Motion Problems

3. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 3 © 2010 Pearson Education, Inc. Angles Two angles whose sum is 90° are called complementary angles. Each angle is called the complement of the other. Two angles whose sum is 180° are called supplementary angles. Each angle is called the supplement of the other. (90 – x)° x°x° complementary angles x°x° (180 – x)° supplementary angles

4. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 4 © 2010 Pearson Education, Inc. Solving Angle Problems Example: Angle A and angle B are complementary angles, and angle A is 21º more than twice angle B. Find the measure of both angles. Step 1: Identify This is complementary problem. We are looking for the measure of the two angles whose sum is 90°. Step 2: Name Let a represent the measure of angle A. B A Continued.

5. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 5 © 2010 Pearson Education, Inc. Solving Angle Problems Example continued: Step 3: Translate Angle A is 21° more than twice the measure of angle B. a = 21 + 2 · m  B a = 21 + 2 (90 – a) Continued. Step 4: Solve Solve the equation. a = 21 + 2 (90 – a) a = 21 + 180 – 2a Distribute. a = 201 – 2a Combine like terms.

6. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 6 © 2010 Pearson Education, Inc. Solving Angle Problems Example continued: a = 201 – 2a 3a = 201 Add 2a to both sides. a = 67 Divide both sides by 3. Step 5: Check The measure of  A is 67°. The measure of  B is 90° – a = 90° – 67° = 23°. Step 6: Answer the Question 67° + 23° = 90° The two complementary angles measure 67° and 23°.

7. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 7 © 2010 Pearson Education, Inc. Solving Triangle Problems Example: Find the measure of Angle C. Step 1: Identify This is an “angles of the triangle” problem. Step 2: Name Let c represent the measure of angle C. Continued. 42° 84° AB C Remember that the sum of the measures of the interior angles of a triangle is 180°.

8. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 8 © 2010 Pearson Education, Inc. Solving Triangle Problems Example continued: Continued. Step 3: Translate The three angles add up to 180°. Step 4: Solve Solve the equation. 84 + 42 + c = 180 126 + c = 180 Combine like terms. c = 54 Subtract 126 from both sides.

9. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 9 © 2010 Pearson Education, Inc. Solving Triangle Problems Example continued: 84° + 42° + 54° = 180° Step 5: Check Is the sum of the three angles equal to 180°? 180° = 180° Step 6: Answer the Question The measure of Angle C is 54°. 42° 84° AB C 54°

10. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 10 © 2010 Pearson Education, Inc. Solving Geometry Problems Example: Julie is making cone-shaped candles. The mold for the candles is 4 in. in diameter and 7 in. high. How many cubic inches of wax does Julie need to buy if she wants to make 50 candles? Step 1: Identify This is a geometry volume problem. We want to find the volume of the cone-shaped candle to determine the amount of wax needed for 50 candles. Step 2: Name Let V represent the volume of one cone. Continued.

11. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 11 © 2010 Pearson Education, Inc. Solving Geometry Problems Example continued: Continued. Step 3: Translate We need to use the formula for the volume of a cone. Step 4: Solve Solve the equation.   3.14; r = 2; h = 7

12. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 12 © 2010 Pearson Education, Inc. Solving Geometry Problems Example continued: This is the amount needed for one candle. 29.3 in. 3  50  1465 in. 3 This is the amount needed for 50 candles. Step 5: Check Step 6: Answer the Question Julie needs to buy approximately 1465 in. 3 of wax to make 50 candles. 29.3 in. 3  50 = 1465 in. 3

13. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 13 © 2010 Pearson Education, Inc. Uniform Motion Objects that move at a constant velocity (speed) are said to be in uniform motion. Uniform Motion Formula If an object moves at an average speed r, the distance d covered in time t is given by the formula d = rt. Rate·Time=Distance Object #1distance 1 Object #2distance 2 The following table is helpful in solving motion problems.

14. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 14 © 2010 Pearson Education, Inc. Uniform Motion Problem Example: Steve jogs at an average rate of 8 kilometers per hour. How long would it take him to jog 14 kilometers? Step 1: Identify This is a uniform motion problem. We are looking for the length of time it would take Steve to jog 14 kilometers. Step 2: Name Let t represent the length of time it would take Steve to jog 14 kilometers. Continued.

15. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 15 © 2010 Pearson Education, Inc. Uniform Motion Problem Example continued: Step 4: Solve Continued. Divide both sides by 8. Step 3: Translate Organize the information in a table. Distance, kmRate, kphTime, hours 148t 14 = 8t d = rt Simplify.

16. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 16 © 2010 Pearson Education, Inc. Uniform Motion Problem Example continued: Step 5: Check t = 1.75 represents the length of time. Step 6: Answer the Question 14 = (8)(1.75) It takes Mark 1.75 hours (or 1 hour and 45 minutes) to run 14 kilometers. d = rt 14 = 14

17. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 17 © 2010 Pearson Education, Inc. Uniform Motion Problem Example: Nina drove her car to Cleveland while Paula drove her car to Columbus. Nina drove 360 kilometers while Paula drove 280 kilometers. Nina drove 20 kilometers per hour faster than Paula on her trip. What was the average speed in kilometers per hour for each driver? Let r = the rate of Paula’s car. Step 1: Identify Continued. Let r + 20 = the rate of Nina’s car. Distance problems can be solved using the formula distance = rate × time (d = rt). The time, t, for each driver was the same. Step 2: Name

18. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 18 © 2010 Pearson Education, Inc. Uniform Motion Problem Continued. Step 4: Solve Example continued: r280Paula r + 20360Nina rd Since the time for each driver was the same, we can set the times equal to each other. Step 3: Translate

19. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 19 © 2010 Pearson Education, Inc. Uniform Motion Problem Example continued: Multiply both sides by r(r + 20). Simplify. Subtract 280r from each side. Paula’s rate was 70 kilometers per hour. Nina’s rate was r + 20 = 90 kilometers per hour. Divide each side by 80. Distribute. Continued.

20. Sullivan & Struve, Elementary and Intermediate Algebra, 2/e2.7 - 20 © 2010 Pearson Education, Inc. Uniform Motion Problem Example continued: Step 5: Check Step 6: Answer the Question Paula’s rate was 70 kilometers per hour. Nina’s rate was 90 kilometers per hour.