Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 2 Equations, Inequalities, and Applications Chapter 2

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide Formulas and Additional Applications from Geometry

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 4 Objectives 1.Solve a formula for one variable, given the values of the other variables. 2.Use a formula to solve an applied problem. 3.Solve problems involving vertical angles and straight angles. 4.Solve a formula for a specified variable. 2.5 Formulas and Additional Applications from Geometry

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 5 36 = 2W P = 2L + 2WCheck: 2 · · 18 = 2.5 Formulas and Additional Applications from Geometry Solving a Formula for One Variable 18 = W Example 1 Find the value of the remaining variable. P = 2L + 2W; P = 52; L = 8 52 = 2 · 8 + 2W 52 = W – = 52

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 6 AREA FORMULAS  Triangle A = ½bh  Rectangle A = LW  Trapezoid A = ½h(b + B) 2.5 Formulas and Additional Applications from Geometry b = base h = height b h L = Length W = Width L W h = height b = small base B = large base b B h Solving a Formula for One Variable

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 7 Example 2 The area of a rectangular garden is 187 in 2 with a width of 17 in. What is the length of the garden? A = LW Check: 17 · 11 = Formulas and Additional Applications from Geometry Using a Formula to Solve an Applied Problem 187 = L · = L The length is 11 in. 17 L

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 8 63 = 9h Example 3 Bob is working on a sketch for a new underwater vehicle (UV), shown below. In his sketch, the bottom of the UV is 10 ft long, the top is 8 ft long, and the area is 63 ft 2. What is the height of his UV? A = ½h(b + B) Check: ½ · 7 · 18 = Formulas and Additional Applications from Geometry Using a Formula to Solve an Applied Problem 63 = ½h(8 + 10) 7 = h 63 = ½h(18) The height of the UV is 7ft h

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide The figure shows two intersecting lines forming angles that are numbered:,,, and. 2.5 Formulas and Additional Applications from Geometry Solving Problems Involving Vertical and Straight Angles 324 Angles and lie “opposite” each other. They are called vertical angles. Another pair of vertical angles is and. Vertical angles have equal measures. Now look at angles and. When their measures are added, we get the measure of a straight angle, which is 180°. There are three other such pairs of angles: and, and, and. and

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 10 –4 = 2x – 40 3x – 4 = 5x – 40 Check: 5 · 18 – 40 = 50° 2.5 Formulas and Additional Applications from Geometry 18 = x Example 4 Find the measure of each marked angle below. –3x Solving Problems Involving Vertical and Straight Angles (3x – 4)°(5x – 40)° Since the marked angles are vertical angles, they have equal measures = 2x Thus, both angles are 3 · 18 – 4 = 50° CAUTION Here, the answer was not the value of x!

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 11 (x + 15)° 6 · 25 – 10 = 140° and = 40° 7x + 5 = 180 6x – 10 + x + 15 = 180 Check: 140° + 40° = 180° 2.5 Formulas and Additional Applications from Geometry x = 25 Example 4 Find the measure of each marked angle below. – 5 Solving Problems Involving Vertical and Straight Angles (6x – 10)° Since the marked angles are straight angles, their sum will be 180°. 7x = 175 Thus, the angles are

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 12 Example 5 Solve A = ½bh for b. A = ½bh 2.5 Formulas and Additional Applications from Geometry Solving a Formula for a Specified Variable 2A = bh The goal is to get b alone on one side of the equation. 2 ·2 · · 2