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

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 3

2.5

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 + 2W Check: 2 · 8 + 2 · 18 =

2.5 Formulas and Additional Applications from Geometry

Solving a Formula for One Variable

2

18 = W

Example 1

Find the value of the remaining variable.

P = 2L + 2W; P = 52; L = 8

52 = 2 · 8 + 2W

52 = 16 + 2W–16 –16

16 + 36 = 52

2

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 6

AREA FORMULAS

TriangleA = ½bh

RectangleA = LW

TrapezoidA = ½h(b + B)

2.5 Formulas and Additional Applications from Geometry

b = baseh = height

b

h

L = LengthW = Width

LW

h = heightb = small baseB = large base

b

Bh

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 in2 with a width of 17 in. What is the length of the garden?

A = LW

Check: 17 · 11 = 187

2.5 Formulas and Additional Applications from Geometry

Using a Formula to Solve an Applied Problem

187 = L · 1717

11 = L

The length is 11 in.

1717

L

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 8

63 = 9h9

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 ft2. What is the height of his UV?

A = ½h(b + B)

Check:

½ · 7 · 18 = 63

2.5 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.

10

8h

9

Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.5 – Slide 9

23

1The 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

32 4

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 .

14

1 32 4

1 2

2 3 3 44 1

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

2

18 = x

Example 4

Find the measure of each marked angle below.

–3x –3x

2

Solving Problems Involving Vertical and Straight Angles

(3x – 4)° (5x – 40)°

Since the marked angles are vertical angles, they have equal measures.

+ 40 + 40

36 = 2x

Thus, both angles are

3 · 18 – 4 = 50°CAUTIONHere, 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°and25 + 15 = 40°

7x + 5 = 180

6x – 10 + x + 15 = 180

Check:

140° + 40° = 180°

2.5 Formulas and Additional Applications from Geometry

7

x = 25

Example 4

Find the measure of each marked angle below.

– 5 – 5

7

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

h

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 = bhh

The goal is to get b alone on one side of the equation.

2 · · 2

2A= b

h