Beginning Algebra, 11th Edition

OBJECTIVES

Formulas and Additional Applications from Geometry

2.5

1 Solve a formula for one variable, given 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.

A formulais an equation in which variables are used to describe a relationship. For

example, formulas exist for finding perimeters and areas of geometric figures, cal-

culating money earned on bank savings, and converting among measurements.

Formulas

Many of the formulas used in this book are given on the inside covers.

OBJECTIVE 1 Solve a formula for one variable, given values of the other

variables. In Example 1,we use the idea of area. The areaof a plane (two-

dimensional) geometric figure is a measure of the surface covered by the figure.

Using Formulas to Evaluate Variables

Find the value of the remaining variable in each formula.

(a)

As shown in FIGURE 7, this formula gives the area of a rec-

tangle with length Land width W. Substitute the given values into

the formula.

Let and

Divide by 10.

6.4=W

64

10

=

10 W

10

64 = 10 W a= 64 L=10.

a=LW

a

a= LW; a= 64, L= 10

EXAMPLE 1

P= 4 s, a= pr^2 , I=prt, F=

9

5

C+ 32

120 CHAPTER 2 Linear Equations and Inequalities in One Variable

NOW TRY
EXERCISE 1
Find the value of the remain-
ing variable.

P=78, a= 12

P= 2 a+ 2 b; Rectangle a = LW

W

L

FIGURE 7

Trapezoid a = h(b + B)

h

B

b

1 2 FIGURE 8

NOW TRY ANSWER

b= 27

Solve for W.

Solve for b.

In this book, adenotes area.

The width is 6.4. Since the given area, the answer checks.

(b)

This formula gives the area of a trapezoid. See FIGURE 8.

Let

Multiply. Distributive property Subtract 135. Combine like terms.

Divide by 5.

The length of the shorter parallel side, b, is 15. This answer checks, since

as required. NOW TRY

1

2

1102115 + 272 = 210,

15 = b

75

5

=

5 b

5

75 = 5 b

210 - 135 = 5 b+ 135 - 135

210 = 5 b+ 135

210 = 51 b+ 272

210 = a=210, h=10, B=27.

1

2

11021 b+ 272

a=

1

2

h 1 b+B 2

a=

1

2

h 1 b+ B 2 ; a =210, B=27, h= 10

101 6.4 2 =64,

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