Chapter 2 Section 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Chapter Chapter 22Section Section 55

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley


Formulas and Applications from Geometry

11

44

33

22

2.52.52.52.5Solve a formula for one variable, given values of the other variables.Use a formula to solve an applied problem.Solve problems involving vertical angles and straight angles.Solve a formula for a specified variable.


Formulas and Applications from GeometryMany applied problems can be solved with formulas. A

formula is an equation in which variables are used to describe a relationship. For example, formulas exist for geometric figures such as squares and circles, for distance, for money earned on bank savings, and for converting English measurements to metric measurements.

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

Slide 2.5 - 3

4 ,P s 2 ,A r ,I prt 932

5F C

Formulas


Objective 11

Slide 2.5 - 4

Solve a formula for one variable, given the values of the other variables.


Solve a formula for one variable, given the values of the other variables.

Given the values of all but one of the variables in a formula, we can find the value of the remaining variable. In Example 1,we use the idea of area. The area of a plane (two-dimensional) geometric figure is a measure of the surface covered by the figure.

Slide 2.5 - 5


Find the value of the remaining variable.

EXAMPLE 1Using Formulas to Evaluate Variables

Solution:

Slide 2.5 - 6

126,P 2 2 ;P L W 25W

226 521 2L 126 20 50 055 L

76

2 2

2L

38L The length of the rectangle is 38.

Once the values of the variables are substituted, the resulting equation is linear in one variable and is solved as in previous sections.


Objective 22

Use a formula to solve an applied problem.

Slide 2.5 - 7


Use a formula to solve an applied problem.Formulas are often used in applied problems. It is a good idea

to draw a sketch when a geometric figure is involved. Examples 2 and 3 use the idea of perimeter. The perimeter of a plane (two-dimensional) geometric figure is the distance around the figure. For a polygon (e.g., a rectangle, square, or triangle), it is the sum of the lengths of its sides. We use the six steps introduced in the previous section.

Slide 2.5 - 8


EXAMPLE 2

Solution:Let L = the length of the field, then L − 50 = the width of the field.

Finding the Dimensions of a Rectangular Field

2 2P L W

The length of the field is 225 m and the width is 175 m.

200 028 5L L

Slide 2.5 - 9

A farmer has 800 m of fencing material to enclose a rectangular field. The width of the field is 50 m less than the length. Find the dimensions of the field.

100800 4 000 110L 900

4 4

4L

225L 505 722 1 5


The longest side of a triangle is 1 in. longer than the medium side. The medium side is 5 in. longer than the shortest side. If the perimeter is 32 in., what are the lengths of the three sides?

EXAMPLE 3

Solution:Let x − 5 = the length of the shortest side,then x = the length of the medium side,and x + 1 = the length of the longest side.

Finding the Dimensions of a Triangle

5 1P x x x

Slide 2.5 - 10

36

3 3

3x

32 4 443 x

12x

The shortest side of the triangle is 7 in., the medium side is 12 in., and the longest side is 13 in.

12 1 13 12 5 7


EXAMPLE 4

Solution:Let b = the base of the triangle.

Finding the Height of a Triangle

1

2A bh

Slide 2.5 - 11

1

12 12

20 12b

1120 4

22b 10b

The base of the triangle is 10 m.

The area of a triangle is 120 m2. The height is 24 m. Find the length of the base of the triangle.


Objective 33

Solve problems involving vertical angles and straight angles.

Slide 2.5 - 12


Solve problems involving vertical angles and straight angles.

Now look at angles and . When their measures are added, we get 180°, the measure of a straight angle. There are three other such pairs of angles: and , and , and and .

Slide 2.5 - 13

The figure below shows two intersecting lines forming angles that are numbered , , , and . Angles and lie “opposite” each other. They are called vertical angles. Another pair of vertical angles is and . In geometry, it is known that vertical angles have equal measures.


EXAMPLE 5

Solution:

Finding Angle Measures

6 29 11 180x x

Slide 2.5 - 14

Find the measure of each marked angle in the figure.

407 40 04 180x 7 140

7 7

x

20x

6 29 420 1 9 20 11 31

The two angle measures are 149° and 31°.

The answer is not the value of x. Remember to substitute the value of the variable into the expression given for each angle.


Objective 44

Solve a formula for a specified variable.

Slide 2.5 - 15


Solve a formula for a specified variable.

Sometimes it is necessary to solve a number of problems that use the same formula. For example, a surveying class might need to solve the formula for the area of a rectangle, A = LW. Suppose that in each problem area (A) and the length (L) of a rectangle are given, and the width (W) must be found. Rather than solving for W each time the formula is used, it would be simpler to rewrite the formula so that it is solved for W . This process is called solving for a specified variable or solving for a literal equation.

In solving a formula for a specified variable, we treat the specified variable as if it were the only variable in the equation and treat the other variables as if they were numbers. We use the same steps to solve the equation for specified variables that we use to solve equations with just one variable.

Slide 2.5 - 16


EXAMPLE 6

Solution:

Solving for a Specified Variable

Slide 2.5 - 17

I prt

I tpr

I p

pr r

rt

p

,I

prt t

I

pror

We use color for a specified variable. Try to circle or underline the variable you are solving for.

Solve for t.


EXAMPLE 7

Solution:


Slide 2.5 - 18

22 2S rh r 22 2hS r r

2 2222 22hS r rr r

22,

2

rh

S

r

22

2

S

rh

r

or

2

2

2 2

2 r

h

r

S r r

Solve for h.


EXAMPLE 8

Solution:


Slide 2.5 - 19

A p prt

tp pA p pr

A

pr pr

p prt

,A

tp

pr

t

A p

pr

or

Solve for t.


Solve for h.

EXAMPLE 9

Solution:


Slide 2.5 - 20

1

2A h b B

1

2A Bh b

12 2

2hA b B

2 h b B

B B

A

b b

2

hA

b B

or

2

,A

b Bh

Chapter 2 Section 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

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Transcript of Chapter 2 Section 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.