Download Solving for a Specified Variable

Chapter 2
Section 5
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2.5
1
2
3
4
Formulas and Applications from
Geometry
Solve 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.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Formulas and Applications from Geometry
Many 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.
9
P  4 s, A  r , I  prt , F  C  32
5
Formulas
2
The formulas used in this book are given on the inside covers.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.5 - 3
Objective 1
Solve a formula for one variable,
given the values of the other
variables.
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Slide 2.5 - 4
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.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.5 - 5
EXAMPLE 1
Using Formulas to Evaluate
Variables
Find the value of the remaining variable.
P  2 L  2W ;
Solution:
P  126,
W  25
126  2L  2  25
126  50  2L  50  50
76 2L

2
2
L  38
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.
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Slide 2.5 - 6
Objective 2
Use a formula to solve an
applied problem.
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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 (twodimensional) 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.
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Slide 2.5 - 8
EXAMPLE 2
Finding the Dimensions of a
Rectangular Field
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.
Solution:
Let
L = the length of the field,
then L − 50 = the width of the field.
L  225
P  2L  2W
225  50  175
800  2  L   2  L  50
800 100  4L 100 100
The length of the field is
900 4L
225 m and the width is

175 m.
4
4
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Slide 2.5 - 9
EXAMPLE 3
Finding the Dimensions of a
Triangle
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?
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.
x  12
12  1  13
12  5  7
P  x   x  5   x  1
32  4  3x  4  4 The shortest side of the triangle is
36 3x
7 in., the medium side is 12 in., and

the longest side is 13 in.
3
3
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Slide 2.5 - 10
EXAMPLE 4
Finding the Height of a Triangle
The area of a triangle is 120 m2. The height is 24 m.
Find the length of the base of the triangle.
Solution:
Let b = the base of the triangle.
1
A  bh
2
1
120  b 24
2
120 12b

12
12
b  10
The base of the triangle is 10 m.
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Slide 2.5 - 11
Objective 3
Solve problems involving
vertical angles and straight
angles.
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Slide 2.5 - 12
Solve problems involving vertical
angles and straight angles.
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.
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
.
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Slide 2.5 - 13
EXAMPLE 5
Finding Angle Measures
Find the measure of each marked angle in the figure.
Solution:
 6x  29   x  11  180
7 x  40  40  180  40 6  20  29  149
20  11  31
7 x 140

The two angle measures
7
7
are 149° and 31°.
x  20
The answer is not the value of x. Remember to substitute the
value of the variable into the expression given for each angle.
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Slide 2.5 - 14
Objective 4
Solve a formula for a specified
variable.
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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.
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Slide 2.5 - 16
EXAMPLE 6
Solving for a Specified Variable
Solve I  prt for t.
Solution:
I  prt
I
prt

pr pr
I
 t,
pr
or
I
t
pr
We use color for a specified variable. Try to circle or underline the
variable you are solving for.
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Slide 2.5 - 17
EXAMPLE 7
Solving for a Specified Variable
Solve S  2rh  2r for h.
2
S  2rh  2r 2
Solution:
S  2r 2  2rh  2r 2  2r 2
S  2r
2rh

2r
2r
2
S  2r
S  2r
 h, or h 
2r
2r
2
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2
Slide 2.5 - 18
EXAMPLE 8
Solving for a Specified Variable
Solve A  p  prt for t.
Solution:
A  p  p  prt  p
A  p prt

pr
pr
A p
 t,
pr
or
A p
t
pr
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.5 - 19
EXAMPLE 9
Solving for a Specified Variable
1
Solve A  h  b  B  for h.
2
1
Solution: A  h  b  B 
2
1
A  2  h b  B   2
2
h b  B 
2A

b  B  b  B 
2A
 h, or
b  B 
2A
h
b  B 
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Slide 2.5 - 20