Download Solve a formula for a specified variable.

Chapter 2
Section 5
2.5 Formulas and Additional Applications from Geometry
Objectives
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.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Definition.
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
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.5-3
Objective 1
Solve a formula for one variable,
given values of the other variables.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.5-4
Solve a formula for one variable, given 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. In the following formula A is the area
of a rectangle with length L and width W.
A  LW
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
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.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.5-6
Objective 2
Use a formula to solve an applied
problem.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.5-7
Use a formula to solve an applied problem.
It is a good idea to draw a sketch when solving an applied
problem that involves a geometric. 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. In the following
formula P is the perimeter of a rectangle with length L and width W.
P  2L  2W
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.5-8
EXAMPLE 2 Finding the Dimensions of a Rectangular Yard
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.
P  2L  2W
800  2  L   2  L  50
800 100  4L 100 100
900 4L

4
4
L  225
225  50  175
The length of the field is 225 m and the width is 175 m.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
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.
P  x   x  5   x  1
32  4  3x  4  4
36 3x

3
3
x  12
12  1  13
12  5  7
The shortest side of the triangle is 7 in., the medium side is 12 in.,
and the longest side is 13 in.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.5-10
EXAMPLE 4 Finding the Height of a Triangular Sail
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.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.5-11
Objective 3
Solve problems involving vertical
angles and straight angles.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
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
,
.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
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
6  20  29  149
7 x  40  40  180  40
20  11  31
7 x 140

7
7
The two angle measures are
x  20
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.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.5-14
Objective 4
Solve a formula for a specified
variable.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
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.
When 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.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
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
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
or
I
t
pr
Slide 2.5-17
EXAMPLE 7 Solving for a Specified Variable
Solve
S  2rh  2r 2 for h.
Solution:
S  2rh  2r
2
S  2r  2rh  2r  2r
2
2
2
S  2r 2 2rh

2r
2r
S  2r 2
 h,
2r
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
or
S  2r 2
h
2r
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
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
or
A p
t
pr
Slide 2.5-19
EXAMPLE 9 Solving for a Specified Variable
Solve
1
A  h b  B 
2
Solution:
for h.
1
A  h b  B 
2
1
A  2  h b  B   2
2
h b  B 
2A

b  B  b  B 
2A
 h,
b  B 
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
or
2A
h
b  B 
Slide 2.5-20