2.5Formulas and
Additional
Applications from
Geometry
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
Slide 2.5-3
Objective 1
Solve a formula for one variable, given
values of the other variables.
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
Slide 2.5-5
CLASSROOM
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.
Slide 2.5-6
Objective 2
Use a formula to solve an applied
problem.
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
Slide 2.5-8
CLASSROOM
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.
Slide 2.5-9
CLASSROOM
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.
Slide 2.5-10
CLASSROOM
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.
Slide 2.5-11
Objective 3
Solve problems involving vertical
angles and straight angles.
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
.
Slide 2.5-13
CLASSROOM
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
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.
Slide 2.5-14
Objective 4
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.
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.
Slide 2.5-16
CLASSROOM
EXAMPLE 6
SolveI
Solving for a Specified Variable
 prt for t.
Solution:
I  prt
I
prt

pr pr
I
 t,
pr
or
I
t
pr
Slide 2.5-17
CLASSROOM
EXAMPLE 7
SolveS
Solving for a Specified Variable
 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
or
S  2r 2
h
2r
Slide 2.5-18
CLASSROOM
EXAMPLE 8
Solve A
Solving for a Specified Variable
 p  prt
for t.
Solution:
A  p  p  prt  p
A  p prt

pr
pr
A p
 t,
pr
or
A p
t
pr
Slide 2.5-19
CLASSROOM
EXAMPLE 9
Solving for a Specified Variable
1
Solve A  h  b  B  for h.
2
Solution:
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 
or
2A
h
b  B 
Slide 2.5-20