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Section 2.5: Formulas and Additional Applications from Geometry
§1 Solve A Formula For A Given Variable
What are some well-known formulas that you know? Maybe you know the one for the perimeter of a rectangle,
P = 2l + 2w, where P represents the perimeter, l represents the length of the rectangle and w represents the
width of the rectangle. We say that this formula is solved for P, because that is the value that is by itself. So, if
we are given the length and width of the rectangle, then we can find the perimeter.
The question then becomes; what if I give you the perimeter and the length. Can you find the width of the
rectangle? For example, what if I say that the perimeter of a certain rectangle is 36 feet and the length of the
rectangle is 13 feet. How would you find the width? It’s simply a matter of plugging in the values that are given
and solving for the given variable.
Here, we would end up with 36 = 2(13) + 2w. We end up with 10 = 2w, or w = 10. Hence the width of this
rectangle is 10 feet. You can think of it as working backwards
Similarly, say we know that the area of triangle is 40 inches squared and the base of the triangle is 10 inches.
What is the height of the triangle? Well, we know that the formula for the area of a triangle is
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A  bh ,
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where A is the area, b is the base and h is the height. Here, we can simply plug these values in. We end up with
40 
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 10  h . We can solve this for h. The answer is the height is 8 inches.
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NOTE: Make sure you are familiar with the more popular formulas used in geometry. You can find most of them
in the inside cover of the book!
PRACTICE
1) Find the value of the remaining variable, given that
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A  h  B  b  , where A = 60, h = 5, and B = 16.
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2) Find the value of the remaining variable, given that
P  2L  2w, where P = 84 and L = 19.
§2 Solving Word Problems Using Formulas
In these examples, the formula is not given. You will first need to determine which formula to use. Then you can
proceed with the 4-step method of solving word problems that we went over in the previous section. Its best to
draw a diagram to get a better idea of how to set up the problem!
Try this one: a farmer has 800 feet of fencing material to enclose a rectangular field. The width of the field is 50
feet less than the length. Find the dimensions of the field.
Since we are dealing with perimeter, we need to use P = 2L + 2W. Note that here the perimeter is given but the
width and length is not. So let’s let x = the length of the field and then x – 50 = the length. Now we can set up
our equation. We end up with
800  2 x  2( x  50). Solve this for x to get that x = 225. Hence the length is
225 meters and the width is 175 meters.
PRACTICE
3) The longest side of a triangle is 1 inch longer than the medium side. The medium side is 5 inches longer than
the shortest side. If the perimeter of the triangle is 32 inches, what are the lengths of the three sides?
§3 Solve Problems Involving Vertical Angles and Straight Angles
First some terminology. Look at the following figure.
When two lines intersect, they create special types of angles. The first type is called vertical angles. You can
think of these as being opposites or across each other. In the figure above, angles 1 and 3 are vertical angles,
and angles 2 and 4 are vertical angles as well. You need to remember the following important property about
vertical angles – they have the same measure!
The second type of angle created is called straight angles. These are two angles which together make up a
straight line. In the figure above there are 4 pairs of straight angles – 1 and 2, 2 and 3, 3 and 4, and 1 and 4.
Straight angles have the property that they add up to 180 degrees. These can also be called supplementary
angles.
NOTE: Remember, a right angle is equal to 90 degrees. If two angles added together equals 90 degrees, we call
these complementary angles.
How can we use these special angles to solve problems? Look at the following figure.
What is the measure of each angle? We can see that these angles are vertical angles. Since vertical angles are
equal, we simply set 6 x  2  8 x  8 and solve for x! The answer is x = 5. Hence the vertical angles have a
measure of 32 degrees.
PRACTICE
4) Find the measure of each marked angle.
a)
b)
c)
§4 Solving A Formula For A Specified Variable
In this section, we deal with formulas again but we don’t use numbers to solve. Instead, we work with the
variables. For example, say we have the formula for the perimeter of a rectangle,
you to solve this for w? What does that mean?
P  2l  2w. What if I asked
When you are asked to solve a formula for a specific variable, you need to isolate the specified variable. You
need to remember a couple things. First, treat the specified variable as if it were the only variable – all the other
variables can be thought of as numbers. What we mean is, we need to re-arrange the terms so that ONLY the
specified variable is isolated. So we need to either add, subtract, multiply or divide the other variables to the
other side. Look at the following:
Note that the same steps as solving an equation were used. PEMDAS is very important here. We don’t want to
move terms around in the wrong order. Remember, the coefficient of the specified variable is always done last.
For example, you may know that the formula to converting temperatures from Celsius to Fahrenheit is
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F  C  32 .What if I asked you to solve this for C?
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First, note that the variable C has a coefficient. Remember, we leave this for last. First, subtract 32 to both sides
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F  32  C . Now we can multiply both sides by the reciprocal of the coefficient to get
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C   F  32  .
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to get
PRACTICE
5) Solve the following for the specified variable:
a)
I  prt for t
b)
A  p  prt for t
c)
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A  h  B  b  for b.
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