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.

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

5F C

Formulas

Slide 2.5-3

Objective 1

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

Slide 2.5-4

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.

Slide 2.5-5

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

A LW

Find the value of the remaining variable.

Solution:

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.

Slide 2.5-6

Using Formulas to Evaluate VariablesCLASSROOM EXAMPLE 1

Objective 2

Use a formula to solve an applied problem.

Slide 2.5-7

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.

2 2P L W

Slide 2.5-8

Use a formula to solve an applied problem.

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

2 2P L W

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

200 028 5L L

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

Slide 2.5-9

Finding the Dimensions of a Rectangular YardCLASSROOM EXAMPLE 2

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.

5 1P x x x

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

Slide 2.5-10

Finding the Dimensions of a TriangleCLASSROOM EXAMPLE 3

Solution:Let b = the base of the triangle.

1

2A bh 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.

Slide 2.5-11

Finding the Height of a Triangular SailCLASSROOM EXAMPLE 4

Objective 3

Solve problems involving vertical angles and straight angles.

Slide 2.5-12

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 .

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.

Slide 2.5-13

Solve problems involving vertical angles and straight angles.

Solution:

6 29 11 180x x

CLASSROOM EXAMPLE 5

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.

Slide 2.5-14

Finding Angle Measures

Objective 4

Solve a formula for a specified variable.

Slide 2.5-15

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

Solve a formula for a specified variable.

Solution:

Solve for t.I prt

I tpr

I p

pr r

rt

p

,I

prt t

I

pror

Slide 2.5-17

Solving for a Specified VariableCLASSROOM EXAMPLE 6

Solution:

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

22 2S rh r Solve for h.

Slide 2.5-18

Solving for a Specified VariableCLASSROOM EXAMPLE 7

Solution:

tp pA p pr

A

pr pr

p prt

,A

tp

pr

t

A p

pr

or

A p prt Solve for t.

Slide 2.5-19

Solving for a Specified VariableCLASSROOM EXAMPLE 8

Solution:

Solve for h. 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

Slide 2.5-20

Solving for a Specified VariableCLASSROOM EXAMPLE 9