Download Chapter 2, Section 5

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2.5 – Slide 1
Chapter 2
Equations, Inequalities, and
Applications
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2.5 – Slide 2
2.5
Formulas and Additional
Applications from Geometry
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2.5 – Slide 3
2.5 Formulas and Additional Applications
from Geometry
Objectives
1.
2.
3.
4.
Solve a formula for one variable, given the 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.
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2.5 – Slide 4
2.5 Formulas and Additional Applications
from Geometry
Solving a Formula for One Variable
Example 1
Find the value of the remaining variable.
P = 2L + 2W; P = 52; L = 8
P = 2L + 2W
52 = 2 · 8 + 2W
Check: 2 · 8 + 2 · 18 =
16 + 36 = 52
52 = 16 + 2W
–16 –16
36 = 2W
2
2
18 = W
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2.5 – Slide 5
2.5 Formulas and Additional Applications
from Geometry
Solving a Formula for One Variable
AREA FORMULAS



Triangle
A = ½bh
Rectangle
A = LW
Trapezoid
A = ½h(b + B)
h
b
L
W
b
h
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B
b = base
h = height
L = Length
W = Width
h = height
b = small base
B = large base
2.5 – Slide 6
2.5 Formulas and Additional Applications
from Geometry
Using a Formula to Solve an Applied Problem
Example 2
The area of a rectangular garden is 187 in2 with a width of
17 in. What is the length of the garden?
L
17
A = LW
187 = L · 17
17
17
11 = L
The length is 11 in.
Check: 17 · 11 = 187
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2.5 – Slide 7
2.5 Formulas and Additional Applications
from Geometry
Using a Formula to Solve an Applied Problem
Example 3
Bob is working on a sketch for a new underwater vehicle
(UV), shown below. In his sketch, the bottom of the UV
is
10 ft long, the top is 8 ft long, and the area is 63 ft2. What
is the height of his UV?
A = ½h(b + B)
8
h
10
63 = ½h(8 + 10)
63 = ½h(18)
63 = 9h
9
9
The height of the UV
is 7ft.
Check:
½ · 7 · 18 = 63
7=h
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2.5 – Slide 8
2.5 Formulas and Additional Applications
from Geometry
Solving Problems Involving Vertical and Straight Angles
1
2
4
3
The figure shows two intersecting lines forming angles that are
numbered: 1 , 2 , 3 , and 4 .
Angles 1 and 3 lie “opposite” each other. They are called
vertical angles. Another pair of vertical angles is 2 and 4 .
Vertical angles have equal measures.
Now look at angles 1 and 2 . When their measures are added,
we get the measure of a straight angle, which is 180°. There
are three other such pairs of angles: 2 and 3 , 3 and 4 , and
4 and 1 .
.
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2.5 – Slide 9
2.5 Formulas and Additional Applications
from Geometry
Solving Problems Involving Vertical and Straight Angles
Example 4
Find the measure of each marked angle below.
(3x – 4)°
(5x – 40)°
CAUTION
Here, the answer was
not the value of x!
Since the marked angles are vertical
angles, they have equal measures.
3x – 4 = 5x – 40
–3x
–3x
–4 = 2x – 40
+ 40
+ 40
36 = 2x
2
2
18 = x
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Thus, both angles
are
3 · 18 – 4 = 50°
Check:
5 · 18 – 40 = 50°
2.5 – Slide 10
2.5 Formulas and Additional Applications
from Geometry
Solving Problems Involving Vertical and Straight Angles
Example 4
Find the measure of each marked angle below.
Since the marked angles are
straight angles, their sum will be
(6x – 10)° (x + 15)°
180°.
Thus, the angles
6x – 10 + x + 15 = 180
are
7x + 5 = 180
6 · 25 – 10 = 140°
–5 –5
and
Check:
7x = 175
25 + 15 = 40°
7
7
140° + 40° = 180°
x = 25
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2.5 – Slide 11
2.5 Formulas and Additional Applications
from Geometry
Solving a Formula for a Specified Variable
Example 5
Solve A = ½bh for b.
The goal is to get b alone on one side of the equation.
2 · A = ½bh · 2
2A = bh
h
h
2A
=b
h
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2.5 – Slide 12