This document contains 10 slides summarizing key concepts and examples for solving formulas and additional applications from geometry. The objectives covered include: solving a formula for one variable; using a formula to solve an applied problem; and solving problems involving vertical angles and straight angles. Several examples are provided to demonstrate how to isolate a specified variable in a formula and apply formulas to geometry problems involving area, triangles, rectangles, trapezoids, and angles.

1. Slide - 1Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
2
Equations,
Inequalities, and
Applications
10

2. Slide - 2Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
1. Solve a formula for one variable, given the
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.
Objectives
10.5 Formulas and Additional Applications
from Geometry

3. Slide - 3Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
36 = 2W
P = 2L + 2W
Check: 2 · 8 + 2 · 18 = 52
Solve a Formula for One Variable
2
18 = W
Example
Find the value of the remaining variable.
P = 2L + 2W; P = 52; L = 8
52 = 2 · 8 + 2W
52 = 16 + 2W
–16 –16
16 + 36 = 52
52 = 52
2

4. Slide - 4Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
AREA FORMULAS
 Triangle
A = ½bh
 Rectangle
A = LW
 Trapezoid
A = ½h(b + B)
b = base
h = height
b
h
L = Length
W = Width
L
W
h = height
b = small base
B = large base
b
B
h
Solve a Formula for One Variable

5. Slide - 5Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
Example
The area of a rectangular garden is 187 in2 with a width of
17 in. What is the length of the garden?
A = LW
Check: 17 · 11 = 187
Use a Formula to Solve an Applied Problem
187 = L · 17
17
11 = L
The length is 11 in.
17
17
L

6. Slide - 6Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
63 = 9h
9
Example
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)
Check:
½ · 7 · 18 = 63
Use a Formula to Solve an Applied Problem
63 = ½h(8 + 10)
7 = h
63 = ½h(18)
The height of the UV
is 7ft.
10
8
h
9

7. Slide - 7Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
2
3
1
The figure shows two intersecting lines forming angles that are
numbered: , , , and .
Solve Problems Involving Vertical and Straight Angles
32 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.
1
4

8. Slide - 8Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
–4 = 2x – 40
3x – 4 = 5x – 40
Check:
5 · 18 – 40 = 50°2
18 = x
Example
Find the measure of each marked angle below.
–3x –3x
2
Solve Problems Involving Vertical and Straight Angles
(3x – 4)° (5x – 40)°
Since the marked angles are vertical
angles, they have equal measures.
+ 40 + 40
36 = 2x
Thus, both angles
are
3 · 18 – 4 = 50°
CAUTION
Here, the answer was
not the value of x!

9. Slide - 9Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
(x + 15)°
6 · 25 – 10 = 140°
and
25 + 15 = 40°
7x + 5 = 180
6x – 10 + x + 15 = 180
Check:
140° + 40° = 180°
7
x = 25
Example
Find the measure of each marked angle below.
– 5 – 5
7
Solve Problems Involving Vertical and Straight Angles
(6x – 10)°
Since the marked angles are
straight angles, their sum will be
180°.
7x = 175
Thus, the angles
are

10. Slide - 10Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
h
Example
Solve A = ½bh for b.
A = ½bh
Solve a Formula for a Specified Variable
2A = bh
h
The goal is to get b alone on one side of the equation.
2 · · 2
2A
= b
h