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a research about sets

Effat University
Project
FALL 2021
ELECTRICAL AND COMPUTER ENGINEERING DEPARTMENT
INSTRUCTOR NAME: Dr. Nema Salem
COURSE NAME: Mathematics for Arch. & Design
COURSE #: GMTH141A
SECTION: 01 and 04
Student Name: ——————————-Student No. : ——————————-Date & Time:
Instructor’s Name:
Dr. Nema Salem
Instructor’s Signature:
GMTH141A
Instructor Name: Dr. Nema Salem
Page 1 of 2
Effat University
Project
FALL 2021
Question (1)
(5 Points)
1. Students are working in team of three.
2. Select an idea/problem
3. Do literature review about it
4. Describe your methodology in solving it
5. Show your results: analytically, graphically, etc.
6. Write a discussion and then conclude your work
Required
1. Submit a report
2. Present, orally, your work
GMTH141A
Instructor Name: Dr. Nema Salem
Page 2 of 2
Set Theory
CHAPTER 2
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Sets and Elements Notation
â–ªGrouping objects together and viewing them as one entity.
â–ªSets are noted by Upper Case Letters, like: P, M, X, R
â–ªElements are noted by Lower Case Letters, like: p, m, x, r
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Sets and Elements
â–ªSet:
â–ª Any well defined collection of objects,
â–ª Objects are called the elements of the set.
â–ªExamples:
â–ª Vowels of the Alphabets: P = {a, e, i, o, u}
â–ª Colors of the Italian Flag: Y = {red, white, green}
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Sets and Elements
â–ªPrime numbers:
A = {2, 3, 5, 7, 11, 13, …….} infinite
▪Letters in the word “hobby”:
R = {h, o, b, y}
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People in a class: { Alice, Maryam, Sophie, Sammy}
Classes offered by a department: {CS 101, CS 202, … }
Colors of a rainbow: { red, orange, yellow, green, blue, indigo, purple }
States of matter { solid, liquid, gas}
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Sets and Elements
We symbolize an element belonging to a set as follows:
Example: A= {1, 7 , 5 , 10 , 14},
Element 10 belongs to set A, it can be expressed as:
10 Є A
If an element does not belong to a set, we use the symbol ∉
13 ∉ A
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Sets and Elements: Exercises
1. A = {1, 3, 5}
{1} ……………. A
{2} ……………..A
2. B = { } = …………..
3. C is the set of even numbers greater than 2 and less than 12.
C = {……………………………..}
{4} …………C
{6 }……… C
{8 }…………C
{10} ………..C
{2} ………….C
{12} …………C
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Infinite Sets
â–ª Infinite Set as odd numbers:
P = {1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 , 17 , ………..}
â–ª Set-builder notation:
P = {x| x is an odd integer}
read as;
“P is the set of x such that x is an odd integer”
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Infinite Sets: Example
• Natural numbers N = {1, 2, 3, …}
• Whole numbers = {0, 1, 2, 3, …}
• Integers Z = {…, -2, -1, 0, 1, 2, …}
• Positive Integers Z+ = {1, 2, 3, 4, …}
• Rational Numbers = Q = {a/b | aZ , bZ, and b0}
= {1.5, 2.6, -3.8, 15, …}
• Real Numbers = {47.3, -12, , …} (rational and irrational numbers)
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Universal Set (U)
â–ª A large set including all the elements, it depends on the objects we
are studying.
Example:
â–ªIf we are studying non-zero positive integers up to 15:
â–ª The Universal set is {1, 2, 3, 4, 5, 6, 7, 8, 9,10, 11, 12, 13, 14, 15}.
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U = Universal Set
U = {0,1, 2, 3, 4, 5, 6, 7, 8, 9}
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
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Empty Set
▪ It is a set that includes no elements at all, it has a symbol Ø
called ‘Phi’ as:
â–ªThe people living on Mars.
â–ªThe people who visited Neptune.
â–ªThe number of astronauts who walked in space without a space suit.
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Power Set
The Power Set of set A, denoted is the set of all subsets.
P({1,2, 3}) = {
,
{1},{2},{3},
{1,2},{1, 3},{2, 3},
{1,2, 3} }
If a set has n elements, then the power set will have 2n elements
Find the Power set of set B = {1,4,8,5}
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Set Builder Notation
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Subsets and Proper Subsets
â–ª Set A is a subset of set B if every element in set A is also an element in
set B. (A is contained in B) 𝐴 ⊆ 𝐵
â–ª If every element of A is in B but they are not equal, then A is a Proper
Subset of B: 𝐴 ⊂ 𝐵
â–ª A = {1, 4, 7}
B = {1, 2, 3, 4, 5, 6, 7}
𝐴 ⊂ 𝐵 : Proper Subset
â–ªA = {1, 4, 7}
𝐴⊆𝐵
B = {1, 4, 7}
Subset
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Not a Subset
If just one element in set A does not exist in set B, then A is Not
a subset of B:
𝐴 ⊄ 𝐵
➢ Example:
If A = {1, 7, 9} and B = {1, 4, 5, 6, 7},
then 𝐴 ⊄ 𝐵
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Notes
â–ª Any set (A) is a subset of itself.
𝐴 ⊆𝐴
â–ª Every set (A) is a proper subset of the Universal set (U).
𝐴 ⊂𝑈
▪The empty set Ø is also a proper subset of every set (A).
∅ ⊂𝐴
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Subset
{3, 4, 5, 6}  {3, 4, 5, 6, 8}
{1, 2, 6}
GMTH 141A DR. NEMA SALEM
 {2, 4, 6, 8}
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Subsets and Proper Subsets: Exercises
Find the relation between sets.
1. A is the set of people living in Phoenix, Arizona, and USA.
B is the set of people living in Arizona and USA
2. C is the set of weekdays
D is the set of Monday, Tuesday, Wednesday, Thursday and Friday
3. A = {1, 2, 3, 5, 8, 9}
B={2, 3, 8, 11, 12, 13}
c={1, 2, 8}
4. A= {a, b, c, d, e}
B={b, c, d}
C={a, d, e, b, c}
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Examples
Answer: a) 10 b) 0, 10 c) -10, 0, -5, 10 d) all except square root of 5 e) Square root of 5 f) all of them
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Venn Diagram
Some elements are common
between A and B
U
U
A
B
All elements of A are
elements in B
U
Disjoint sets:
No common elements
U
U
A
B
A
B
A
B
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Set Operations/Algebra: Union
â–ª Union of two sets A and B (A U B) is the set of all the elements in A or
B.
U
â–ª A = {2 , 3 , 5 , 7 }
â–ª B = {10 , 3 , 5 , 1}
A
B
â–ª A U B = {1 , 2 , 3 , 5 , 7 , 10 }
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Set Operations/Algebra: Intersection
➢Intersection of two sets A and B (A ∩ B) is the set of all the elements
in A and B.
UU
â–ªA = {2,3,5,7}
â–ª B = {10,3,5,1}
AA
B
▪ A ∩ B = {3,5}
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Set Operations/Algebra: Complement
â–ªComplement of a set A (Ac) is the set of all the elements which belong to U but
do not belong to A.
â–ª U = {1,2,3,4,5,6,7,8,9,10}
U
â–ª A = {3,5,1}
â–ª Ac = {2,4,6,7,8,9,10}
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Set Operations/Algebra: Difference
â–ª Set Difference of two sets A and B “(A – B)” is the set of all the elements that
belong to A but not to B.
â–ª A= {3,4,5,6,7}
U
â–ª B = {3,5,1}
â–ª A – B = {4,6,7}
A
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B
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Example
If A = {1,2,3,4,5,6,7} and B
= {6,7} are two sets.
Then, the difference of set
A and set B is given by;
A – B = {1,2,3,4,5}
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Example
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Examples
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Set Properties
GMTH 141A
DR. NEMA SALEM
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DeMorgan’s Law
GMTH 141A
Dr. Nema Salem
Fall 2021
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/TB
U
Consider the following Venn Diagram :
GMTH 141A
DR. NEMA SALEM
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DeMorgan’s Law
GMTH 141A
Dr. Nema Salem
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/TB
GMTH 141A
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Exercises
GMTH 141A DR. NEMA SALEM
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Sketch Venn diagram for:
1. U={1, 2, 3, 4, 5}
â—¦ A={1, 2, 3}
◦ AC= ………….
2. U={1, 2, 3, 4, 5, 6}
A={1, 2, 3}
B= ……………. As disjointed sets
3. U={1, 2, 3, 4, 5, 6}
A={1, 2, 3, 4}
B= ……………. As Proper sets
4. U={1, 2, 3, , 5, 6,7,8,10}
X = {1, 2, 6, 7}
Y={1, 3, 4, 5, 8}
X U Y = …………………….
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5. U={1, 2, 3, 5, 6,7,8,9,10}
X = {1, 6, 9}
Y={1, 3, 5, 6, 8, 9}
X U Y = …………………….
6. U={1, 2, 3, 4, 5}
A = {1, 2,3}
B = …………..equal subsets
7. U={1, 2, 3, 4, 5, 6}
A = {2,3,4}
B={3,4,5}
A  B= ………………
A  B =…………………….
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8. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11},
A = {2, 3, 4, 6, 7, 8, 10}
B = {1, 3, 5, 7, 9}
C = {2, 3, 4}
9. Sketch Venn diagram and show the relations for:
oU is the set of whole numbers from 1 till 15
â—¦ A is the set of multiple of 3 within U
â—¦ B is the set of prime numbers within U
â—¦ C is the set of odd numbers within U
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10. Given Venn diagram, evaluate each of the following relations
a. 1. AB
b. 2. AB
U
c. 3. (AB)C
d. 4. ACB
e. 5. ABC
B
20
A
2
6
7
21
4
5
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10
17
19
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GMTH 141A
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❑List all the subsets of the set Q = {x, y, z}
✓Answer: The subsets of Q are { }, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}
and {x, y, z}
❑Q = {x, y, z}. How many subsets will Q have?
✓Answer: Number of subsets = 23 = 8
❑Draw a Venn diagram to represent the relationship between the sets. A = {1,
3, 5} and B = {1, 2, 3, 4, 5}
✓Answer:
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Let P = {3, 5, 7, 9, 11}
Q = {9, 11, 13} R = {3, 5, 9} and S = {13, 11}
Write Yes or No for the following:
(a) R ⊂ P
(b) Q ⊂ P
(c) R ⊂ S
(d) S ⊂ Q
(e) S ⊂ P
(f) P ⊄ Q
(g) Q ⊄ R
(h) S ⊄ Q
Answer
Write all the subsets for the following.
(a) {3}
(b) {6, 11}
Answer: (a) ∅ , {3}
(c) ∅
(b) ∅, {6}, {11}, {6,11} c) ∅
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GMTH 141A
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Shade the following in Venn diagram
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Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, and C = {2, 3, 5, 7} three subsets of the
universal set U = {1, 2, 3, 4, 5, 6, 7, 8}.
Draw Venn diagram to represent the subsets A, B, C and the Universal set U.
GMTH 141A DR. NEMA SALEM
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GMTH 141A DR. NEMA SALEM
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Handout #
Mathematics for Architecture and
Design – GMTH 141A
Chapter Review R1
Geometry
1
Geometry
Points, lines, surfaces, solids
GMTH
BUSINESS
141A
INFOGRAPHIC
Math
for Arch
& Design
Perimeter and Area
Volume
Composite Figures
Solid Figures
2
Geometry is concerned with the various aspects of size, shape
and space.
In this unit, you will explore the concepts of angles, shapes,
area, and volume.
3
/tb
3
Points
A point is a location in space. It has no size.
It is named by a capital letter.
.
Read it as: “Point A”
Write it as: “A”
A
4
Lines
A line is a series of points that extends in two opposing
directions without end. Lines have no thickness.
Q
S
Read it as: “Line QS” or “Line SQ”
Write it as: QS or SQ
5
Segments
A segment is a part of a line with two endpoints.
R
B
Read it as: “Segment RB” or “Segment BR”
Write it as:
RB
or
BR
6
Rays
A ray consists of an endpoint and all the points of a line on one
side of the endpoint.
C
D
Read it as: “Ray CD” (the order does matter)
Write it as:
CD
7
J
Parallel Lines- lines that never cross or become
further apart from each other.
L
JK â•‘ LM
K
M
Read: “Line JK is parallel to
line LM”
8
Intersecting Lines- lines that cross each other.
O
Q
OP intersects
R
P
QR
9
Collinear and Non-collinear Points
•!”#$%&’%()%’)*+'”$’%(+’&),+’-#$+’)*+’.)–+/’.”–#$+)*’0″#$%&1
•23′)’&#$4-+’-#$+’.)$$”%’5+’/*)6$’%(*”74(‘)–‘%(+’0″#$%&8’%(+$’%(+’
0″#$%&’)*+’$”$9.”–#$+)*1
C D E F
Collinear points
U
V
W
X
Noncollinear points
10
Planes
A plane is a flat surface with no thickness that extends
without end in all directions.
Intersecting lines are lines that cross
(intersect) at exactly one point.
Parallel lines are lines that do not cross (they
have no points in common).
Skew lines are not parallel and they do not
intersect. They lie in different planes.
11
Handout #
Perimeter and Area
12
Handout #
Volume
GMTH 141-A – Mathematics for Architecture & Design
13
/tb
13
Handout #
Most Common Types of Shapes
14
/tb
14
Handout #
15
Handout #
16
/tb
16
Handout #
17
Handout #
18
Handout #
19
/tb
19
Handout #
Cylindrical
Design
20
/tb
20
Handout #
Pyramids
Design
21
/tb
21
Dimensions of length, area and
volume
length
area
volume
length
mm, cm, m, km, inch,
foot
two dimensions
length ×
length
mm2, cm2, m2, hectare,
km2, square inch,
square foot
three
dimensions
length ×
length ×
length
mm3, cm3, m3, km2,
cubic inch, cubic foot
one dimension
22
Rectangles
The perimeter of a rectangle with length l and width w can be written as:
l
Perimeter = 2l + 2w
or
w
Perimeter = 2(l + w)
The area of a rectangle is given as:
Area = lw
23
Squares
When the length and the width of a rectangle are equal we call it a square. A
square is just a special type of rectangle.
The perimeter of a square with length l is given as:
Perimeter = 4l
l
The area of a square is given as:
Area = l2
24
Shapes
made
from
rectangles
How can we find the area of the shaded shape?
9 cm
4 cm
A
We can think of this shape
as being made up of one
rectangle cut out of another
rectangle.
Label the rectangles A and B.
11 cm
B
6 cm
Area A = 9 × 11 = 99 cm2
Area B = 4 × 6 = 24 cm2
Total area = 99 – 24 = 75 cm2
25
The area of a triangle
h
b
The area of a triangle with base b and perpendicular height h is given by:
1
Area of a triangle =
bh
2
26
The area of a triangle
h
b
Any side of the triangle can be taken as the base, as long as the height is
perpendicular to it:
b
h
b
h
27
The area of a triangle
Suppose that instead of the height of a triangle, we are given
the base, one of the sides and the included angle.
We can use trigonometry to find the area of the triangle:
A
c
B
b
a
Area of triangle ABC =
C
1
ab sin C
2
28
The area of a parallelogram
The area of any parallelogram can be found using the formula:
Area of a parallelogram = base × perpendicular height
perpendicular
height
base
Or using letter symbols,
A = bh
29
The area of a parallelogram
What is the area of this parallelogram?
4.9 cm
4.5 cm
We can ignore
this length
7 cm
Area of a parallelogram = bh
= 7 × 4.5
= 31.5 cm2
30
The area of a trapezium
The area of any trapezium can be found using the formula:
Area of a trapezium =
a
1
(sum of parallel sides) × height
2
perpendicular
height
b
Or using letter symbols,
A=
1
(a + b)h
2
31
The area of a trapezium
What is the area of this trapezium?
1
Area of a trapezium =
(a + b)h
2
6m
9m
=
1
(6 + 14) × 9
2
1
=
× 20 × 9
2
14 m
= 90 m2
32
The area of a trapezium
What is the area of this trapezium?
1
Area of a trapezium =
(a + b)h
2
=
9m
4m
12 m
1
(9 + 4) × 12
2
1
=
× 13 × 12
2
= 78 m2
33
Cubes and cuboids
A cuboid is a 3-D shape with edges of different lengths. All of its
faces are rectangular or square.
How many faces does a cuboid have?
6
Face
How many edges does a cuboid have?
12
Edge
Vertex
How many vertices does a cuboid
8
have?
A cube is a special type of cuboid with edges of equal length. All of its faces are
square.
34
Length around the edges
Suppose we have a cuboid of length 5 cm, width 4 cm and height 3 cm. What
is the total length around the edges?
Imagine the cuboid as a hollow wire frame:
The cuboid has 12 edges.
3 cm
4 edges are 5 cm long.
4 edges are 4 cm long.
4 cm
4 edges are 3 cm long.
5 cm
Total length around the edges = 4 × 5 + 4 × 4 + 4 × 3
= 20 + 16 + 12
= 48 cm
35
Length around the edges
To find the length around the edges of a cuboid of length l, width w and height h we can use the formula:
Length around the edges = 4l + 4w + 4h
or
Length around the edges = 4(l + w + h)
To find the length around the edges of a cube of length l we can use the formula:
Length around the edges = 12l
36
Surface area of a cuboid
To find the surface area of a cuboid, we calculate the total area of all of the faces.
A cuboid has 6 faces.
The top and the bottom of the cuboid have
the same area.
37
Surface area of a cuboid
To find the surface area of a cuboid, we calculate the total area of all of the faces.
A cuboid has 6 faces.
The front and the back of the cuboid have
the same area.
38
Surface area of a cuboid
To find the surface area of a cuboid, we calculate the total area of all of the faces.
A cuboid has 6 faces.
The left hand side and the right hand side
of the cuboid have the same area.
39
Formula for the surface area of a cuboid
We can find the formula for the surface area of a cuboid as follows.
l
h
w
Surface area of a cuboid =
2 × lw
Top and bottom
+ 2 × hw
Front and back
+ 2 × lh
Left and right side
Surface area of a cuboid = 2lw + 2hw + 2lh
40
Surface area of a cube
How can we find the surface area of a cube of length l?
All six faces of a cube have the
same area.
The area of each face is l × l = l2
Therefore,
l
Surface area of a cube = 6l2
41
Volume of a cuboid
The following cuboid is made out of interlocking cubes.
How many cubes does it contain?
42
Volume of a cuboid
We can work this out by dividing the cuboid into layers.
The number of cubes in each layer
can be found by multiplying the
number of cubes along the length
by the number of cubes along the
width.
3 × 4 = 12 cubes in each layer
There are three layers altogether
so the total number of cubes in the
cuboid = 3 × 12 = 36 cubes
43
Volume of a cuboid
We can find the volume of a cuboid by multiplying the area of the base by the height.
The area of the base
= length × width
So,
height, h
Volume of a cuboid
= length × width × height
width, w
length, l
= lwh
44
Volume of a cube
How can we find the volume of a cube of length l?
The length, width and height of a
cube are the same.
Therefore:
Volume of a cube
l
= (length of one edge)3
= l3
45
Volume of shapes made from cuboids
What is the volume of this 3-D shape?
3 cm
We can think of this shape as
two cuboids joined together.
3 cm
4 cm
Volume of the green cuboid
= 6 × 3 × 3 = 54 cm3
6 cm
Volume of the blue cuboid
= 3 × 2 × 2 = 12 cm3
Total volume
5 cm
= 54 + 12 = 66 cm3
46
Prisms
A prism is a 3-D shape that has a constant cross-section along its length.
For example, this hexagonal prism has the same hexagonal
cross-section throughout its length.
This is called a
hexagonal prism
because its crosssection is a hexagon.
47
Volume of a prism The volume of a prism is found by multiplying the area of its
cross-section A by its length l (or by its height if it is
standing on its cross-section).
l
h
A
A
V = Al
or
V = Ah
48
Volume of a prism
What is the volume of this triangular prism?
7.2 cm
4 cm
5 cm
Area of cross-section = ½ × 5 × 4 = 10 cm2
Volume of prism = 10 × 7.2 = 72 cm3
49
Volume of a prism
What is the volume of this prism?
12 m
7m
4m
3m
5m
Area of cross-section = 7 × 12 – 4 × 3 = 84 – 12 = 72 m2
Volume of prism = 72 × 5 = 360 m3
50
Surface area of a prism
Here is the net of a triangular prism.
What is its surface area?
10 cm
13 cm
260
60
200
12 cm
260
20 cm
We can work out the area
of each face and write it in
the diagram of the net.
60
Total surface area
= 60 + 60 + 200 + 260 + 260
= 840 cm2
51
Pyramids
A pyramid is a 3-D shape whose base is usually a polygon
but can also be a shape with curved edges. The faces
rising up from the base meet at a common vertex or apex.
The most common pyramids are:
A tetrahedron
or triangular
pyramid.
A square-based
pyramid
A cone
52
Volume of a pyramid
The volume of a pyramid is found by multiplying the area of its base A by its perpendicular height h and
dividing by 3.
Apex
slant height
h
A
base
Volume of a pyramid =
V=
1
3
1
3
× area of base × height
Ah
53
Volume of a pyramid
What is the volume of this rectangle-based pyramid?
Area of the base = 5 × 3
= 15 cm2
8 cm
Volume of pyramid =
3 cm
5 cm
=
1
3
1
3
Ah
× 15 × 8
= 40 cm3
54
Surface area of a pyramid
Here is the net of a regular tetrahedron.
What is its surface area?
Area of each face = ½bh
= ½ × 6 × 5.2
= 15.6 cm2
5.2 cm
Surface area = 4 × 15.6
= 62.4 cm2
6 cm
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Volume of a cylinder
A cylinder is a special type of prism with a circular cross-section.
Remember, the volume of a prism can be found by multiplying the area of the
cross-section by the height of the prism.
r
The volume of a cylinder is given by:
Volume = area of circular base × height
h
or
V = πr2h
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Surface area of a cylinder
To find the formula for the surface area of a cylinder we can draw its net.
r
h
?
2Ï€r
How can we find the width of the curved face?
The width of the curved face is equal to the
circumference of the circular base, 2Ï€r.
Area of curved face = 2Ï€rh
Area of 2 circular faces = 2 × πr2
Surface area of a cylinder = 2Ï€rh + 2Ï€r2
or
Surface area = 2Ï€r(h + r)
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Volume of a cone
A cone is a special type of pyramid with a circular base.
Remember, the volume of a pyramid can be found by multiplying the
area of the base by the height and dividing by 3.
The volume of a cone is given by:
Volume =
h
1
3
× area of circular base × height
or
r
V=
1
3
Ï€r2h
58
Volume and surface area of a sphere
A sphere is a 3-D shape whose surface is always the same distance from the
center. This fixed distance is the radius of the sphere.
For a sphere of radius r :
r
Volume =
4
3
Ï€r3
and
Surface area = 4Ï€r2
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Most Common Types of Angles
n
n
The most common unit for measuring angles is the degree.
The major types of angles are acute angle, right angle, obtuse angle
and straight angle.
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Types of Triangles: Angles
n
Note: The sum of the measures of the angles of any triangle is 180°.
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Polygons
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Handout #
Sept. 4!
Composite Figures
!”#$%&$'()* +(,-.*”(‘”+$.%*/”+.$%”)0$”$.”%$.*”+(,-.*’1
2$”+(3/”)4*”5.*5″$+”5″#$%&$'()*”+(,-.*6
7(3/”)4*”5.*5′”$+”*5#4″+(,-.*”)4*3″5// )4*%”-&1
2$”+(3/”)4*”5.*5″$+”5″’45/*/”.*,($38″9$-“3**/”)$”‘-:).5#) )4*”5.*5’1
Area of square:
A = lw = 6(6) = 36 ft2
Area of circle:
A = pr2
A = p(3)2 = 9p ft2
Area of semicircle =
½ (9p) = 4.5p ft2
Total area of figure:
Add areas of square and
semicircle:
A = 36 + 4.5p cm2
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63
Handout #
Area of Composite Figure
6cm
8cm
12cm
8cm

4cm
4cm
64
Handout #
12cm
6cm
8cm
+
8cm
4cm
1
4cm
A = bh
= 12 ´ 8
= 96 cm 2
1
A = (b1 + b2 )h
2
1
= ´ ( 6 + 4) ´ 8
2
= 40cm 2
4cm
bh
A=
2
4´ 4
=
2
= 8cm 2
Total Area = 96+40-8= 128cm2
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Handout #
Total Area
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Handout #
Shaded Area
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Handout #
Shaded Area
68
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68
Handout #
Area Shaded
69
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69
Handout #
Area Shaded
70
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Handout #
71
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71
Handout #
Surface Area
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Handout #
Surface Area
73
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Handout #
Volume of Prisms
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Handout #
Volume of Prisms
75
Handout #
Composite Solid Figures
76
Handout #
Composite Solid Figures
77
Handout #
Composite Solid Figures
78
Handout #
Composite Solid Figures
GMTH 141-A – Mathematics for Architecture & Design
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79
Handout #
Pyramid & Cone
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80
Handout #
Prism
81
Handout #
Volume of Pyramids
82
Handout #
Volume of Cylinders
83
Handout #
Volume of a Cone
84
Handout #
Find the surface area of the sphere
S = 4pr2
Surface area of a sphere
= 4p(4.5)2
Replace r with 4.5.
≈ 254.5
Simplify.
85
Handout #
Example
A. Find the surface area of the hemisphere.
Find half the area of a sphere with the radius of
3.7 millimeters. Then add the area of the great circle.
GMTH 141-A – Mathematics for Architecture & Design
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Handout #
Surface area of a
hemisphere
Replace r with 3.7.
≈ 129.0
Use a calculator.
Answer: about 129.0 mm2
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