Download Welcome to our sixth seminar!

Welcome to our sixth seminar!
We’ll begin shortly
Line segments
A point is a spot in space that has no length, width or
height. Think of it as a place holder.
A line segment is a line that ends at two points.
For example:
B
A
This line segment is named AB or AB
Rays
Rays are lines with only one end point. You can think of
them as starting at a point and then extending into
infinity in one direction. For example:
D
C
This ray is called AB or AB
Angles
An angle is made up of two rays, lines, or line segments
which start at the same endpoint (called the vertex).
For example:
A
B
S
k
m
T
K
J
N
a
C
R
L
This angle is named
‫ے‬ABC ,‫ے‬CBA, ‫ے‬B
or ‫ ے‬k
M
The highlighted angle is named
‫ے‬JNL, ‫ ے‬LNJ, or ‫ ے‬a
This angle is named
‫ ے‬RST, ‫ے‬TSR, ‫ے‬S
or ‫ ے‬m
Types of angles
1800
3600
900
Line segment
One full rotation=
3600
<
Straight angle =
1800
900
An acute angle is one that
measures less than 900
Right angle =
900
> 900
An obtuse angle is one that
Measures more than 900
More types of angles
Angles whose sum is 1800 are
called supplementary angles
Angles whose sum is 900 are
called complementary angles.
Examples:
600 +300 =900
600
300
320
320 +580 =900
580
1340
460 +1340 =1800
460
Note that if you know one angle you can calculate the other:
a
380
The sum of a and 38 is 180
a + 38 = 180
Subtract 38 from both sides:
a = 180 – 38
a = 1420
Planes (a 2 dimensional surface)
Pairs of lines
Parallel lines have the same distance between them at each point.
They never intersect and the angle between them is 00
Perpendicular lines intersect at 900.
b
a
c
d
a
Intersecting lines meet at a point at an angle not 90 . The
vertical angles (opposite) are equal to each other. The sum
of the adjacent angles (next to each other) is 180 . Here
Vertical angles are a,c, and b,d. Adjacent angles are a,b;
b,c; c,d; and a,d.
b
c d
e f
g
h
A transversal line is one that intersection two other lines.
Corresponding angles are those on the same side of the
Two lines. Here they are a,e; c,g; b,f; and d,h
Alternate interior angles are those that are opposite
Interior. Here they are c,f and d,e.
If the two lines are parallel then
Corresponding lines are equal and alternate interior angles
Are also equal.
A few examples; solve for all angles
c and 35 are vertical angles and equal so c = 350
a and c are adjacent angles so
a + c= 1800
a +35 = 1800
a = 1800 -350
a = 1450 and b = 1450 (opposite)
Summary: a = 1450 , b = 1450 , c =350
350
a
b
c
1220
d
e
f
h
i
j
k
1220 and i are corresponding angles; i = 1220
1220 and e; I and j are vertical angles; e=1220 , j=1220
d and 1220 are adjacent angles so
d + 1220 = 1800
d = 1800 – 1220
d = 580
d and h are vertical angles so h = 580
d, f and h,k are vertical angles so f=580 and k = 580
Summary: d=580 , e=1220 , f=580 , h=580 , i=1220 ,
j=1220 ,k = 580
Polygons (closed 2-D figures)
Types of triangles
Similar figures (same shape, different sizes)
A = A'
B = B'
C = C'
AB BC AC


A'B' B'C' A'C'
Example
Find ZX and ZY
ZX 12
=
5
3
ZX
=4
5
5 ZX
= 4(5)
5
ZX = 20 units
ZY 12
=
4
3
ZY
=4
4
4 ZY
= 4(4)
4
ZY = 16 units
Area and perimeter formulas
Perimeter
The perimeter (P) of a polygon (a two-dimensional shape with at least
three sides) is the sum of all of the sides. In other words it is the
distance around the shape. Don’t forget to include units; they will be
in length (like m, ft, in, etc).
What is the perimeter of this trapezoid (a
four sided shape with unequal sides)?
P = the sum of the sides
P = 28 + 15 + 8 + 12
P = 43 + 20
P = 63 units
NOTE: we ignored the extraneous
information (the height). If there are no
given units write “units”
A dollar bill has a width of 2.56in and a length of 6.14
inches. Find the perimeter.
The perimeter of a rectangle (which has two equal sides)
can be found by using the formula:
P = 2L + 2W
Where L is length and W is width
Here W = 6.14in and L = 2.56in
P = 2(6.14) + 2(2.56)
Multiply:
P = 12.28 + 5.12
P = 17.4 in
The perimeter of the dollar bill is 17.4 inches.
Example
If the radius of this circle is 4 inches,
what is the perimeter?
Use C = 2 r
C = 2(3.14)(4in)
C = 6.28(4in)
C = 25.13 in
Round to the
nearest whole number:
C = 25 in
A circle with a radius of 4 in has a perimeter of 25 in
Step one: find the area of the rectangle
A = LW
L = 3ft W = 8ft
A r = (3ft)(8ft)
A r = 24ft 2
This figure contains a
rectangle which is 8
by 3 ft and a triangle
with a height of 8 ft
and a base of 6 – 3 or
3ft.
Add these two areas to
get the total
Step two: find the area of the triangle:
1
A = bh
b = 3ft h = 8ft
2
1
A t =  3ft  (8ft) half of 8 is 4
2
A t = 4ft(3ft)
A t = 12ft 2
Step three: Add
A total  A r + A t
A total = 24ft 2 + 12ft 2
A total = 36ft 2
The area of this figure is 36ft 2
Volume and formula problems
V=πr 2h
Surface area: the total amount of area on the surface of a three
dimensional figure. The units are the same as area: length squared.
Here are some of the common formulas for finding surface area:
Rectangular solid:
S = the sum of all of the sides
For this diagram the surface area is:
S = 2lw + 2lh + 2wh
Or in this second picture add the areas:
S = 2A + 2B + 2C
Explaining the cylinder
Here is a picture of the
surfaces of a cylinder
that is enclosed. I has
the area of each end and
the rectangle that
surrounds it.
If it was not enclosed or had
only one end enclosed
you remove those parts
from the equation.
Examples
Find the surface area of a closed
cylinder with a height of 16 inches
and a diameter of 12 inches. The
radius is half of 12. r = 6 in. round to
the nearest tenth.
S = 2 r 2 + 2 rh
r = 6in
h = 16in
S = 2(3.14)(6in) 2 + 2(3.14)(6in)(16in)
square and multiply:
S = 2(3.14)(36in 2 )  2(3.14)(96in 2 )
S = 6.28(36in 2 )  6.28(96in 2 )
S = 226in 2 + 602.88in 2
S = 828.88in 2
Round to the nearest tenth:
S = 828.9in 2
The surface area of the object is 828.9in 2
The surface are of a sphere is given by S = 4 r 2
S = 4(3.14)(14cm) 2
S = 4(3.14)(196cm 2 )
S = 12.47(196cm 2 )
S = 2463cm 2
The surface area of this sphere is 2,463cm 2 .
and r = 14cm
Volume: the amount of ‘stuff’ enclosed in a three
dimensional object. Here is an example of what it looks
like. The units are length cubed.
This object has 72 "little unit
cubes inside of it because its
volume is 72units3 . We'll work
this in a moment.
4 3
The volume of a sphere is given by: V =  r
3
r = 14cm
4
V = (3.14)(14cm)3
3
4
V = (3.14)(2744cm3 )
3
12.56
3
V=
2744cm


3
34,464.64 3
V=
cm
3
V = 11,488.21cm3
The volume of this sphere is 11,488,21cm3 .
Examples
We found the surface area of this
object, now let’s find the volume.
The volume of a rectangular solid is given by V = LWH
Let the units be cm's.
L = 8cm
W = 3cm
H = 4cm
V = (8cm)(3cm)(4cm)
V = (24cm 2 )(4cm)
V = 96cm3
The volume of this object is 96cm3
Pyramid
1
The volume of a triangular pyramid is V = lwh
3
l=2500ft w = 2500ft
h = 2000ft
This is a triangular
pyramid that is
2000ft tall and
each base length
is 2500ft. What is
the area?
1
2
2500ft

  2000ft 
3
1
V =  6, 250, 000ft 2   2000ft 
3
1
V = 12,500, 000, 000ft 3 
3
V = 4,166,666,667ft 3
V=
The volume of the pyramid is 4,166,666,667ft 3 .
Euler’s formula:
# of vertices - # edges + #face = 2
Example:
If #vertices = 11, # faces = 5, find the # edges
# of vertices - # edges + #face = 2
11 - E + 2 = 2
13 - E = 2
13 - E – 13 = 2 – 13
-E = -11
E = 11
There are 11 edges
Converting square and cube units
Convert 15.6 ft 2 to yd 2
 1yd 
2
15.6 ft  

12
ft


= 15.6 ft 2
2
1yd 2

144 ft 2
 0.108 yd 2
15.6 ft 2 is 0.108 yd 2
Convert 78.33 yd 3 to ft 3
3
 12 ft 
78.33 yd 3  

1
yd


3
1728
ft
 78.33 yd 3 
1 yd 3
 135354 ft 3
78.33 yd 3 is 135,354 ft 3
Thank you for attending!