Solids
Surface Area and Volume
Chapter 12
D. Prescott
BEHS
Polyhedron
• Definition:
•
A 3-D region
bounded by polygons
Classification
• All polyhedra are either convex or
nonconvex (concave)
• Convex polyhedron – any 2 of its surface
points can be connected by a segment that
lies entirely inside or on the polyhedron
Nonconvex
Regular polyhedron
All of its polygonal surfaces are:
• congruent to each other and
• equilateral and equiangular
Platonic Solids are regular polyhedra
Terms:
All “sides” are called faces. They are
always in the shape of a polygon.
An edge is a segment formed
by the intersection of 2 faces.
A vertex is a point where
3 or more edges meet.
Euler’s Theorem
• The number of faces (F) plus the number
of vertices (V) is always equal to two more
than the number of edges (E).
• F+V=E+2
5+6=9+2
11 = 11
Cross Section
• Definition – the intersection of a plane and
a solid
Click Below
for Prisms or Pyramids
• Prism Presentation
• Pyramid Tutorial
Prisms
Measurement of Prisms
Including Surface Area and Volume
Table of Contents (Prisms)
I. Prisms
• Definition
• Basic Terms for Prisms
– Faces, Edges, Height
• Classification of Prisms
– Base Shape, Right, Oblique
• Surface Area of Prisms
– Definition, Lateral Area
– Total Area, Example
• Volume of Prisms
– Definition
– Formula, Example
Prism – a polyhedron that is
composed of 2 congruent
parallel faces and
parallelograms for the
remaining faces.
BASES
In every prism there are
always 2 faces that are
congruent polygons lying in
parallel planes. These faces
are called bases.
In this figure the 2 bases are trapezoids
located in the front and back.
Names of Prisms
Prisms are named by the
shape of their bases.
Octagonal prism
Trapezoidal prism
Rectangular prism
 FACES 
In any prism the lateral faces
are parallelograms.
In any right prism the lateral
faces are rectangles.
EDGES
The intersection of any 2
faces is a segment that is
called an edge. If it is a
side of a base, then it is
called a base edge.
The intersection of 2 lateral
faces is called a lateral edge.
Height of a Prism
The altitude of a
prism is a segment
joining the 2 bases
and is perpendicular
to both.
The height of a prism
is called an altitude.
Right Prisms
If the lateral faces are
rectangles, then the
height will be congruent
to a lateral edge.
Any prism that has rectangular lateral
faces is called a right prism.
Oblique Prisms
• If the lateral faces are not rectangles, then the
altitude will not be congruent to a lateral edge.
• Any prism that does not have rectangular lateral
faces is called an oblique prism.
Classify the Prisms
Right Pentagonal Prism
Right Rectangular Prism
Surface Area of Prisms
The surface area of a
prism can be found by
finding the area of each
face and then adding them
together. However, this
technique can be quite
time consuming.
This octagonal prism has 8 lateral faces and 2
bases for a total of 10 faces. You would have
to find the area of 10 polygons and then add
the 10 numbers together.
Formula for Lateral Area
Face 1: e1•h
h
e2
e1
e4
Face 2: e2•h
Face 3: e3•h
e3
Face 4: e4•h
Lateral Area = e1•h + e2•h + e3•h + e4•h
LA = (e1+ e2+ e3+ e4)•h
LA = p •h
Total Surface Area
h
e2
e1
e4
e3
p=perimeter of the base
H=Height of the prism
B= Area of the base
The total surface area(TSA) is
equal to the lateral area(LA)
plus the area of the Base(B)
times 2. The Bases are congruent
polygons.
TSA = LA + 2B
Or
TSA = pH + 2B
Find the Total Surface Area
5cm =H
3cm
7cm
7
3
3
7
TSA =
p•H + 2B
TSA = 20cm •5cm + 2(21cm2)
TSA = 100cm2 + 42cm2
TSA = 142cm2
p = 7+3+7+3
p = 20cm
B = 7cm •3cm
B = 21cm2
Find the surface area of the
triangular prism:
TSA = p•H + 2B
8cm
20cm =H
6cm
First find the perimeter and area
of the base:
8cm
6cm
p=a+b+c
p = 6 + 8 + 10 c2 = a2 + b2
p = 24cm
c2 = (6)2 + (8)2
c2 = 36 + 64
1
c2 = 100
B  bh
2
c = 10
1
B  (6)(8)
2
2
B  24cm
2
B=24cm
p=24cm,
and H =20cm
TSA = p•H + 2B
S = (24cm)•(20cm) + 2(24cm2)
TSA = 480cm2 + 48cm2
8cm
20cm
6cm
TSA = 528cm2
Surface Area of Cylinders
Proof of formula based on prisms formula:
TSA = p•H + 2B
(prisms)
TSA  ch  2 B
TSA  2rh  2r
2
Net Diagrams can be used to find the
surface area of a 3-D figure:
A net diagram is a 2-dimensional sketch
made by making imaginary cuts along some
of the solid’s edges and “unfolding” it.
Total Surface Area of Cylinders
TSA=2πrh + 2πr2
r
r
h
2
Area = length · width
Area = 2πr·h
h
2πrh
C  2r
r 2
Find the Total Surface Area
TSA  2rh  2r
2
S

2

(
10
)(
12
)

2

(
10
)
10cm
12cmTSA  240  200
2
TSA  440 cm
2
Pyramids
Measurement of Pyramids including
Volume and Surface Area
by Donna B. Prescott
Table of Contents (Pyramids)
II. Pyramids
• Historical Connections
– Mayan Pyramids
– Egyptian Pyramids
• Geometrical Terms
– Faces, Vertex, and Regular
Pyramids
– Classification
– Slant Height and Altitude
•Right Triangles in
Regular Pyramids
–Formed with
the Altitude
–Formed with
the Slant Height
•Surface Area
–Lateral Area
–Total Area
•Volume
•Credits
•References
Mexico’s Pyramids
The ancient
Mayans also
built pyramids. It
is believed that
their pyramids
had celestial
significance
(GeoCities.com).
Egyptian Pyramids
“The ancient Egyptians built pyramids as
tombs for the pharaohs and their
queens”(British Museum, 1999).
Geometrical Terms
Vertex
All pyramids have triangular
lateral faces which intersect
in a single point called the
vertex of the pyramid.
Pyramids are classified
according to the shape of the
base. Regular pyramids have
regular polygons for their
bases.
Classify the following pyramids.
Triangular pyramid
Hexagonal pyramid
Altitude and Slant Height
The length of the segment
joining the vertex (V) and the
center of the base (A) is the
altitude of the pyramid.
V
A
B
It is perpendicular to the base
in a right pyramid.
The length of the segment
from the vertex (V) to the
midpoint of a base edge (B) is
called slant height.
Right Triangles in
Regular Pyramids
Given: A regular square
pyramid with base edges
8cm and altitude 12cm
V
How long is the slant height?
A
B
8
l
12
4
4  12  l
2
2
16  144  l
160  l
2
4 10cm
2
2
l=slant
height
Another Right Triangle
V
Find the length of a lateral edge.
4 10   4
2
8
160  16  c
B
E 4
V
c
E
2
4 10
4
B
c
2
176  c 2
176  c
4 11  c
2
2
Formula for Surface Area
V
TSA = LA + B
TSA = ½ pl + B
l=slant height
p =perimeter of the base
B = area of the Base
LA = ½ e1l + ½ e2l + ½ e3l + ½ e4l
LA = ½(e1 + e2 + e3 + e4)l
LA = ½ pl
Find the Area
V
l = 4 10
TSA= ½ pl + B
TSA= ½·(32)(4 10) + 64
8cm
TSA=(64 10 + 64)cm2
8cm
p = 4(8)
p = 32cm
B = 8(8)
B = 64cm2
Terminology in Cones
Vertex
Cone’s
height
Slant
height
20 in
l
h
Cones also have a slant
height, the cone’s height,
and a vertex.
Right Triangles in Cones:
h  6  20
2
h  36  400
2
h  364
h  2 91
2
6 in
2
2
h
20 in
6 in
Surface Area of a Cone
TSA= ½ pl + B (pyramids)
TSA= ½ cl + B
TSA= ½ (2πr)l + πr2
TSA= (½ ·2)πrl + πr2
TSA= πrl + πr2
Find the surface area:
TSA= πrl + πr2
TSA= π(6)(20) + π(6)2
TSA = 120π + 36π
TSA = 156π in2
20 in
6 in
Section 12.4 Volume of Prisms and Cylinders
Page 743
Mrs. Prescott, BEHS
VOLUME OF SOLIDS
Volume of Prisms
h
e1
e2
e4
e3
Volume of a prism is
measured in cubic units,
and measures the space
inside of the figure.
A cubic unit is like an ice cube which
1 measures 1 unit in length, width, and
1 1 height. Finding the volume would be
like counting the number of ice cubes
in a stack of ice trays.
Volume Formula
The number of cubes that
would take to fill up the
h bottom row of the figure
can be determined by
e
2
e1
e3
finding the area of the
e4
base(B).
The number of rows is equal to the height of the prism.
Volume is equal to the area of the prism’s base(B)
times the prism’s height(H).
V= BH
Find the Volume of the Prism
5cm
Base = 21cm2 3cm
7cm
3cm
7cm
Height = 5cm
V= BH
V = 21cm2•5cm
V=105cm3
Find the Volume:
1
B  bh
2
1
B  (5cm)(10cm)
2
B  25cm 2
18cm = height
10cm
5cm
V= BH
V  (25cm )(18cm)
2
V  450cm3
Volume of A Cylinder
V= BH
V  r h
2
Base is a Circle
B = _________
 r2
Find the Volume
V r h
2
10cm
14cm
V   (10cm) (14cm)
2
V   (100cm )(14cm)
2
V  100 14 cm  cm
2
V  1400 cm
3
Volume of a Pyramid
A pyramid can hold onethird the volume of a prism
with the same base area and
height.
1
V   BH
3
The formula for finding the volume of a pyramid
is one-third times the area of the base times the
pyramid’s altitude.
Given: A regular square pyramid with
base edges 8cm and altitude 12cm
1
V  BH
12cm
3
8cm
H = 12cm
B=
64cm2
1
V  (64cm 2 )(12cm)
3
3
1  64  12cm
V 
3
V  256cm
3
Volume of a Cone
1
Cone' s Volume  of cylinder ' s volume
3
1 2
V  r h
3
h
r
Find the Cone’s Volume
First find the cone’s height
h  6  20
2
h  36  400
2
20 in
20in
2
h  364in
h  2 91in
2
6 in
2
2
h
20in
6 in
1 2
Cone' s Volume  r h
3
20 in
6 in
h  2 91
1
2
V   (6in ) (2 91in )
3
1
2
V  (36in )( 2 91in )
3
3
V  12  (2 91) in
V  24 91 in
3
Spheres
• Definition
– The set of all points that are a
given distance from a set point
Interesting Facts
The Planet Earth, our home, is
nearly a sphere, except that it is
squashed a little at the poles.
It is a spheroid, which means it just misses
out on being a sphere because it isn't perfect
in one direction (in the Earth's case: NorthSouth)
Largest Volume for Smallest Surface
Put another way it can contain
the greatest volume for a fixed
surface area
If you blow up a balloon it naturally
forms a sphere because it is trying to
hold as much air as possible with as
small a surface as possible. The sphere
appears in nature whenever a surface
wants to be as small as possible.
Examples include bubbles and water
drops.
Terms in spheres:
•
•
•
•
Center O
Radius OD
Diameter CE
Chord AB
A
E
B
O
C
D
Great Circle & Hemisphere
In geometry a
hemisphere is an exact
half of a sphere.
It also refers to half of the
Earth, such as the
"Northern Hemisphere”
A Great Circle is a cross section that passes
through the sphere’s center and divides the
sphere into 2 congruent halves.
Formulas for Spheres
How are they derived?
TSA  4r
2
Formulas for Spheres
2
Volume of Sphere  of Volume of Cylinder
3
Formulas for Spheres
2
Volume of Sphere  of Volume of Cylinder
3
2
2
V   r h
3
2
2
V   r (2r )
r
3
4 3
h
V  r
3
Formulas for Spheres
The total surface area of a sphere:
TSA  4r
The volume of a sphere:
4 3
V  r
3
2
Given: r = 10cm, find the sphere’s
surface area and volume.
TSA  4r
2
S  4 (10cm)
2
S  4  100cm
2
TSA  400 cm
2
Given: r = 10cm, find the sphere’s
surface area and volume.
4 3
V  r
3
4
3
4
3
V



1000
cm
V   (10 cm)
3
3
4000
3
V
cm
3
Given: the sphere’s surface area is
800π m2, find its radius.
TSA  4r
2
800 m  4r
2
4π
4π
200 m  r
2
2
200 m  r
2
10 2 m  r
2
2