Chapters 5: 3-Dimensional Shapes and
Surface Area
Definition
Simple
Definition
Simple
Closed
Definition
Simple
Closed
Definition
A simple closed surface has exactly one interior, no holes and is
hollow.
Definition
Simple
Closed
Definition
A simple closed surface has exactly one interior, no holes and is
hollow.
Definition
A solid is a simple closed surface taken with its interior.
Definition
Simple
Closed
Definition
A simple closed surface has exactly one interior, no holes and is
hollow.
Definition
A solid is a simple closed surface taken with its interior.
Definition
A polyhedron is a solid where each face is a polygon.
Parts of a Polyhedron
Naming Polyhedron
Shape of base
Naming Polyhedron
Shape of base
Whether or not the faces are perpendicular to the base (Right)
Naming Polyhedron
Shape of base
Whether or not the faces are perpendicular to the base (Right)
Type of polyhedron
Nets
What is a net?
Nets
What is a net?
Definition
A net is a two dimensional representation of a three dimensional
object. Nets are not unique with respect to the orientation of the faces
of the figure.
Prisms
Prisms
Properties
two bases, which are necessarily parallel
Prisms
Properties
two bases, which are necessarily parallel
lateral faces are all parallelograms
Prisms
Properties
two bases, which are necessarily parallel
lateral faces are all parallelograms
if a right prism, the bases are perpendicular to the lateral faces
Prisms
Properties
two bases, which are necessarily parallel
lateral faces are all parallelograms
if a right prism, the bases are perpendicular to the lateral faces
if a right prism, the lateral faces are rectangles
Pyramids
Pyramids
Properties
one base
Pyramids
Properties
one base
lateral faces are all triangles, meeting at the apex
Pyramids
Properties
one base
lateral faces are all triangles, meeting at the apex
if a right pyramid, the altitude is perpendicular to the center of
the base
Pyramids
Properties
one base
lateral faces are all triangles, meeting at the apex
if a right pyramid, the altitude is perpendicular to the center of
the base
if a right pyramid, the lateral faces are isosceles triangles
Cylinders
Cylinders
Properties
two bases that are necessarily parallel but not necessarily circles
Cylinders
Properties
two bases that are necessarily parallel but not necessarily circles
one lateral face, which when ‘opened up’ is a parallelogram
Cylinders
Properties
two bases that are necessarily parallel but not necessarily circles
one lateral face, which when ‘opened up’ is a parallelogram
if bases are directly over each other, meaning that the lateral face
is perpendicular to the bases, then we have a right cylinder
Cylinders
Properties
two bases that are necessarily parallel but not necessarily circles
one lateral face, which when ‘opened up’ is a parallelogram
if bases are directly over each other, meaning that the lateral face
is perpendicular to the bases, then we have a right cylinder
if a right cylinder, the lateral face is a rectangle
Cylinders
Properties
two bases that are necessarily parallel but not necessarily circles
one lateral face, which when ‘opened up’ is a parallelogram
if bases are directly over each other, meaning that the lateral face
is perpendicular to the bases, then we have a right cylinder
if a right cylinder, the lateral face is a rectangle
Note: If not right, then the figure is called oblique.
Cones
Cones
Properties
one base, not necessarily a circle
Cones
Properties
one base, not necessarily a circle
one lateral face, which when ‘opened up’ is a portion of a circle
Cones
Properties
one base, not necessarily a circle
one lateral face, which when ‘opened up’ is a portion of a circle
the point at the top is the apex
Cones
Properties
one base, not necessarily a circle
one lateral face, which when ‘opened up’ is a portion of a circle
the point at the top is the apex
if the altitude is perpendicular to the base, we have right cone
Spheres
Spheres
Properties
one face
Spheres
Properties
one face
no edges
Spheres
Properties
one face
no edges
every point on sphere is equidistant from the center
Spheres
Properties
one face
no edges
every point on sphere is equidistant from the center
gray circle is called the great circle
Euler’s Formula
There is a relationship between the number of vertices, faces and
edges in any simple polyhedron. The shape of the base doesn’t matter
- the formula will hold.
Using your nets for polyhedra, count the number of faces, edges and
vertices and look for a relationship.
Euler’s Formula
There is a relationship between the number of vertices, faces and
edges in any simple polyhedron. The shape of the base doesn’t matter
- the formula will hold.
Using your nets for polyhedra, count the number of faces, edges and
vertices and look for a relationship.
Euler’s Formula
Euler’s Formula
For a simple polyhedra, F + V = E + 2.
Platonic Solids
Definition
A Platonic solid is a regular, convex polyhedron with all faces
congruent and the same number of faces meeting at each vertex.
Only five polyhedra are knows to satisfy this definition.
Tetrahedron
Hexahedron
Octahedron
Dodecahedron
Icosahedron
Example
Example
What is the length of the longest rod we can fit in a right circular
cylinder with radius 3 inches and height 8 inches?
Example
Example
What is the length of the longest rod we can fit in a right circular
cylinder with radius 3 inches and height 8 inches?
3 in
8 in
Solution
3 in
3 in
D
8 in
Solution
3 in
3 in
D
8 in
So what we have is a right triangle with legs 6 and 8 inches long.
Solution
3 in
3 in
D
8 in
So what we have is a right triangle with legs 6 and 8 inches long.
So, by the Pythagorean theorem, we have
62 + 82 = D2 ⇒ D2 = 100 ⇒ D = 10
So, the length of the longest rod is 10 inches.
Surface Area
Who can define surface area?
Surface Area
Who can define surface area?
Definition
The surface area of a solid figure is the sum of all of the areas of the
faces, whether polygons or curved figures.
Surface Area
Who can define surface area?
Definition
The surface area of a solid figure is the sum of all of the areas of the
faces, whether polygons or curved figures.
Definition
The lateral surface area of a solid figure is the sum of the areas of the
faces, not including the bases.
Prisms
What is the formula for the surface area of a right prism? Can we
generalize the formula so that it applies to all right prisms?
Prisms
What is the formula for the surface area of a right prism? Can we
generalize the formula so that it applies to all right prisms?
Formula
Right Prism
The surface area of a right prism is 2AB + PB h.
Pyramids
Can we do similar for right pyramids?
Pyramids
Can we do similar for right pyramids?
Formula
Right Pyramid
The surface area of a right pyramid is AB + 12 PB l.
Cylinders
We said cylinders are not polyhedra. Why?
Cylinders
We said cylinders are not polyhedra. Why?
Can we generalize the surface area formula of the right cylinder?
Cylinders
We said cylinders are not polyhedra. Why?
Can we generalize the surface area formula of the right cylinder?
Formula
Right Circular Cylinder
The surface area of a right circular cylinder is 2πr2 + 2πrh.
Cones
And finally, our cones. Can we do the same for these?
Cones
And finally, our cones. Can we do the same for these?
Formula
Right Circular Cone
The surface area of a right circular cone is πr2 + πrl where l is the
slant height.
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Face
πrl
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Face
πrl
If the lateral face was a whole circle, we would have an area of πl2 .
But ... we don’t. So what part do we have?
Justification for the Cone Formula
There are two parts to the formula
Base
πr2
This is the area of a circle.
Lateral Face
πrl
If the lateral face was a whole circle, we would have an area of πl2 .
But ... we don’t. So what part do we have?
πl2 ×
2πr
r
= πl2 × = πrl
2πl
l
Sphere
Formula
Sphere
The surface area of a sphere is 4πr2 .
Sphere
Formula
Sphere
The surface area of a sphere is 4πr2 .
Lateral SAcylinder = 2πrh = 2πr × 2r = 4πr2
Surface Area of a Cone
Example
Find the surface area of the cone if the net corresponding to the lateral
surface area of the cone has a radius of 3 inches and this portion of the
net has a central angle of 180◦ .
Surface Area of a Cone
Example
Find the surface area of the cone if the net corresponding to the lateral
surface area of the cone has a radius of 3 inches and this portion of the
net has a central angle of 180◦ .
This is
1
2
of a circle with radius 3 inches, so the lateral surface area is
1
9
π(3)2 = π in2
2
2
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =
3
in
2
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =
Therefore, the base has an area of
3
in
2
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =
3
in
2
Therefore, the base has an area of
2
3
9
π
= π in2
2
4
Base of the Cone
Now the dimensions of the base isn’t given. We know it is a circle,
and we know it’s circumference must be as long as the rolled up
lateral face. So, the circumference must be
1
2π(3) = 3π in
2
Now, we need the radius of the base so that we can find the area.
3π = 2πr ⇒ r =
3
in
2
Therefore, the base has an area of
2
3
9
π
= π in2
2
4
So, the total surface area is
9
9
27
π in2 + π in2 = π in2
2
4
4
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Inner ring: We pretend we have a full cylinder and are finding it’s
surface area.
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Inner ring: We pretend we have a full cylinder and are finding it’s
surface area.
SAlat = 2πrh = 2π(25)(10) = 500π mm2
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Inner ring: We pretend we have a full cylinder and are finding it’s
surface area.
SAlat = 2πrh = 2π(25)(10) = 500π mm2
Outer ring: Again, we pretend we have a full cylinder.
SAlat = 2πrh = 2π(35)(10) = 700π mm2
Surface Area of a Ring
Example
Suppose you have a ring that you want to get replated. The jeweler
needs to know the surface area of the ring in order to determine how
much to charge you. If the ring has an inner radius of 25 mm and the
ring is 10 mm thick and is 10 mm high, find the surface area.
Inner ring: We pretend we have a full cylinder and are finding it’s
surface area.
SAlat = 2πrh = 2π(25)(10) = 500π mm2
Outer ring: Again, we pretend we have a full cylinder.
SAlat = 2πrh = 2π(35)(10) = 700π mm2
Now the circles on top and bottom ...
Solution (cont.)
Outer circle: The outer radius is 35 mm, so the area is
Solution (cont.)
Outer circle: The outer radius is 35 mm, so the area is
π(35)2 = 1225π mm2
The inner radius is 25 mm, so the area is
Solution (cont.)
Outer circle: The outer radius is 35 mm, so the area is
π(35)2 = 1225π mm2
The inner radius is 25 mm, so the area is
π(25)2 = 625π mm2
So, the area of the disc is
Solution (cont.)
Outer circle: The outer radius is 35 mm, so the area is
π(35)2 = 1225π mm2
The inner radius is 25 mm, so the area is
π(25)2 = 625π mm2
So, the area of the disc is
1225π mm2 − 625π mm2 = 600π mm2
Putting these together we see
SA = 2 × 600π mm2 + 700π mm2 + 500π mm2 = 2400π mm2