Naming Geometric Solids
Dice: Cube
Soup Can: Cylinder
Basketball: Sphere
One of these is classified as a Polyhedron while two are not. What is a polyhedron?
A polyhedron is a three-dimensional figure formed by polygons.
As you recall, the sides of a polygon must be segments, no curves!
This means the cube, created by six squares is a polyhedron and the cylinder and sphere are not.
Polyhedra are divided into two basic categories, prisms and pyramids. The polygons that create the sides are
called faces and determine the names. The faces meet at edges. The edges meet at vertices which we often
call corners.
A prism has two matching polygonal faces parallel to each other, called bases, connected by rectangles or
parallelograms. The most common prism is the rectangular prism which looks like a box.
All prisms are named by the matching polygon surfaces.
Examples:
Pentagonal Prism
Rectangular Prism
Pyramids have one polygonal face, the base, joined by triangles coming together at a common vertex.
All pyramids are named using the base.
Examples:
Square Pyramid
Triangular Pyramid
Hexagonal Pyramid
Non-polyhedron solids include shapes like cones, cylinders, and spheres.
These solids have curved edges instead of segments where faces and bases meet.
There is a relationship between the number of faces (F), number of vertices (V), and number of edges (E) of a
polyhedron. What do you see?
POLYHEDRON
Rectangular
Prism
Triangular
Pyramid
Octagonal
Prism
Hexagonal
Pyramid
FACES
6
VERTICES
8
EDGES
12
4
4
6
10
16
24
7
7
12
RELATIONSHIP?
For each polyhedron, the number of faces (F) + number of vertices (V) = number of edges (E) + 2.
This is called
Euler’s Theorem.
F+V=E+2
Using Euler’s Theorem:
Example 1: A solid has 20 faces and 30 edges, how many vertices does it have?
Solution:
F+V=E+2
Substitute 20 for the number of faces and 30 for
20 + V = 30 + 2 number of edges. Then solve.
20 + V = 32
V = 12
Example 2: A soccer ball resembles a polyhedron with 32 faces; 20 are regular hexagons and 12 are regular
pentagons. If you constructed a similar polyhedron, how many vertices would it have?
Solution:
F+V=E+2
32 + V = 90 + 2
32 + V = 92
V = 60
Substitute 32 for the number of faces. The number of
edges needs to be computed. Since there are 20
hexagons, 20(6) = 120. There are 12 pentagons 12(5) =
60. There are not 180 edges in the polyhedron since the
edges are shared. Each edge will be shared by two
polygons so we divide 180 by 2 arriving at 90. Substitute
90 for E and solve.
Cross Sections
Visualize a plane slicing through a solid. The intersection of that plane and the solid is called a cross section,
much like slicing a stick of butter. You can slice the butter to get a section exactly like the end and have a
square. Leaning the knife can result in a triangle, rectangle, or trapezoid. Try it!
Examples:
For each of the solids below, describe the shape formed by the intersection of a horizontal plane and the solid.
a. Cube
b. Cylinder
c. Sphere
Solutions:
a. square
b. circle
c. circle * Unless the plane intersects exactly with the top or bottom of the sphere. If the plane
intersects exactly with the top or bottom of the sphere the intersection will be a point.
Forming Solids
Were you able to sketch a pattern that can be cut out, folded, and formed to create the cube? These
patterns are the nets for the solids and unique for each shape.
The net for a rectangular prism could look like this:
The yellow rectangles that serve as
the ends of the prism could be moved
so the net would look like this:
This means that the net can be drawn
more than one way. A net for this
prism needs two of each rectangle
arranged so congruent rectangles will
be parallel once folded and matching
edge lengths would be able to meet.
Either of these could be folded to
form rectangular prisms like the one
below.
A rectangular pyramid would need a
rectangle for the base and four triangles
for the lateral sides.
Here’s one possible net.
The net for a cylinder will be made using
a rectangle and two congruent circles.
All other pyramids would have similar
nets with the base in the middle surrounded
by triangles on each side.
Platonic Solids
There are five special solids formed using regular polygons.
Cube:
6 square faces
Tetrahedron:
4 equilateral triangle faces
Octahedron:
8 equilateral triangle faces
Dodecahedron:
12 regular pentagonal faces
Icosahedron:
20 equilateral triangle faces
Lateral Area, Surface Area and Volume of Prisms & Cylinders
Prisms can be right or oblique. If the lateral faces of the prism are rectangles resulting in lateral edges that are
perpendicular to the bases, the prism is right. If the lateral faces are parallelograms the lateral edges will meet
the bases to form angles that are not perpendicular and the prism will be oblique. This is important because the
altitude or height of a prism will always be determined by measuring the distance between the two bases. If the
prism is oblique, the height will not be one of the actual edges.
width
height
width
length
height
length
FORMULAS FOR RIGHT PRISMS:
The Lateral Area of any polyhedron is the sum of its lateral faces.
The formula to compute the lateral area for right prisms is LA = Ph where P represents the perimeter of the
base and h is the height of the prism
The Surface Area of any polyhedron is the sum of the areas of its faces.
To determine the surface area of right prisms SA = 2B + Ph where B is the area of the base, P represents the
perimeter of the base, and h is the height of the prism.
Volume is the number of cubic units contained in the interior of any of polyhedron and for right or oblique
prisms V = Bh where B is the area of the base and h is the height of the prism.
Example:
Determine the Lateral Area, Surface Area and Volume for prism below.
Solutions:
12 cm
10 cm
LA = PhThe perimeter of the base is 36 since
= (36)(12)
10 + 10 + 8 + 8 = 36.
2
= 432 cm
The height is 12.
SA = 2B + Ph
The perimeter was determined to be
= 2(80) + (36)(12) 36, the height is 12. Since
= 160 + 432
the base is a rectangle the
8 cm
2
= 592 cm
area of the base is 10 x 8.
(continued on the next page)
V = Bh or V = lwh
Replacing length, width and height
= 10(8)(12)
with the appropriate values
3
3
= 960 cm
we find the volume is 960 cm .
(We used length x width since the base is a rectangle and A = length x width.)
The formulas for Right Cylinders are:
LA = 2  rh
2
SA = 2  rh + 2  r
2
V = r h
r is the radius of the cylinder and h is the height.
r is the radius of the cylinder and h is the height.
r is the radius of the cylinder and h is the height.
Why? Because a cylinder is just a special type of prism with a circular base.
When the base formulas (perimeter of the base or area of the base) are replaced with the formulas we have for
the circle, we derive specific formulas we can use just for cylinders.
Example:
solve.
Solutions:
LA = 2  (6)(4)
= 48 
= 48(3.14)
2
= 150.72 cm
Substitute the radius and height
values for the variables and
SA = 2  rh + 2  r
Substitute the radius and height
2
= 2  (6)(4) + 2  (4)
values for the variables and
= 48  + 32 
solve.
= 80 
= 80(3.14)
2
= 251.2 cm
2
V = r h
2
=  (6) 4
= 144 
= 144(3.14)
3
= 452.16 cm
2
Substitute the radius and height
values for the variables and
solve.