12.1 Pythagoras & Volume
The student will learn about:
the Pythagorean Theorem,
and the volumes of basic
shapes.
1
1
Early Beginnings
In ancient times the
special relationship
between a right
triangle and the
squares on the
three sides was
known.
2
Early Beginnings
OR
3
Early Beginnings
Indeed, the Assyrians had knowledge of
the general form before 2000 b.c.
The Babylonians had knowledge of all of the
Pythagorean triples and had a formula to
generate them.
(3,4,5)
( 5, 12, 13)
( 7, 24, 25)
( 8, 15, 17)
( 9, 40, 41)
(11, 60, 61) (12, 35, 37) (13, 84, 85)
(16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65)
(36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)
4
Pythagorean Theorem 1
Pythagorean dissection proof.
b
a
a
b
a
a
b
b
=
b
a
c
c
a
b
a
c
c
b
c
b
a
5
Pythagorean Theorem 2
Bhaskara’s dissection proof.
c
a
c
b
a
c2 = 4 · ½ · a · b + (b – a)2
b
b
a
b
c
a
c
Careful: You need to show largest and
center quadrilateral are squares.
6
Pythagorean Theorem 3
a
b
a
b
a
c
Start with a square of sides
length a + b.
c
c
b
c
b
Construct the four congruent
right triangles.
a
The quadrilateral formed by the four hypotenuses
form a square. Why?
(a + b) 2 = c 2 + 4 · ½ · a · b
Pythagorean Theorem 4
Garfield’s dissection proof.
a
b
c
a
c
b
½ (a + b) · (a + b) = 2 · ½ · a · b + ½ ·
c2
8
http://www.usna.edu/MathDept/mdm/pyth.html
9
Pythagorean Theorem 6
Euclid’s proof.
10
Pythagorean Theorem 6
Ratio-proportion proof.
C
a
b
A
S x b
: 
H b c
y
x
B
c
L y a
: 
H a c
11
Extensions
Semicircles
Prove it for homework.
12
Extensions
Golden Rectangles
Prove it for homework.
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THE GENERAL EXTENSION TO PYTHAGORAS'
THEOREM: If any 3 similar shapes are drawn on
the sides of a right triangle, then the area of the
shape on the hypotenuse equals the sum of the
areas on the other two sides.
14
Postulates - Hilbert
Unit Postulate - The volume of a rectangular
parallelepiped is the product of the altitude and
the area of the base.
Cavalieri’s Principle – next slide please.
15
Cavalieri’s Principle for Volmue
If, in two solids of equal altitude, the sections
made by planes parallel to and at the same
distance from their respective bases are always
equal, then the volumes of the two solids are
equal
16
Prisms.
A prism is the figure formed when the
corresponding vertices of two congruent
polygons, lying in parallel planes are joined.
The lines joining the corresponding vertices are
called lateral edges. The congruent polygons are
called the bases, and the other surfaces are
called the lateral faces, or as a group, the lateral
surfaces.
17
Prisms.
If the lateral edges are perpendicular to the
plane of the bases, the prism is a right prism;
otherwise, it is an oblique prism.
18
Volume of a Prism.
Theorem. The volume of any prism is the
product of the altitude and the area of the base.
Use Cavalieri’s Principle and the unit postulate.
19
Pyramids.
Given a polygonal region R in a plane E, and a
point V not in E. the pyramid with base R and
vertex V is the union of all segments VQ for
which Q belongs to R.
V
The altitude of the
pyramid is the
perpendicular distance
from V to E.
R
E
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Theorem.
Theorem. If two pyramids have the same
altitude and the same base area, and their bases
lie in the same plane, then they have the same
volume.
Use Cavalieri’s Principle.
21
Theorem.
Theorem. The volume of a triangular pyramid
is one-third the product of its altitude and its
base.
In the figure the prism
is divided into three
pyramids of equal
volume.
22
Theorem.
Theorem. The volume of any pyramid is onethird the product of its altitude and its base.
Use Cavalieri’s Principle on a triangular pyramid
with area the same base as the given pyramid.
23
Truncated Pyramids.
Volume of a truncated pyramid is that of the
full pyramid minus the pyramid cut off the top.
24
Cylinders.
Use Cavalieri’s Principle a cylinder may be
treated the same as a prism. Hence the volume
of a cylinder is the product of its altitude and
the area of its base.
25
Cones.
Use Cavalieri’s Principle a cone may be treated
the same as a pyramid. Hence the volume of a
cone is one-third the product of its altitude and
the area of its base.
26
Sphere
The Problem is to find a shape with a
known volume that compares to a sphere –
What do you thinks works?
Cavalieri’s Method
Hemisphere only!
r 2  h2
r
h
2π(r – h)
VHS = VP =
r
r
h
h
1 2
2
r 2π r = π r 3
3
3
P
h
r
S
2π r
4
Hence the volume of the sphere is  r 3
3
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Sphere
4 3
The volume of the sphere is
r
3
The surface area of the sphere is 4π r 2.
The derivative of the volume!
Platonic Solids
You should be able to find the surface area of
all five of these solids.
You should be able to find the volume of the
first three of these solids.
Summary.
• We learned about volume postulates.
• We learned Cavalieri’s Principle for volume.
• We learned about the volume of basic shapes.
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Assignment: 12.1