1 12.1 pythagoras & volume the student will learn about: the pythagorean theorem, 1 and the...
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12.1 Pythagoras & Volume
The student will learn about:
the Pythagorean Theorem,
1
and the volumes of basic shapes.
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Early Beginnings
In ancient times the special relationship between a right triangle and the squares on the three sides was known.
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Early Beginnings
OR
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Indeed, the Assyrians had knowledge of the general form before 2000 b.c.
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Early Beginnings
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)
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Pythagorean Theorem 1Pythagorean dissection proof.
a
a
a
a
b
b
bb
cc
c
c
c
=
a
a
a
a
b
b
b
b
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Pythagorean Theorem 2Bhaskara’s dissection proof.
c
a
a
a
bb
b
b
a
c c
c
c2 = 4 · ½ · a · b + (b – a)2
Careful: You need to show largest and center quadrilateral are squares.
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Pythagorean Theorem 3
a
a
a
a
b
b
b
b
cc
cc
Start with a square of sides length a + b.
Construct the four congruent right triangles.
The quadrilateral formed by the four hypotenuses form a square. Why?
(a + b) 2 = c 2 + 4 · ½ · a · b
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Pythagorean Theorem 4Garfield’s dissection proof.
a
b
b
ac
c
½ (a + b) · (a + b) = 2 · ½ · a · b + ½ · c2
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Pythagorean Theorem 6Euclid’s proof.
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Pythagorean Theorem 6Ratio-proportion proof.
ab
Ac
C
Bx y
S x b:
H b c
L y a:
H a c
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ExtensionsSemicircles
Prove it for homework.
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ExtensionsGolden 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.
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Postulates - Hilbert
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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.
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Cavalieri’s Principle for Volmue
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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
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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.
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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.
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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.
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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.
The altitude of the pyramid is the perpendicular distance from V to E. R
E
V
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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.
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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.
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Theorem.
Theorem. The volume of any pyramid is one-third 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.
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Truncated Pyramids.
Volume of a truncated pyramid is that of the full pyramid minus the pyramid cut off the top.
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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.
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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.
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Sphere
The Problem is to find a shape with a known volume that compares to a sphere – What do you thinks works?
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Cavalieri’s Method
r
r h
r h
h
2π r
2π(r – h)
P
Hemisphere only!
VHS = VP = r 2 2π r = π r 3 1
3
2
3
S
rh
22 hr
Hence the volume of the sphere is 3r3
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Sphere
The derivative of the volume!
The volume of the sphere is 3r3
4
The surface area of the sphere is 4π r 2.
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Platonic Solids
You should be able to find the volume of the first three of these solids.
You should be able to find the surface area of all five of these solids.
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Summary.
• We learned about volume postulates.
• We learned about the volume of basic shapes.
• We learned Cavalieri’s Principle for volume.
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Assignment: 12.1