Title: Pythagoras
1Pythagoras
2What is Pythagoras Theorem?
a2 b2 c2
- The famous British mathematician and science
writer Jacob Bronowski considered this the most
important mathematical result of all times
because it connects space with numbers in a
dramatic way
3Was Pythagoras the Only One Who Knew?
- We have evidence that the Babylonians knew this
relationship some 1000 years earlier. - Plimpton 322, a Babylonian mathematical tablet
dated back to 1900 B.C., contains a table of
Pythagorean triplets.
4Following that
There have been many (well over 300) proofs of
the theorem Liu Hui ? Leonardo Da Vinci ?
President J.A. Garfield
Shearing Similarity Dissection
Three main types of proofs
5Shearing
6Shearing
(ABCD) (AB1C1D)
(ABC) (ABC1)
7Shearing
(ABCD) 2(ABC1)
8Shearing
AC2 BC2 AB2
This is proved by showing,
(ACQP)
(BRSC)
(AXYB)
9Shearing
ACQP is a square. Therefore,
AC2 (ACQP)
10 (ABP) ½(base)(height)
½(AP)(AC) 2(ABP) (AP)(AC)
(ACQP)
2(ABP) (ACQP)
base
11Shearing
ACQP is a square.
Therefore, AC2 (ACQP)
1. 2(ABP) (ACQP)
12Y
U
X
V
(AXC) ½(base)(height)
½(AX)(AV) 2(AXC) (AX)(AV)
(AXUV)
2(AXC) (AXUV)
13Shearing
ACQP is a square.
Therefore, AC2 (ACQP)
1. 2(ABP) (ACQP)
2. 2(AXC) (AXUV)
14Shearing
15 AP AC
AB AX
?PAB 90O ?BAC
?CAX 90O ?BAC
?PAB ?CAX
ABP and AXC are congruent triangles.
(SAS)
(ABP) (AXC)
16Shearing
Y
ACQP is a square.
Therefore, AC2 (ACQP)
U
B
X
V
1. 2(ABP) (ACQP)
2. 2(AXC) (AXUV)
A
C
3. (ABP) (AXC)
2(ABP) 2(AXC) (ACQP)
(AXUV)
Q
P
17 Shearing
Y
U
B
(ACQP) (AXUV)
X
V
A
C
Q
P
18Shearing
BRSC is a square. Therefore, BC2
(BRSC)
19 (ABR) ½(base)(height)
½(BR)(BC) 2(ABR) (BR)(BC)
(BRSC)
2(ABR) (BRSC)
20Shearing
BRSC is a square. Therefore, BC2
(BRSC)
1. 2(ABR) (BRSC)
21 (YBC) ½(base)(height) ½
(YB)(BV) 2(YBC) (YB)(BV)
(VUYB)
2(YBC) (VUYB)
22Shearing
BRSC is a square. Therefore, BC2
(BRSC)
V
1. 2(ABR) (BRSC)
2. 2(YBC) (VUYB)
23(No Transcript)
24 BR BC
BA BY
ABP and AXC are congruent triangles. (SAS)
?ABR 90O ?CBA
?YBC 90O ?CBA
(ABR) (YBC)
?ABR ?YBX
25Shearing
BRSC is a square. Therefore, BC2
(BRSC)
1. 2(ABR) (BRSC)
2. 2(YBC) (VUYB)
V
3. (ABR) (YBC)
2(ABR) 2(YBC) (BRSC)
(VUYB)
26Shearing
(BRSC) (VUYB)
V
27Shearing
(ACQP) (AXUV)
(BRSC) (VUYB)
(AXUV) (VUYB) (AXYB)
(ACQP) (BRSC) (AXYB)
AC2 BC2 AB2
28Similarity
29Similarity
3
2
4
1
2
5
1
2
6
1
?1 ?3 90
Therefore, ?3 ?2
?4 ?2 90
Therefore, ?5 ?2
Therefore, ?4 ?1
Therefore, ?6 ?1
30All the 3 triangles have the same corresponding
angles. Therefore, they are SIMILAR.
31Similarity
By Similarity, b/c x/b
x b2/c
32Similarity
By Similarity, a/c y/a
y a2/c
33Similarity
c x y
b2/c a2/c
c2 b2 a2
34Dissection
35Dissection
Let us begin with a right-angled triangle with
sides a, b, and c (where c is the hypotenuse).
?C90?
Angle sum of a triangle 180
?A?B90
36Dissection
Let us rotate the triangle 90, 180 and 270 to
get four congruent triangles.
37Let us now rearrange the 4 triangles...
Area of big square c2
38The triangles fit neatly to form the square
because
Each angle of the square is formed of ?A?B
a
b
From the 1st slide ?A?B 90
39Finally, a little calculation
c2 area of triangles area of small square
c2 4(½ab) (a-b)2
c² 2ab (a² 2ab b²)
c² a² b²