Let ABC be a righ-angled triangle,angle A of which measures one right angle.
Let BC=a,CA=b andAB=c
Construction:- A square PQRS ,whose side
is b+c,is drawn.PointsE,F,G,H are taken on sides PQ,QR,RS and SP respectively such that PE=QF=SH=RG=b.JoinE &F,F & G,G & H and H & E.
Proof:-We now get four right -angled triangles having "a" as hypotenuse and b and c as other two sides.The remaining part of the figure is a^2
Therefore,
(b+c)^2
=a^2+4*bc/2[area of the square being a^2 and area of each of the four triangles being (1/2)*bc]
=>b^2+c^2+2bc=a^2+2bc
=>b^2+c^2=a^2 {Proved)
2007-05-25 22:42:36
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answer #1
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answered by alpha 7
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1. First draw a big square.
2. Then draw another (smaller) square inscribed in the big one (inscribed means having the each of the vertices on one side of the big square).
3.After you draw all this, you'll see that the big square was divided in 4 triangles having a 90 degrees angle and in a square (the small one you've drawn).
4. Now denote the sides of any of the 4 triangles formed by a, b and c, with a and b being the sides found on the sides of the big square (the ones that form a 90 degrees angle).
5. The area of each triangle (all 4 have the same area because they are congruent) is ab/2, so the area of all 4 triangles is 4*ab/2 = 2ab.
6. The area of the central square (the small one drawn by you) is c^2.
7. You can observe that the side of the big square equals a+b (just take a look at the sides of the 4 triangles formed). Therefore, the area of the big square will be (a+b)^2.
8. But the area of the big square equals the are of the small square, plus the area of the 4 right-angled triangles, so you can write that:
(a+b)^2 = 2ab+c^2
==> (a+b)(a+b) = 2ab + c^2
==> a^2+2ab+b^2 = 2ab + c^2
==> a^2 + b^2 = c^2,
which is exactly the Pythagorean theorem. :)
Sorry if my explanations are sometimes lengthy and unnecessary, but I don't know what's your math level. :)
2007-05-25 22:46:37
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answer #2
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answered by Anonymous
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1)
A Proof of the Pythagorean Theorem From Heron's Formula
Let the sides of a triangle have lengths a,b and c. Introduce the semiperimeter p = (a + b + c)/2 and the area S. Then Heron's formula asserts that
S2 = p(p - a)(p - b)(p - c)
W.Dunham analyzes the original Heron's proof in his Journey through Genius.
For the right triangle with hypotenuse c, we have S = ab/2. We'll modify the right hand side of the formula by noting that
p - a = (- a + b + c)/2, p - b = (a - b + c)/2, p - c = (a + b - c)/2
It takes a little algebra to show that
(a + b + c)(- a + b + c)(a - b + c)(a + b - c)
= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)
For the right triangle, this expression is equal to 16S2 = 4a2b2. So we have
4a2b2= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)
Taking all terms to the left side and grouping them yields
(a4 + 2a2b2 + b4) - 2a2c2 - 2b2c2 + c4 = 0
With a little more effort
(a2 + b2)2 - 2c2(a2 + b2) + c4 = 0
And finally [(a2 + b2) - c2]2 = 0
2) check out this website:
http://jwilson.coe.uga.edu/EMT668/emt668.student.folders/HeadAngela/essay1/Pythagorean.html
2007-05-26 02:58:16
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answer #3
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answered by bob 2
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Consider a triangle with sides a b and c, with c as the hypotenuse and angle opposite c a right angle,
now make 4 copies of it, the original, rotated 90 degrees, rotated 180 degrees and rotated 270 degress.
Now put these 4 together to form a square of side c, there'll be a square hole in the middle with side (a-b)
Now, the area of the large square, A =
area of small square + area of 4 identical triangles
this means,
c^2 = (a-b)^2 + 4(a.b / 2) = a^2 - 2ab + b^2 +2ab
= a^2 + b^2
which was to be proved!!! Amazing eh? Also I remember there being a page on cut-the-knot.org which had almost all the proofs of the Pythagorean theorem you'd find on the net!!
Hope this helps!!
2007-05-26 00:25:03
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answer #4
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answered by yasiru89 6
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3^2+4^2=5^2
9+16=25
2007-05-26 01:03:26
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answer #5
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answered by atheist kid 3
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Why does everyone need to struggle? No proof can be understood properly with just words and no diagrams. Go to the website below and ckeck out the proofs. If you have any difficulty understanding, email me.
http://en.wikipedia.org/wiki/Pythagoras_theorem#Proofs
This site hes only a few proofs, but if you want nearly all of them, including a cool java applet proof, this is your place:
http://www.cut-the-knot.org/pythagoras/index.shtml
2007-05-26 00:36:31
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answer #6
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answered by Akilesh - Internet Undertaker 7
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Everest isn't the talles mountain... Mauna Kea is.
2016-04-01 09:02:51
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answer #7
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answered by Anonymous
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The formula is a^2 + b^2 = c^2
*The "a" variable represents the shortest side of a triangle.
*The "c" variable represents the longest side of a triangle.
2007-05-26 05:26:12
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answer #8
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answered by ♪♥Annie♥♪ 6
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a and b are the lengths of the sides adjacent to the right angle. c is the length of the hypotenuse. x is the angle formed by the a length side and the hypotenuse. The other angle would be pi/2 - x.
a/c = cos(x)
b/c = sin(x)
(a/c)^2 = cos^2(x)
(b/c)^2 = sin^2(x)
1 = sin^2(x) + cos^2(x) = (a^2 + b^2)/c^2
c^2 = a^2 + b^2
2007-05-25 23:01:33
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answer #9
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answered by jcsuperstar714 4
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all i know is a squared plus b squared equals c squared
hope i helped
2007-05-26 01:46:26
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answer #10
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answered by soccerful 3
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