Proof 1 (To understand)
See the picture below:
A triangle of side a (base), b (altitude) and c hypotenuse is shown with squares made outer of each sides. This say that square next to the side C is equal to squares next to a and b. This can be only Proven if calculated separately.
Proof 2 (Algebraically)
See the image below:
It's a blue square and a yellow square in it. Upper side of blue square is named a and b partially. So its length is a+b as well as all sides. And yellow square's length is c.
Now we have to find out total area of blue square. By applying length X breadth theory, it is (a+b)(a+b) or (a+b)^2
(a+b)^2=a^2+2ab+b^2
Now calculate the area of same blue square in other way, i.e it's covered with four triangles abc and a yellow square c.
Calculating so its area would equal to sum of these five shapes.
So,
(area of triangle abc X 4 ) + (area of yellow square)=area of blue square
area of triangle is 1/2 (ab)
such four triangles' area is 1/2 (ab) four times. i.e, half of ab four times becomes 2ab
(half of ab+half of ab+half of ab+half of ab)=2ab
Now we have got the blue square's area in two ways and so both of it would be equal.
i.e, a^2+2ab+b^2=2ab+c^2
removing the 2ab from two sides it reaches to:
a^2+b^2=c^2
Proof 3 (Geometrically)
See the image below:
Two colorful squares are seen. Both include 4 triangles and on white square. The white area in first square is C^2. In the next image, triangles are rearranged after cutting it separately. The white area should be C^2, but it's equal to B^2+A^2. So these two should be equal i.e, A^2+b^2=c^2
For more details:
See the picture below:
A triangle of side a (base), b (altitude) and c hypotenuse is shown with squares made outer of each sides. This say that square next to the side C is equal to squares next to a and b. This can be only Proven if calculated separately.
Proof 2 (Algebraically)
See the image below:
It's a blue square and a yellow square in it. Upper side of blue square is named a and b partially. So its length is a+b as well as all sides. And yellow square's length is c.
Now we have to find out total area of blue square. By applying length X breadth theory, it is (a+b)(a+b) or (a+b)^2
(a+b)^2=a^2+2ab+b^2
Now calculate the area of same blue square in other way, i.e it's covered with four triangles abc and a yellow square c.
Calculating so its area would equal to sum of these five shapes.
So,
(area of triangle abc X 4 ) + (area of yellow square)=area of blue square
area of triangle is 1/2 (ab)
such four triangles' area is 1/2 (ab) four times. i.e, half of ab four times becomes 2ab
(half of ab+half of ab+half of ab+half of ab)=2ab
Now we have got the blue square's area in two ways and so both of it would be equal.
i.e, a^2+2ab+b^2=2ab+c^2
removing the 2ab from two sides it reaches to:
a^2+b^2=c^2
Proof 3 (Geometrically)
See the image below:
Two colorful squares are seen. Both include 4 triangles and on white square. The white area in first square is C^2. In the next image, triangles are rearranged after cutting it separately. The white area should be C^2, but it's equal to B^2+A^2. So these two should be equal i.e, A^2+b^2=c^2
For more details:
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