what is pithogerous theoram and in what kind of triangle it is used
omshukla1:
square of hypotenious=square of base + square of perpendicular
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square of hypotenuse is equal to square of base plus square of perpendicular
it is used in right angled triangle
it is used in right angled triangle
u should give full equation
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The theorem states that:
"The square on the hypotenuse of a right triangle is equal to the sum of the squares on the two legs" (Eves 80-81).
This theorem is talking about the area of the squares that are built on each side of the right triangle.
Accordingly, we obtain the following areas for the squares, where the green and blue squares are on the legs of the right triangle and the red square is on the hypotenuse.
area of the green square is
area of the blue square is
area of the red square is
From our theorem, we have the following relationship:
area of green square + area of blue square = area of red square or
As I stated earlier, this theorem was named after Pythagoras because he was the first to prove it. He probably used a dissection type of proof similar to the following in proving this theorem.
Pythagoras' Proof
"Let a, b, c denote the legs and the hypotenuse of the given right triangle, and consider the two squares in the accompanying figure, each having a+b as its side. The first square is dissected into six pieces-namely, the two squares on the legs and four right triangles congruent to the given triangle. The second square is dissected into five pieces-namely, the square on the hypotenuse and four right triangles congruent to the given triangle. By subtracting equals from equals, it now follows that the square on the hypotenuse is equal to the sum of the squares on the legs" (Eves 81).
Consider the following figure.
The area of the first square is given by (a+b)^2 or 4(1/2ab)+ a^2 + b^2.
The area of the second square is given by (a+b)^2 or 4(1/2ab) + c^2.
Since the squares have equal areas we can set them equal to another and subtract equals. The case (a+b)^2=(a+b)^2 is not interesting. Let's do the other case.
4(1/2ab) + a^2 + b^2 = 4(1/2ab)+ c^2
Subtracting equals from both sides we have
concluding Pythagoras' proof.
"The square on the hypotenuse of a right triangle is equal to the sum of the squares on the two legs" (Eves 80-81).
This theorem is talking about the area of the squares that are built on each side of the right triangle.
Accordingly, we obtain the following areas for the squares, where the green and blue squares are on the legs of the right triangle and the red square is on the hypotenuse.
area of the green square is
area of the blue square is
area of the red square is
From our theorem, we have the following relationship:
area of green square + area of blue square = area of red square or
As I stated earlier, this theorem was named after Pythagoras because he was the first to prove it. He probably used a dissection type of proof similar to the following in proving this theorem.
Pythagoras' Proof
"Let a, b, c denote the legs and the hypotenuse of the given right triangle, and consider the two squares in the accompanying figure, each having a+b as its side. The first square is dissected into six pieces-namely, the two squares on the legs and four right triangles congruent to the given triangle. The second square is dissected into five pieces-namely, the square on the hypotenuse and four right triangles congruent to the given triangle. By subtracting equals from equals, it now follows that the square on the hypotenuse is equal to the sum of the squares on the legs" (Eves 81).
Consider the following figure.
The area of the first square is given by (a+b)^2 or 4(1/2ab)+ a^2 + b^2.
The area of the second square is given by (a+b)^2 or 4(1/2ab) + c^2.
Since the squares have equal areas we can set them equal to another and subtract equals. The case (a+b)^2=(a+b)^2 is not interesting. Let's do the other case.
4(1/2ab) + a^2 + b^2 = 4(1/2ab)+ c^2
Subtracting equals from both sides we have
concluding Pythagoras' proof.
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