Pythagoras Problem Solving

Pythagoras Problem Solving-15
The theorem can be generalized in various ways, including higher-dimensional spaces, to spaces that are not Euclidean, to objects that are not right triangles, and indeed, to objects that are not triangles at all, but n-dimensional solids.

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You may want to watch the animation a few times to understand what is happening.

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The large square is divided into a left and right rectangle.

A triangle is constructed that has half the area of the left rectangle.

This argument is followed by a similar version for the right rectangle and the remaining square.

Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. Let A, B, C be the vertices of a right triangle, with a right angle at A.

Let ABC represent a right triangle, with the right angle located at C, as shown on the figure.

Draw the altitude from point C, and call H its intersection with the side AB.

Then another triangle is constructed that has half the area of the square on the left-most side.

These two triangles are shown to be congruent, proving this square has the same area as the left rectangle.


Comments Pythagoras Problem Solving

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