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Pythagoras theorem says that for[br]any right angle triangle area,

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see the area of the square of[br]the hypotenuse equals A plus B.

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The some of the areas of the[br]other two squares.

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But how could we prove this?

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Let's start by making a second[br]copy of the diagram.

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On the left hand copy will[br]draw this square. It just

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surrounds the square on the[br]hypotenuse.

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And on the right-hand copy will[br]draw this square.

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Again, it just surrounds the[br]squares on the other two sides.

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Now these two knew squares we've[br]drawn are both the same size.

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In each case, the side of the[br]new square has length little A

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plus littleby. The sum of the[br]length of the two shorter sides

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of the triangle.

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But now.

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Look at the areas of[br]these two new squares.

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The one on the left is made up[br]of square, see.

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Plus 4 copies.

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Of the triangle.

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The one on the right is made up[br]of squares A&B.

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Plus 4 copies.

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Of the triangle.

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But they're both the same area.

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C plus four triangles.

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Equals area A plus area B[br]plus four triangles.

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So area C equals area A[br]plus area be.

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And that's Pythagoras theorem.