What a clever little piece of mathematics this is. It is a much
neater proof of Pythagoras Theorem than the one I was shown at
school. There were also some very well laid out solutions with
clear explanations, so well done.
Complete solutions were received from:
Aftab Hussain - whose solution is given below,
Michael Brooker (home educated),
Andrei Lazanu (School number 205, Bucharest),
Charles Blackham (Shrewsbury House School).
Here is Aftab's solution:
Area of the square = (a+b) 2 (square of sides a+b)
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Area of Square = a2 + 2ab + b2 |
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Area of the Trapezium = Area of square divided by 2 (rotational
symmetry)
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Area of Trapezium = |
a2 + 2ab + b2
2
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|
|
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Area of Trapezium = |
a2
2
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+ ab + |
b2
2
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Area of Trapezium as a sum of areas of triangles
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Area of Trapezium = |
ab
2
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+ |
ab
2
|
+ |
c2
2
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|
|
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Area of Trapezium = |
ab + ab + c2
2
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|
|
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Area of Trapezium = |
2ab + c2
2
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|
|
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Area of Trapezium = ab + |
c2
2
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Proof
Area of Trapezium derived from square = Area of Trapezium as a
sum of areas of three triangles.
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|
a2
2
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+ ab + |
b2
2
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= b + |
c2
2
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(subtracting ab both sides) |
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|
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a2
2
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+ ab + |
b2
2
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- ab = |
c2
2
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(multiplying 2 both sides) |
|
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a2 + b2 = c2 (Phythagoras Theorem) |
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