How do you show that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides ?

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How do you show that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides ?

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Answer 1 · 2017-01-28T01:44:41

See explanation...

Explanation:

Consider this diagram:

enter image source here

Each of the triangles is a right angled triangle with sides $a, b, c$ $a, b, c$ and area $\frac{a b}{2}$ $(ab)/2$ (since it is half of an $a \times b$ $axxb$ rectangle)

The area of the large square is:

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2 = a^2+2ab+b^2$

The area of the smaller square plus the area of the triangles is:

$c^{2} + 4 (\frac{a b}{2}) = c^{2} + 2 a b$ $c^2+4((ab)/2) = c^2+2ab$

These two expressions must be equal. So we have:

$a^{2} + 2 a b + b^{2} = c^{2} + 2 a b$ $a^2+2ab+b^2 = c^2+2ab$

Subtracting $2 a b$ $2ab$ from both sides, we find:

$a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$

This is Pythagoras' Theorem:

In a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

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How do you show that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides ?

1 Answer

Explanation:

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