- #1

julian

Gold Member

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Say the hypotenuse is of length ##c## and the other two sides are of length ##a## and ##b##. First of all, we obviously have:

##a^2 + b^2 > a^2 \quad## and ##\quad a^2 + b^2 > b^2##.

By Pythagoras' theorem ##c^2 = a^2 + b^2##. Using this in the above relations we obtain:

##c^2 > a^2 \quad## and ##\quad c^2 > b^2##.

Now ##c^2## is the area of a square with sides of length ##c##, and ##a^2## is the area of a square with sides of length ##a##. The relation ##c^2 > a^2## says the square of area ##c^2## is bigger than the square of area ##a^2##, therefore, we must have

##c > a.##

Similarly, ##c^2 > b^2## implies

##c > b##.

And we have proved that the hypotenuse is the longest side.

Q.E.D.

Is this correct? Thanks.

Also, I want to ask what other proofs are there that the hypotenuse is the longest side of a right angled triangle?