How would I find the area of a 45-45-90 triangle with one side of length 73?

1 Answer
Nov 25, 2015

AB = 73 sqrt2, BC = AC = 73

Explanation:

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Now i hope you are aware of the Pythagoras Theorem which states;

AB^2 = BC^2 + AC^2

Additionally
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This is the definition of a Pythagorean triplet where m>n

Now notice BC = AC in the above image

For this scenario we have special formula;

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H = Lsqrt2

So this is a nice template for this type of a triangle;

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For the side provided to be the hypotenuse it has to be a multiple of sqrt2 which clearly it isn't;

So we can make out first assumption

BC = AC = 73

Now its just substitution;

73^2 + 73^2 = x^2

sqrt(73^2 + 73^2) = x

x = sqrt(2*73^2)

=>AB = 73 sqrt2