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

Nov 25, 2015

$A B = 73 \sqrt{2} , B C = A C = 73$

Explanation:

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

$A {B}^{2} = B {C}^{2} + A {C}^{2}$

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;

$H = L \sqrt{2}$

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

For the side provided to be the hypotenuse it has to be a multiple of $\sqrt{2}$ which clearly it isn't;

So we can make out first assumption

$B C = A C = 73$

Now its just substitution;

${73}^{2} + {73}^{2} = {x}^{2}$

$\sqrt{{73}^{2} + {73}^{2}} = x$

$x = \sqrt{2 \cdot {73}^{2}}$

$\implies A B = 73 \sqrt{2}$