I don't disagree with any of the posts that got 16 as the answer, but some of them were not the easiest to follow. Here is how I did it.

formula for a right isosceles triangle is

1:1:\sqrt{2} so the perimeter for the most basic (easiest to figure out) triangle of this type would be

1+1+\sqrt{2}. If we let d = one side (NOT THE HYPOTNUSE!!) we get the formula d + d + d

\sqrt{2} With the other equation, we don't put the 1 in front of the square root of 2, but it's "implied" so the d is required in the second forumla/equation.

We know the perimeter is 16 + 16

\sqrt{2}, so

d + d +d

\sqrt{2} = 16 + 16

\sqrt{2}2d +d

\sqrt{2} = 16 + 16

\sqrt{2}FACTOR OUT d

d(2+

\sqrt{2}) = 16 + 16

\sqrt{2}divide both sides by (2 +

\sqrt{2}) and we get

d =

\frac{16 + 16\sqrt{2}}{2 + \sqrt{2}}Now, if we have a radical in the bottom, how do we get rid of it? We cant just muliply by

\sqrt{2} because when we do 2 *

\sqrt{2} we would still be creating a radical in the denominator.

Remember this rule: (x + y)(x - y) =

x^2 - y^2? This is how we get rid of complex radicals (not sure if that is the proper word for it, but i think they're complex because they're a pain in the a$$ to get rid of.

so

(2 + \sqrt{2})(2 - \sqrt{2}) = 2^2 - \sqrt{2}^2 which = 4 - 2 = 2

Now multiple the numerator by the (2 -

\sqrt{2} also, because it has to be even or it changes the problem/answer.

16 + 16\sqrt{2}) ( 2 - \sqrt{2})32 - 16\sqrt{2} + 32\sqrt{2} - 16\sqrt{2}^2becomes

32 - 16\sqrt{2} + 32\sqrt{2} - 32becomes

-16\sqrt{2} + 32\sqrt{2}becomes

16\sqrt{2}Don't forget this is all over 2 in the denominator.

16 / 2 = 8 so we now have d =

8\sqrt{2}Remember that we have the ratio

1:1:\sqrt{2} and d = the side, not the hypotnuse, so we must take the value of d and multiply it by

\sqrt{2} to get the final value for the hypotnuse.

\sqrt{2} *

\sqrt{2} * 8 = 2 * 8 = 16

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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