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