GPREP - Perimeter

boy141 wrote:Ossa, I cannot understand where you got
"16 √2(2+√2)" especially the "(2+√2)" part because you said multiply both sides by "√2" how did you gain a 2 in "(2+√2)"?

This is where I am in this math problem:
I am attempting to find "a"

a + a + a√2 =16 +16√2
2a + a√2 = 16 + 16√2
a(2 + √2) = 16(1 + √2)

now multiply both sides by √2

a√2(2 + √2) = 16 √2(1+√2)

now distribute the √2 into the parentheses ON THE RIGHT SIDE ONLY:
a√2(2 + √2) = 16(√2 + 2) -- note the second term because √2√2 = 2
the terms in parentheses are now identical, so we have a√2 = 16.
since a√2 is the hypotenuse, there's no need to isolate 'a'; you have your answer right there. the hypotenuse is 16.

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if you didn't figure that out, you can also isolate 'a' by division, as in any other algebra problem in which you need to remove a coefficient from something. the only thing that's different is that the coefficient here is a binomial (2 + √2), so you have to know how to divide by those.

a(2 + √2) = 16 + 16√2
therefore
a = (16 + 16√2) / (2 + √2)
multiply top and bottom by the CONJUGATE of 2 + √2, which is 2 - √2:
a = (16 + 16√2)(2 - √2) / (2 + √2)(2 - √2)
by the difference of squares formula, the bottom is 4 - 2 = 2.
the top multiplies out to 32 - 16√2 + 32√2 - 32, or 16√2.
therefore, a = 16√2 / 2 = 8√2.
the hypotenuse is therefore 8√2 times √2, or 16.

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the easiest way to solve this problem is to memorize an additional 45-45-90 template, though, as remarked in this post.