If the perimeter of a isosceles right triangle is 16 + 16*sqrt(2), what is the length of its hypotenuse?

a) 8
b) 16
c) 4*sqrt(2)
d) 8*sqrt(2)
e) 16*sqrt(2)

Is this question == > 16 + 16*sqrt(2)
_________________

GMAT Club Error Log V3.0 - with Timer
GMAT Club Error Log
Spoilt Error Log V0.0.9 - with Timer
Awesome Error Log - By Spoilt and Trianglock

LET a be the side of the the triangle

16 + 16 \sqrt{2}= 2a + a \sqrt{2}
16 + 16 \sqrt{2} = a(2 + \sqrt{2})
so
a = (16 + 16 \sqrt{2}) / (2 + \sqrt{2})

After rationalizing by (2-\sqrt{2})

we get a = 8\sqrt{2}

FROM THIS WE CAN FIND HYPOTENUSE = 8\sqrt{2} * \sqrt{2} = 16

HTH

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
_________________

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

Find out what's new at GMAT Club - latest features and updates

Spoilt,

There is not necessarily anything wrong with it, but the start of your method actually skips some steps that most people won't realize are necessary.

You see that if you have 8\sqrt{2} + 8\sqrt{2} as the base and height of the triangle, then that would give you 16 for your hypotnuse and a perimeter of 16 + 16\sqrt{2}. Most people are not going to realize that. We are going to see 16\sqrt{2} and think...ok, the ratio is 1:1:\sqrt{2}so the one with the \sqrt{2} must be part of the hypotnuse. You were able to look at the problem and in seconds see the correct answer. That's impressive, but not necessarily a "method" for the rest of us to use.
_________________

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

Find out what's new at GMAT Club - latest features and updates