GMAT PREP ?

[thesithlords]

The perimeter of a certain isosceles right triangle is 16 + 16(Sq. rt. 2). What is the length of the hypotenuse?

8
16
4(sq rt 2)
8(sq rt 2)
16(sq rt 2)

My best guess is answer E. Please Explain

[xilef]

we know that perimeter of isosceles is 2x+x(sqrt2)

so:

2x+x(sqrt2)=16+16(sqrt2)
6x^2-768=0
x^2-128=0
x=8(sqrt2) so hypotenuse = 8(sqrt2)(sqrt2) = 16

Answer B

[lunarpower]

this is one of those cases in which the gmat is just trying to be a bastard and trick you into picking the wrong answer based on common prejudices. in particular, you have the following common prejudice: you think that only the hypotenuse of this triangle can contain the quantity √2 in it, and so you just automatically assume that 16√2 represents the hypotenuse.

here's the problem with that: if the hypotenuse is 16√2, then each of the sides has to be 16, which means that the perimeter has to be 32 + 16√2, contradicting what you're told.

but ... if the legs of the triangle have √2 in them, then the hypotenuse won't (because the √2 will turn back into a whole number when it's multiplied by another √2).

--

just try plugging in numbers.

if you try your answer (e), then, as mentioned above, the hypotenuse is 16√2 and, therefore, each leg must be 16. this gives the wrong perimeter.

if you try hypotenuse = 16, then each leg is 16/√2 = 8√2 (post if you don't know how to get that). that gives the right perimeter, so it's the correct answer.

--

This is one of those problems where ESTIMATION can really save the day: you should memorize the fact that root(2) is approximately equal to 1.4.

Then your equation - in its original form - becomes:
16 + about 22.4 = 2a + about 1.4a
--> about 38.4 = about 3.4a
--> a equals a little less than 12
--> hypotenuse = a times 1.4 = a little less than 16.8.

Looks like 16 is the best choice.

--

if you're looking for a really mechanical way to solve, then you can always do this:

16 + 16√2 = 2a + a√2
16 + 16√2 = a(2 + √2)
so
a = (16 + 16√2) / (2 + √2)
you can rationalize the denominator by multiplying by its conjugate, (2 - √2), making the denominator into a difference of squares: (2 + √2)(2 - √2) = 4 - 2 = 2.
therefore
a = (16 + 16√2)(2 - √2) / 2
multiply out --> = (32 - 16√2 + 32√2 - 32) / 2
= 16√2 / 2
= 8√2

so hypotenuse = 8√2 x √2 = 8 x 2 = 16