Isosceles Right Triangle

[KTsincere]

Hi everyone,
I just took the GMAT prep test #1 from MBA.com and I was going over the questions I got wrong and I was wondering if anyone can point out what I'm not seeing about this Isosceles Right Triangle question.

The Perimeter of a certain isosceles right triangle is 16+16√2. What is the length of the hypotenuse of the triangle?
A) 8
B) 16
C) 4√2
D) 8√2
E) 16√2

I know the ratio is 1:1:√2 and I know the perimeter of a polygon is all the sides added together but I still can't figure out why my answer doesn't gel with theirs!
Any feedback appreciated...

[tmmyc]

Leg of isosceles right triangle: x
Hypotenuse of isosceles right triangle: x*sqrt(2)

Perimeter: x+x+x*sqrt(2) = 16+16*sqrt(2)

Solve for x:

From here, I recommend plugging in numbers for x using the answer choices as guides. You will find that 8*sqrt(2) works.

But if you want to see the math, here it is.

x+x+x*sqrt(2) = 16+16*sqrt(2)
->2x+x*sqrt(2) = 16+16*sqrt(2)
-->x*[2+sqrt(2)] = 16*[1+sqrt(2)]
--->x = 16*[1+sqrt(2)] / [2+sqrt(2)]
---->x = {16*[1+sqrt(2)]*[2-sqrt(2)]} / {[2+sqrt(2)]*[2-sqrt(2)]}
----->x = [16*(2 + sqrt(2) - 2)] / (4 - 2)
------>x = 16*sqrt(2) / 2
------->x = 8*sqrt(2)

Finally, solve for the hypotenuse.
Hypotenuse of isosceles right triangle: x*sqrt(2) -> 8*sqrt(2)*sqrt(2) -> 16

[RonPurewal]

tricky tricky.

you fell for what is commonly known as the 'sucker answer' in this problem: you just saw a number with a root(2) in front of it, and assumed that that number must be the hypotenuse.

here's the problem (which is 'why your answer doesn't gel with theirs'):
if the hypotenuse of the triangle is 16root(2), then each of the legs is 16. therefore, the perimeter of the triangle would be 32 + 16root(2), not 16 + 16root(2) as required.

once you make that realization, there are two routes:

1) (fast way) realize intuitively that the other number must be the hypotenuse
if you have good number sense about these kinds of things, this is the fastest solution.
when you have one of these triangles, you know that the expressions for the legs and the hypotenuse will not be like terms.
because you've just ruled out the possibility that the 16root(2) is the hypotenuse, you're left with the conclusion that the 16 is the hypotenuse.

2) (slow way) perform calculations, as in the above post.

--

p.s.
how'd you get those square root symbols in there?

[KTsincere]

Thanks for the assistance!
The square root sign came from a handout I was given from an old class, copy & paste and voilà... √

[RonPurewal]

Thanks for the assistance!
The square root sign came from a handout I was given from an old class, copy & paste and voilà... √

that's very cool.

i'm old enough that handouts from my old classes are on this stuff called 'paper'. crazy how things change... :D

[RonPurewal]

the easiest way to solve this problem, however, is to plug in the answer choices.

in order, the choices are
4√2
8
8√2
16
16√2

start with the middle choice.
if the hypotenuse is 8√2, then the legs are each 8, and so the perimeter is 8 + 8 + 8√2, or 16 + 8√2.

this is too small, so the first three answers are gone.

if the hypotenuse is 16, then the legs are each 16/√2, which reduces to 8√2. therefore, the perimeter is 8√2 + 8√2 + 16, or 16 + 16√2.
so the correct answer is (d).

alternatively, once you've killed (a)(b)(c), you can look at (e) and realize that it's exactly twice as long as (c). that will create a triangle with exactly twice the perimeter of (c), or 32 + 16√2, which is too big. that leaves only (d).