The perimeter of a certain isosceles right triangle is 16 +1

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by Natalllllia » Tue Dec 02, 2008 7:12 pm
thank you , dmateer25

now i got it

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by lind0923 » Tue Mar 17, 2009 3:59 pm
Hi Ron and dmateer,
I know this is an old post...... but just referencing it and I don't understand how to simplify 16/root 2 to 8*root 2. Maybe I'm just being dense, but I would appreciate any help you can give me.
thanks!!

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by Stuart@KaplanGMAT » Tue Mar 17, 2009 4:08 pm
lind0923 wrote:Hi Ron and dmateer,
I know this is an old post...... but just referencing it and I don't understand how to simplify 16/root 2 to 8*root 2. Maybe I'm just being dense, but I would appreciate any help you can give me.
thanks!!
Hi!

It's considered mathematically "unclean" to leave a root sign in the denominator of a fraction. On the GMAT, you'll almost never see an answer in that form. To clean up the fraction, we do what's known as "rationalizing the denominator".

To do so, we multiply the top and bottom of the fraction by the root we want to eliminate. In this case we multiply by root2/root2. So:

16/root2 * root2/root2

=16*root2/root2*root2

=16*root2/2

=8*root2
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by minhphuoc » Tue Jan 12, 2010 7:42 pm
Dear all,

I think up a solution:

As some of you said, 2a + a sqrt(2) = 16 + 16 sqrt(2)
--> a sqrt(2) x [1 + sqrt(2)] = 16 x [1 + sqrt(2)]
--> a sqrt(2) = 16, and we already know a sqrt (2) is hypotenuse.

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by lunarpower » Thu Jan 21, 2010 6:30 pm
minhphuoc wrote:Dear all,

I think up a solution:

As some of you said, 2a + a sqrt(2) = 16 + 16 sqrt(2)
--> a sqrt(2) x [1 + sqrt(2)] = 16 x [1 + sqrt(2)]
--> a sqrt(2) = 16, and we already know a sqrt (2) is hypotenuse.
this is very nicely done, although the factoring on the left is certainly not obvious.
Ron has been teaching various standardized tests for 20 years.

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by san2009 » Thu Jun 10, 2010 8:39 am
lunarpower wrote:
airan wrote:The perimeter of a certain isosceles right triangle is 16 +16 sqrt(2). What is the length of the hypotenuse of the triangle

1. 8
2. 16
3. 4 sqrt(2)
4. 8 sqrt(2)
5. 16 sqrt(2)
two ways for you to solve this problem much more easily than with the rather insane algebra:

(1) plug in the answer choices
according to the 45-45-90 template, the hypotenuse of an isosceles right triangle is √2 times each of the legs. therefore, each leg is the hypotenuse divided by √2.
if you plug in choice (b), hypotenuse = 16, you'll find that leg = 16/√2 = 8√2 (post back if you don't know how to simplify this). therefore, perimeter = 2(leg) + hypotenuse = 16√2 + 16.
this is WAY easier than going through with all the algebra.
also, you shouldn't have to do that much plugging. if you plug in (a) first, then you'll get exactly half the desired perimeter, proving that (b) is correct. and if you plug in (c), (d), or (e), then you'll get, respectively, 8 + 4√2, 16 + 8√2, or 32 + 16√2, any of which will allow you to deduce that all three are incorrect.

(2) memorize TWO templates for the 45-45-90 triangle
everyone knows the 1-1-√2 template, which has perimeter 2 + 1√2.
you can also memorize an additional template: √2-√2-2.
this template is just √2 times the original template, but the multiples aren't obvious. this template has perimeter 2 + 2√2.
if you memorize this template, then you'll immediately recognize 16 + 16√2 as exactly eight times its perimeter, so that your triangle is 8√2-8√2-16.
Ron, thanks for the above. Have one quick question. when i first saw the 16 + 16sqrt2, I equated it to X + X + Xsqrt2 -- the standard sides of a 45-45-90 triangle. then, i figured the 2x must equal 16 and Xsqrt2 must equal 16sqrt2. But I realized that by equating the two i was getting conflicting answers for the sides -- at which point I got apprehensive and guessed. Now, since this a 45-45-90 why doesn't it abide by the x:x:Xsqrt2 rule? what am I missing here??

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by lunarpower » Fri Jun 11, 2010 2:41 am
san2009 wrote:Now, since this a 45-45-90 why doesn't it abide by the x:x:Xsqrt2 rule? what am I missing here??
heh.
it does, actually; it's just that "x" is not an integer in this case.

in this case, x is 8√2.

therefore, the two x's are 8√2 + 8√2 = 16√2, so there's where that comes from.

the x√2 is (8√2)√2, which works out to 16.

so there you have it -- the 16√2 is actually the 2x part, and the 16 is the x√2 part.

by the way, i suspect that the entire motivation behind this question was precisely what it successfully did here: to thwart students who will just look at it and expect it to fit an easy memorized template.
Ron has been teaching various standardized tests for 20 years.

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by san2009 » Fri Jun 11, 2010 3:51 am
gotcha. so basically the main takeaway is: one has to be super careful about equating different parts of the LHS and RHS of an equation. I'd rather just do the algebra or use estimation in the future.

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by lunarpower » Fri Jun 11, 2010 5:17 am
san2009 wrote:gotcha. so basically the main takeaway is: one has to be super careful about equating different parts of the LHS and RHS of an equation. I'd rather just do the algebra or use estimation in the future.
that's a good takeaway.

in fact, "one has to be super careful about ________ on the gmat" is pretty much going to be true no matter what is placed in the "________".
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by san2009 » Fri Jun 11, 2010 7:40 am
hah!
that's the part the wears me out.
thanks ron!

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by lunarpower » Mon Jun 14, 2010 3:52 am
san2009 wrote:hah!
that's the part the wears me out.
thanks ron!
maybe so, but you want to keep in mind that this is actually the entire point of the whole quantitative section!

in particular, the entire test is designed to give better scores to students who think conceptually about the material, understanding its limitations, exceptions, and applications.
the quant is specifically designed to be hostile to people who try to memorize their way through the test, and/or who try to solve every problem as though it conforms to a routine template.
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by san2009 » Mon Jun 14, 2010 4:29 am
Point taken. Thank for your help/guidance Ron!

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by Anaira Mitch » Fri Dec 30, 2016 2:56 am
Let the hypotenuse be x, then the length of the leg is x/root2.
x+2x/root2=16+16*root2
x+root*x=16+16*root2
So, x=16

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by Brent@GMATPrepNow » Fri Dec 30, 2016 7:58 am
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
An IMPORTANT point to remember is that, in any isosceles right triangle, the sides have length x, x, and x√2 for some positive value of x.

Note: x√2 is the length of the hypotenuse, so our goal is to find the value of x√2

From here, we can see that the perimeter will be x + x + x√2

In the question, the perimeter is 16 + 16√2, so we can create the following equation:
x + x + x√2 = 16 + 16√2,
Simplify: 2x + x√2 = 16 + 16√2
IMPORTANT: Factor x√2 from the left side to get : x√2(√2 + 1) = 16 + 16√2
Now factor 16 from the right side to get: x√2(√2 + 1) = 16(1 + √2)
Divide both sides by (1 + √2) to get: x√2 = 16

Answer = B

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by [email protected] » Fri Dec 30, 2016 10:39 am
Hi All,

There's an additional discussion of this prompt here:

https://www.beatthegmat.com/hypotenuse-o ... 76892.html

There's a Number Property about Isosceles triangles that will help you to avoid most of the wrong answers in this question...

The ratio of sides in an Isosceles Triangle is: x, x, and x√2

If, for example, one of the "short sides" was 5, then we'd have 5, 5, 5√2 and the perimeter would be 10 + 5√2. Notice that the 10 and 5 are different numbers....That will ALWAYS happen IF the short side is an integer and the hypotenuse has a √2 attached to it....

We're given a perimeter of 16 + 16√2; here, both numbers are 16, so we are NOT dealing with the above scenario. The "short sides" CANNOT be integers and the hypotenuse CANNOT have a √2 attached to it. Eliminate C, D and E.

From there, we can TEST A and B to find the match.

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