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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)

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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 22 Sep 2005, 20:02
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A
B
C
D
E

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The perimeter of a certain isosceles right triangle is \(16 + 16 \sqrt{2}\). What is the length of the hypotenuse of the triangle?


(A) \(8\)

(B) \(16\)

(C) \(4\sqrt{2}\)

(D) \(8\sqrt{2}\)

(E) \(16\sqrt{2}\)
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 23 Sep 2005, 01:44
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It's B

16+16(rt2)=2a+a(rt2) (where a is the lenght of each leg)

16+16(rt2)=a(2+rt2)

(16+16(rt2))/(2+rt2)=a

Hypoteneuse of isosceles rigth triangle is a(rt2)

H=a(rt2)=((16+16(rt2)(rt2))/(2+rt2)

H=(16(rt2)+16x2))/(2+(rt2))

H=16(rt2+2)/(rt2+2)=16
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 23 Sep 2005, 02:03
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Let the length of the opposite and adjacent = a
Length of hypotenuse = h = sqrt(a^2+a^2) = sqrt(2) * a

So a = h/sqrt(2) = (sqrt(2)/2) * h

Perimeter = 2a + h
= 2 * (sqrt(2)/2) * h + h
= h (sqrt (2) + 1 )

Also given
Perimeter = 16 + 16 sqrt (2) = 16 (sqrt (2) + 1 )

So h = 16
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 23 Sep 2005, 07:44
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I also get B

If the length of one of the two equal sides is x, then the other side is also x and the hypotenuse is xrt2.

we know that P= 16+16rt2

to get the perimeter you add the lengths of the three sides so either the equal sides are 16/2 = 8 and the other side is 8rt2 (which it isn't)

or the equal sides are 16rt2/2 = 8rt2 and the hypotenus is 16.

2(8rt2) = 16rt2 so it's pretty obvious that the hypotenuse is 16.
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 10 Apr 2006, 11:45
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y^2 = x^2+x^2 => x = y/sqrt(2)

2x+y=16+16sqrt(2)
ysrrt(2)+y=16+16sqrt(2)
y=16
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 11 Apr 2006, 01:10
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gmatacer wrote:
y^2 = x^2+x^2 => x = y/sqrt(2)

2x+y=16+16sqrt(2)
ysrrt(2)+y=16+16sqrt(2)
y=16


I looked at it in a more simple manner....

Simply kept in mind that two sides must be equal and that the sum of the two sides must be more than the length of the remaining side....
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 11 Apr 2006, 01:19
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Put each sides to be X
Then hyp = xsqrt(2)

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


Solving for x, we get x = 8Sqrt(2)

Hence, Hyp = 16
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 29 Nov 2006, 06:38
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This is actually more simple than you think....

Remember the rule that the sum of any two sides of the triangular must not be less than the length of the remaining side....

Thus isosceles triangular has two sides equal in length... so the sum of those two sides must be more than the remaining side...

You are given a perimeter, you can trace down the answer quickly.... and without calculations...
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 29 Nov 2006, 07:04
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My method for this is simple....

two equal sides
hypotenuse

Perimeter=16+16sqrt2

the sum of the two equal sides cannot be less than the hypotenuse.

Thus hypotenuse must be less than the sum of the other two equal sides:

Perimeter=16+16sqrt2.... the lower value in this equation is the hypotenuse, thus answer is 16

B
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 03 May 2007, 03:40
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We know that an isocele right triangle has the following sides dimension: x, x and xsqrt(2) and we are looking for xsqrt(2).

Lets calculate x:

x+x+xsqrt(2) = 16 + 16sqrt(2)
x (2+sqrt(2)) = 8 (2+2sqrt(2))
x = 8 (2+2sqrt(2)) / (2+sqrt(2))

Lets caculte xsqrt(2):

xsqrt(2) = 8 (2+2sqrt(2)) sqrt(2) / (2+sqrt(2))
xsqrt(2) = 8 (2sqrt(2)+4) / (2+sqrt(2)) = 16
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 03 May 2007, 09:56
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above720 wrote:
How would you solve?

The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?


x + x + x√2 = 16 + 16√2
2x + x√2 = 16 + 16√2
x√2 (1 + √2) = 16 (1 + √2)
x√2 = 16
x = 16/√2 = 8√2

h^2 = x^2 + x^2
h^2 = 2x^2
h^2 = 2 (8√2)^2
h^2 = 2x2x8x8
h = 2 x 8 = 16
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post Updated on: 07 Jul 2007, 22:02
plaguerabbit wrote:
The perimeter of a certain isosceles right triangle is 16 + 16sqrt(2). What is the length of the hypotenuse of the triangle?

(A) 8
(B) 16
(C) 4sqrt(2)
(D) 8sqrt(2)
(E) 16sqrt(2)

how do you solve this without backsolving?


this is a problem that repeats every so often in this forum:

since its a isosceles right triangle, then the ratio of its angels is 45:45:90 and the ratio of its sides is 1:1:sqrt(2).

let x be the longer side (i.e sqrt(2)) then the shorter side will be x/sqrt(2).

16+16*sqrt(2) = x/sqrt(2)+x/sqrt(2)+x

16+16*sqrt(2) = 2x/sqrt(2) + x ---> multply by sqrt(2)

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

x = 16

this is the hypotenuse !

16 is the answer (B)

:-D

Originally posted by KillerSquirrel on 07 Jul 2007, 17:56.
Last edited by KillerSquirrel on 07 Jul 2007, 22:02, edited 2 times in total.
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 08 Jul 2007, 10:05
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plaguerabbit wrote:
The perimeter of a certain isosceles right triangle is 16 + 16sqrt(2). What is the length of the hypotenuse of the triangle?

(A) 8
(B) 16
(C) 4sqrt(2)
(D) 8sqrt(2)
(E) 16sqrt(2)

how do you solve this without backsolving?


Guys, there is a simpler approach. On this board, with all the practice that everyone's doing, we are all so focused on the various nuances of the GMAT, so this should jump out at you.

We know that the triangle has to be x to x to xroot2. But when we try to make it work, it simply doesn't make sense. I mean, if the sides were an integer and the hypotenuse were the same integer times root 2, then the perimeter would have to just be 2x + xroot2. But it's not that. It's x + xroot2. So something's wrong.

You should instantly think - maybe the hypotenuse is the integer. It's the only other way the GMAT has ever really made these things hard.

So now we can eliminate C, D, and E. And since we're left with just 8 or 16, in this case, plugging in isn't so tough, and we get to 16 in about 31 seconds.

This is essentially what Squirrel was saying. Realize it's backwards ahead of time. But then just get the right answer. Remember, the GMAT doesn't award points for slickness of the math, it awards points for right answers in the shortest amount of time.
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 04 Jun 2009, 22:01
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As mentioned in the subject, this is a 45:45:90 isosceles triangle. As per properties of 45:45:90 triangle, the sides are in the ratio of 1:1: √2 (check out OG, or you can try this with couple of numbers)

So, let the sides be x, x, and x√2 (where x is the length of legs and x√2 is hypotenuse). We are supposed to find length of hypotenuse, i.e. x√2

Perimeter =
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16

B is correct answer
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 05 Jun 2009, 02:28
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Perimeter =
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16

B
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 30 Apr 2010, 14:55
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Himalayan wrote:
above720 wrote:
How would you solve?

The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?


x + x + x√2 = 16 + 16√2
2x + x√2 = 16 + 16√2
x√2 (1 + √2) = 16 (1 + √2)
x√2 = 16
x = 16/√2 = 8√2

h^2 = x^2 + x^2
h^2 = 2x^2
h^2 = 2 (8√2)^2
h^2 = 2x2x8x8
h = 2 x 8 = 16


much easier to solve, at least saves few seconds:

if x=8√2, => 2x=2*8√2=16√2, Perimeter-2x= hypotenuse, or 16 + 16√2-16√2=16
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Re: Isosceles Triangle  [#permalink]

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New post Updated on: 02 Jul 2010, 20:33
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Okay, so since this is an isosceles triangle, let's assume the equal sides are x and x and the hypotenuse is

\(\sqrt{x^2 + x^2}\) = \(\sqrt{2}x\)

Perimeter = \(x + x + \sqrt{2}x\) = \(2x+\sqrt{2}x\) = \(\sqrt{2}x (\sqrt{2} + 1)\)

This is equal to 16 + 16 \(\sqrt{2}\)

So equating we get:

\(\sqrt{2}x (\sqrt{2} + 1)\) = 16 + 16 \(\sqrt{2}\) = 16\((\sqrt{2} + 1)\)

Solving this and canceling the \((\sqrt{2} + 1)\) on both sides, we get:

\(\sqrt{2}x=16\)

So we have \(x = \frac{16}{\sqrt{2}}\)

From the first step, we know that the hypotenuse is:\(\sqrt{2}x\) = \(\frac{16}{\sqrt{2}}*\sqrt{2}\) = 16

So I'd say the answer is B.

Hope this helps.

Originally posted by whiplash2411 on 02 Jul 2010, 20:30.
Last edited by whiplash2411 on 02 Jul 2010, 20:33, edited 3 times in total.
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 22 Nov 2010, 19:14
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Hey guys,

Nice solution - just one thing I like to point out on these:

Think like the testmaker!

If I were writing the test and knew that everyone studies the 45-45-90 ratio as: 1, 1, sqrt 2

I would make a living off of making the shorter sides a multiple of sqrt 2 so that the long side is an integer. People aren't looking for that! And they often won't trust themselves enough to calculate correctly...they'll look at the answer choices and see that 3 of them are Integer*sqrt 2, and they'll think they screwed up somehow because the right answer "should" have a sqrt 2 on the end.

So...keep in mind that with the Triangle Ratios:

45-45-90
x: x: x*sqrt 2

30-60-90
x : x*sqrt3 : 2x

One of the easiest tricks up the GMAT author's sleeve is to make x equal to a multiple of the radical so that the radical appears on the side you're not expecting and the integer shows up where you think it shouldn't!

Also, as you go through questions like these, ask yourslf "how could they make that question a little harder" or "how could they test this concept in a way that I wouldn't be looking for it". Training yourself to look out for unique cases, from the testmaker's perspective, helps you to get a real mastery of the GMAT from a high level.
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 28 Apr 2011, 04:43
Let Hypotenuse = x

Sides = x/root(2)

So 2*x/root(2) + x = 16 +16root(2)

x + xroot(2) = 16 + 16root(2)

Answer - B
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 11 May 2011, 06:19
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Please point out what am I doing wrong:

Let the non-hypotenuse sides be \(x\). Therefore, hypotenuse= \(\sqrt{2}x\)

\(So,Perimeter= 2x+\sqrt{2}x= 16+16\sqrt{2}\)

\(=> x(2+\sqrt{2})= 8(2+\sqrt{2})\)

Comparing both sides, \(x=8\), Therefore, \(hypotenuse= 8\sqrt{2}\)

This answer is also included in the answer choices which makes me sure that I'm doing a very silly mistake somewhere, just cant see where! Please guys, please point out that small mistake that is eluding my eyes!
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) &nbs [#permalink] 11 May 2011, 06:19

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