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

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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 19 Sep 2019, 01:40
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rishabhmishra1993 wrote:
Bunuel wrote:
rishabhmishra1993 wrote:
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) \(16\sqrt{2}\)

guy please clear my doubt: i solved this way, I don't know what I did wrong. Please help me:-
x+x+\(x\sqrt{2}\)=16+\(16\sqrt{2}\)
2x+\(x\sqrt{2}\)= 16+\(16\sqrt{2}\)
x(2+\(\sqrt{2}\))=8(2+\(2\sqrt{2}\))
x=8 and x=\(8\sqrt{2}\)



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

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

Factor out \(x\sqrt{2}\) from the left hand side: \(x\sqrt{2}(1+\sqrt{2})=16+16\sqrt{2}\);

Factor out 16 from the right hand side: \(x\sqrt{2}(1+\sqrt{2})=16(1+\sqrt{2})\);

Reduce by \(1+\sqrt{2}\): \(x\sqrt{2}=16\);

Hope it's clear.

but what's wrong with what i did in this question?


First of all, we have a linear (first degrees) equation and it cannot give two solutions. Next, \(x=8\) does not satisfy \(x(2+\sqrt{2})=8(2+2\sqrt{2})\).
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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 Updated on: 07 Jul 2007, 22:02
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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?


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 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 07 Mar 2015, 19:36
So we know that the proportions of a isosceles right triangle are x:x:x\sqrt{2}, since the perimeter is 16+16*sqrt(2), then
2x+x*sqrt(2)=16+16\sqrt{2}. we can solve for x and use that information to figure out what the hypotenuse is

First, let's isolate x, so we get x=16+16*sqrt(2)/(2+sqrt(2))
we can eliminate the root from the denominator by multiplying both the numerator and the denominator by 2-sqrt(2), and we end up with:
16*sqrt(2)/2=x or x=8*sqrt(2)

This is one of the options, because the GMAT is trying to trick you into picking it. However, we are looking for the hypotenuse and this is one of the sides, so we have to multiply it by sqrt(2) and we get 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 05 Nov 2017, 20:33
I am little confuse here. given that this is a Isosceles right angle triangle which can be either 30-60-90 triangle or 45-45-90.
I don't understand,why it should be intuitive to consider 45-45-90 triangle ?

would appreciate your response. thanks !
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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 Nov 2017, 22:55
jyotipes21@gmail.com wrote:
I am little confuse here. given that this is a Isosceles right angle triangle which can be either 30-60-90 triangle or 45-45-90.
I don't understand,why it should be intuitive to consider 45-45-90 triangle ?

would appreciate your response. thanks !


An isosceles triangle is a triangle with at least two sides of the same length. An isosceles triangle also has at least two angles of the same measure; namely, the angles opposite to the two sides of the same length. So, 30-60-90 is NOT an isosceles triangle.

You should go through the basics once more.

Basics to Brush-Up Fundamentals




23. Geometry




24. Coordinate Geometry




25. Triangles




26. Polygons




27. Circles




28. Rectangular Solids and Cylinders




29. Graphs and Illustrations




For more check Ultimate GMAT Quantitative Megathread
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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 Mar 2018, 17:46
coffeeloverfreak wrote:
The perimeter of a certain isosceles right triangle is \(16 + 16 \sqrt{2}\). What is the length of the hypotenuse of the triangle?

\((A) \hspace{5} 8\)

\((B) \hspace{5} 16\)

\((C) \hspace{5} 4\sqrt{2}\)

\((D) \hspace{5} 8\sqrt{2}\)

\((E) \hspace{5} 16\sqrt{2}\)

Main Idea: Equate the expression for the sum of the sides to the value given

Details: Let the equal sides be x. So hypotenuse is x*sqrt(2)

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

Simplifying we have x= 16(1+ sqrt(2)) /(2+sqrt(2))

So x*sqrt(2) = 16.

Hence B.
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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 09 Apr 2019, 03:28
coffeeloverfreak wrote:
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) \(16\sqrt{2}\)

If the legs are x and x, then Hypotenuse = \(\sqrt{2}x\)
\(x+x+\sqrt{2}x=16+16 \sqrt{2}\)
\(x(2+\sqrt{2})=16(1+\sqrt{2})\)
\(x(\sqrt{2}*\sqrt{2} +\sqrt{2})=16(1+\sqrt{2})\)
\(x\sqrt{2}(\sqrt{2} +1)=16(1+\sqrt{2})\)
=>\(\sqrt{2}*x =16\)
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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 19 Sep 2019, 00:02
coffeeloverfreak wrote:
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) \(16\sqrt{2}\)

guy please clear my doubt: i solved this way, I don't know what I did wrong. Please help me:-
x+x+\(x\sqrt{2}\)=16+\(16\sqrt{2}\)
2x+\(x\sqrt{2}\)= 16+\(16\sqrt{2}\)
x(2+\(\sqrt{2}\))=8(2+\(2\sqrt{2}\))
x=8 and x=\(8\sqrt{2}\)
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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 19 Sep 2019, 01:22
rishabhmishra1993 wrote:
coffeeloverfreak wrote:
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) \(16\sqrt{2}\)

guy please clear my doubt: i solved this way, I don't know what I did wrong. Please help me:-
x+x+\(x\sqrt{2}\)=16+\(16\sqrt{2}\)
2x+\(x\sqrt{2}\)= 16+\(16\sqrt{2}\)
x(2+\(\sqrt{2}\))=8(2+\(2\sqrt{2}\))
x=8 and x=\(8\sqrt{2}\)



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

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

Factor out \(x\sqrt{2}\) from the left hand side: \(x\sqrt{2}(1+\sqrt{2})=16+16\sqrt{2}\);

Factor out 16 from the right hand side: \(x\sqrt{2}(1+\sqrt{2})=16(1+\sqrt{2})\);

Reduce by \(1+\sqrt{2}\): \(x\sqrt{2}=16\);

Hope it's clear.
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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 19 Sep 2019, 01:26
Bunuel wrote:
rishabhmishra1993 wrote:
coffeeloverfreak wrote:
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) \(16\sqrt{2}\)

guy please clear my doubt: i solved this way, I don't know what I did wrong. Please help me:-
x+x+\(x\sqrt{2}\)=16+\(16\sqrt{2}\)
2x+\(x\sqrt{2}\)= 16+\(16\sqrt{2}\)
x(2+\(\sqrt{2}\))=8(2+\(2\sqrt{2}\))
x=8 and x=\(8\sqrt{2}\)



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

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

Factor out \(x\sqrt{2}\) from the left hand side: \(x\sqrt{2}(1+\sqrt{2})=16+16\sqrt{2}\);

Factor out 16 from the right hand side: \(x\sqrt{2}(1+\sqrt{2})=16(1+\sqrt{2})\);

Reduce by \(1+\sqrt{2}\): \(x\sqrt{2}=16\);

Hope it's clear.

but what's wrong with what i did in this question?
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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 Sep 2019, 11:36
coffeeloverfreak wrote:
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) \(16\sqrt{2}\)



Iso right triangle ratio is x : x : x√2
We are given P = 16 + 16√2 = 2x + x√2
In a triangle 2 sides must be more than the 3rd side, so 2x= 16√2 and x√2 = 16
You can solve the first one... x=8√2 * √2 for hyp length = 16, ... second one is the answer directly.
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Re: The perimeter of a certain isosceles right triangle is 16 +  [#permalink]

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