GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 26 Feb 2020, 05:34

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 61508
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

19 Sep 2019, 00:40
1
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) $$16\sqrt{2}$$

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 $$x(2+\sqrt{2})=8(2+2\sqrt{2})$$.
_________________
Director
Joined: 08 Jun 2005
Posts: 890
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

Updated on: 07 Jul 2007, 21:02
1
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 !

Originally posted by KillerSquirrel on 07 Jul 2007, 16:56.
Last edited by KillerSquirrel on 07 Jul 2007, 21:02, edited 2 times in total.
Veritas Prep Representative
Joined: 26 Jul 2010
Posts: 425
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

22 Nov 2010, 18:14
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.
_________________
Brian

Curriculum Developer, Instructor, and Host of Veritas Prep On Demand

Save \$100 on live Veritas Prep GMAT Courses and Admissions Consulting

Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews
Intern
Joined: 21 Apr 2014
Posts: 38
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

07 Mar 2015, 18:36
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)
_________________
Eliza
GMAT Tutor
bestgmatprepcourse.com
Intern
Joined: 20 Sep 2016
Posts: 19
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

05 Nov 2017, 19:33
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 !
Math Expert
Joined: 02 Sep 2009
Posts: 61508
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

05 Nov 2017, 21:55
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

For more check Ultimate GMAT Quantitative Megathread
_________________
Director
Joined: 17 Dec 2012
Posts: 620
Location: India
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

08 Mar 2018, 16:46
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.
_________________
Srinivasan Vaidyaraman
Magical Logicians

Holistic and Holy Approach
Director
Joined: 17 Dec 2012
Posts: 620
Location: India
The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

09 Apr 2019, 02:28
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 = $$\sqrt{2}x$$
$$x+x+\sqrt{2}x=16+16 \sqrt{2}$$
$$x(2+\sqrt{2})=16(1+\sqrt{2})$$
$$x(\sqrt{2}*\sqrt{2} +\sqrt{2})=16(1+\sqrt{2})$$
$$x\sqrt{2}(\sqrt{2} +1)=16(1+\sqrt{2})$$
=>$$\sqrt{2}*x =16$$
_________________
Srinivasan Vaidyaraman
Magical Logicians

Holistic and Holy Approach
Intern
Joined: 16 Aug 2019
Posts: 8
The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

18 Sep 2019, 23:02
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) $$16\sqrt{2}$$

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=$$8\sqrt{2}$$
Math Expert
Joined: 02 Sep 2009
Posts: 61508
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

19 Sep 2019, 00:22
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) $$16\sqrt{2}$$

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.
_________________
Intern
Joined: 16 Aug 2019
Posts: 8
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

19 Sep 2019, 00:26
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) $$16\sqrt{2}$$

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?
Senior Manager
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 442
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

### Show Tags

30 Sep 2019, 10:36
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.
Non-Human User
Joined: 09 Sep 2013
Posts: 14142
Re: The perimeter of a certain isosceles right triangle is 16 +  [#permalink]

### Show Tags

09 Oct 2019, 02:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: The perimeter of a certain isosceles right triangle is 16 +   [#permalink] 09 Oct 2019, 02:18

Go to page   Previous    1   2   [ 33 posts ]

Display posts from previous: Sort by