GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 16 Nov 2019, 19:08

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
avatar
Joined: 27 Aug 2005
Posts: 285
Location: Montreal, Canada
The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 22 Sep 2005, 20:02
18
1
141
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

58% (02:01) correct 42% (02:00) wrong based on 1338 sessions

HideShow timer Statistics

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) \(16\sqrt{2}\)
Most Helpful Expert Reply
GMAT Club Legend
GMAT Club Legend
User avatar
V
Joined: 12 Sep 2015
Posts: 4064
Location: Canada
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 19 Apr 2016, 18:14
16
6
Quote:
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

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Image
Most Helpful Community Reply
Intern
Intern
avatar
Joined: 22 Sep 2005
Posts: 1
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 23 Sep 2005, 01:44
17
10
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
General Discussion
Director
Director
User avatar
Joined: 03 Jun 2009
Posts: 649
Location: New Delhi
WE 1: 5.5 yrs in IT
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 04 Jun 2009, 22:01
10
8
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
_________________
Director
Director
avatar
Joined: 06 Feb 2006
Posts: 670
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 11 Apr 2006, 01:10
7
2
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....
Manager
Manager
avatar
Joined: 06 Aug 2005
Posts: 163
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 23 Sep 2005, 02:03
5
2
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
Director
Director
User avatar
Joined: 26 Feb 2006
Posts: 658
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 03 May 2007, 09:56
4
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
Manager
Manager
avatar
Joined: 17 Aug 2005
Posts: 132
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 23 Sep 2005, 07:44
3
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.
Manager
Manager
User avatar
Joined: 24 Jan 2006
Posts: 188
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 10 Apr 2006, 11:45
3
y^2 = x^2+x^2 => x = y/sqrt(2)

2x+y=16+16sqrt(2)
ysrrt(2)+y=16+16sqrt(2)
y=16
Director
Director
avatar
Joined: 06 Feb 2006
Posts: 670
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 29 Nov 2006, 06:38
3
1
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...
Director
Director
avatar
Joined: 06 Feb 2006
Posts: 670
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 29 Nov 2006, 07:04
3
2
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
SVP
SVP
User avatar
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1821
Concentration: General Management, Nonprofit
GMAT ToolKit User
Re: Isosceles Triangle  [#permalink]

Show Tags

New post Updated on: 02 Jul 2010, 20:33
3
1
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.
Intern
Intern
avatar
Joined: 04 Oct 2006
Posts: 49
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 03 May 2007, 03:40
2
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
CIO
CIO
User avatar
Joined: 09 Mar 2003
Posts: 412
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 08 Jul 2007, 10:05
2
2
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.
VP
VP
User avatar
Status: mission completed!
Joined: 02 Jul 2009
Posts: 1203
GPA: 3.77
GMAT ToolKit User Reviews Badge
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 30 Apr 2010, 14:55
2
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
_________________
VP
VP
User avatar
Joined: 29 Apr 2003
Posts: 1225
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 11 Apr 2006, 01:19
1
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
Senior Manager
Senior Manager
User avatar
Joined: 16 Jan 2009
Posts: 288
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

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

B
_________________
Lahoosaher
Senior Manager
Senior Manager
avatar
Status: Up again.
Joined: 31 Oct 2010
Posts: 458
Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 11 May 2011, 06:19
1
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!
_________________
My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 8177
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 19 Apr 2016, 21:00
1
Flexxice wrote:
The perimeter of a certain isosceles triangle is 16 + 16 * square root of 2. What is the length of the hypotenuse of the triangle?

A) 8

B) 16

C) 4 * square root of 2

D) 8 * square root of 2

E) 16 * square root of 2



Firstly you have missed out onright angle in "certain isosceles triangle"..
secondly..
16 +16\(\sqrt{2}\) should tell us that the two equal sides are either 8 or 8\(\sqrt{2}\)..
If it is 8, perimeter = 8+8+8\(\sqrt{2}\), which is not the case here..
if it is 8\(\sqrt{2}\), P = 8\(\sqrt{2}\)+8\(\sqrt{2}\)+8\(\sqrt{2}*\sqrt{2}\) = 16 + 16\(\sqrt{2}\).. which is the P
so side is 8\(\sqrt{2}\) and HYP = 16
B
OA is being edited accordingly
_________________
Senior Manager
Senior Manager
avatar
Joined: 23 Apr 2015
Posts: 281
Location: United States
Concentration: General Management, International Business
WE: Engineering (Consulting)
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

Show Tags

New post 18 Aug 2016, 22:57
1
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?

\((A) \hspace{5} 8\)
\((B) \hspace{5} 16\)
\((C) \hspace{5} 4\sqrt{2}\)
\((D) \hspace{5} 8\sqrt{2}\)
\((E) \hspace{5} 16\sqrt{2}\)


Hence the answer is B) 16
Attachments

triangle1.PNG
triangle1.PNG [ 14.17 KiB | Viewed 2917 times ]

GMAT Club Bot
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)   [#permalink] 18 Aug 2016, 22:57

Go to page    1   2    Next  [ 33 posts ] 

Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne