Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 27 Aug 2005
Posts: 279
Location: Montreal, Canada

The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
22 Sep 2005, 20:02
Question Stats:
58% (02:01) correct 42% (02:01) wrong based on 1386 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) \(16\sqrt{2}\)
Official Answer and Stats are available only to registered users. Register/ Login.




GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4217
Location: Canada

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
19 Apr 2016, 18:14
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√2From 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




Intern
Joined: 22 Sep 2005
Posts: 1

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
23 Sep 2005, 01:44
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




Director
Joined: 03 Jun 2009
Posts: 627
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
04 Jun 2009, 22:01
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
Joined: 06 Feb 2006
Posts: 633

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

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
23 Sep 2005, 02:03
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
Joined: 26 Feb 2006
Posts: 624

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
03 May 2007, 09:56
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
Joined: 17 Aug 2005
Posts: 119

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
23 Sep 2005, 07:44
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
Joined: 24 Jan 2006
Posts: 163

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
10 Apr 2006, 11:45
y^2 = x^2+x^2 => x = y/sqrt(2)
2x+y=16+16sqrt(2)
ysrrt(2)+y=16+16sqrt(2)
y=16



Director
Joined: 06 Feb 2006
Posts: 633

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
29 Nov 2006, 06:38
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
Joined: 06 Feb 2006
Posts: 633

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
29 Nov 2006, 07:04
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
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1815
Concentration: General Management, Nonprofit

Re: Isosceles Triangle
[#permalink]
Show Tags
Updated on: 02 Jul 2010, 20:33
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
Joined: 04 Oct 2006
Posts: 47

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
03 May 2007, 03:40
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
Joined: 09 Mar 2003
Posts: 409

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
08 Jul 2007, 10:05
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
Status: mission completed!
Joined: 02 Jul 2009
Posts: 1186
GPA: 3.77

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
30 Apr 2010, 14:55
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, Perimeter2x= hypotenuse, or 16 + 16√216√2=16
_________________



VP
Joined: 29 Apr 2003
Posts: 1192

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
11 Apr 2006, 01:19
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
Joined: 16 Jan 2009
Posts: 272
Concentration: Technology, Marketing
GPA: 3
WE: Sales (Telecommunications)

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
05 Jun 2009, 02:28
Perimeter = 2x + x√2 = 16 +16√2 or, x√2(√2 + 1) = 16(1 + √2) or, x√2 = 16 B
_________________



Senior Manager
Status: Up again.
Joined: 31 Oct 2010
Posts: 453
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
11 May 2011, 06:19
Please point out what am I doing wrong: Let the nonhypotenuse 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/from620to710mygmatjourney114437.html



Math Expert
Joined: 02 Aug 2009
Posts: 8344

Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)
[#permalink]
Show Tags
19 Apr 2016, 21:00
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 on right 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
Joined: 23 Apr 2015
Posts: 274
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
18 Aug 2016, 22:57
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 [ 14.17 KiB  Viewed 3049 times ]




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 ]



