Author 
Message 
TAGS:

Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 526
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

The perimeter of a right isoscles triangle is 32+32 [#permalink]
Show Tags
04 Feb 2012, 14:50
4
This post received KUDOS
5
This post was BOOKMARKED
Question Stats:
60% (01:11) correct 40% (01:21) wrong based on 258 sessions
HideShow timer Statistics
The perimeter of a right isoscles triangle is 32 + 32 \(\sqrt{2}\). What is the length of the hypotenues of the triangle? A) 16 B) 8 \(\sqrt{3}\) C) 32 D) 8 \(\sqrt{2}\) E) 16\(\sqrt{2}\) How come answer is E ? This is how I am trying to solve Isoscles Right angle triangle mean has x:x:x\(\sqrt{2}\) as its sides. Therefore, x+x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\) 2x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\). I am stuck after this.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Math Expert
Joined: 02 Sep 2009
Posts: 43901

Re: Perimeter of a right isoceles triangle [#permalink]
Show Tags
04 Feb 2012, 15:12
5
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
enigma123 wrote: The perimeter of a right isoscles triangle is 32 + 32 \(\sqrt{2}\). What is the length of the hypotenues of the triangle?
A) 16 B) 8 \(\sqrt{3}\) C) 32 D) 8 \(\sqrt{2}\) E) 16\(\sqrt{2}\)
How come answer is E ?
This is how I am trying to solve
Isoscles Right angle triangle mean has x:x:x\(\sqrt{2}\) as its sides. Therefore, x+x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\) 2x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\). I am stuck after this. In a right triangle where the angles are 45°, 45°, and 90° the sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles. Now, if we proceed the way you suggest: Let the side of this triangle be \(x\) then, \(P=32+32\sqrt{2}=x+x+\sqrt{2}x\) > \(32+32\sqrt{2}=x(2+\sqrt{2})\) >\(x=\frac{32+32\sqrt{2}}{2+\sqrt{2}}\) > muliptly both nominatro and denominator by \(2\sqrt{2}\) (to rationalize the denominator): \(x=\frac{(32+32\sqrt{2})(2\sqrt{2})}{(2+\sqrt{2})(2\sqrt{2})}=\frac{64+64\sqrt{2}32\sqrt{2}64}{42}=\frac{32\sqrt{2}}{2}=16\sqrt{2}\). So, hypotenuse is \(16\sqrt{2}*\sqrt{2}=32\) Answer: C. P.S. Of course you can also solve this problem with approximation method or plugin method using the answer choices.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 169
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)

Re: The perimeter of a right isoscles triangle is 32+32 [#permalink]
Show Tags
05 Feb 2012, 03:03
Ans is C and not E 2a+b = 32+32\sqrt{2} and 2*a^2 = b^2 on solving we get b = 32



Manager
Joined: 24 Sep 2009
Posts: 105

The perimeter of a certain isosceles right triangle is 16 + [#permalink]
Show Tags
13 Mar 2012, 10:12
The perimeter of a certain isosceles right triangle is 16 + \(16\sqrt{2}\). What is the length of the hypotenuse in the triangle?
A. 8 B. 16 C. \(4\sqrt{2}\) D. \(8\sqrt{2}\) E. \(16\sqrt{2}\)
Got this one wrong. Don't know how?



Senior Manager
Joined: 23 Oct 2010
Posts: 375
Location: Azerbaijan
Concentration: Finance

Re: Perimeter of an Iscoceles Right Triangle [#permalink]
Show Tags
13 Mar 2012, 10:45
2a+c=16+16sqrt2 if c=16, then 16sqrt2=2a or a=8sqrt2 lets check it out c^2=(8sqrt2)^2+(8sqrt2)^2=64*4 c=8*2=16 ans is B
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth



Intern
Joined: 20 Feb 2012
Posts: 42

Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]
Show Tags
13 Mar 2012, 12:32
1
This post received KUDOS
Given isosceles right triangle So two angles = 45 Let the side opposite to Angle 45 is x Then hypotenuse2 = x^2 + x^2 Hypotenuse = √2 x



Manager
Joined: 24 Sep 2009
Posts: 105

Re: The perimeter of a right isoscles triangle is 32+32 [#permalink]
Show Tags
13 Mar 2012, 14:49
Idiot me found the length of one of the two other sides and chose that as the answer.
SMDH.



Intern
Joined: 04 Apr 2013
Posts: 1

The perimeter of a certain isosceles right triangle is [#permalink]
Show Tags
01 May 2013, 04:56
bangalorian2000 wrote: zz0vlb wrote: The perimeter of a certain isoceles right triangle is 16 + 16\sqrt{2}. what is the hypotenuse of the triangle? A. 8 B. 16 C. 4\sqrt{2} D. 8\sqrt{2} E. 16\sqrt{2} Source: GMAT Prep let one side be x hypotenuse is x\sqrt{2} perimeter = 2x + x\sqrt{2} = 16+16\sqrt{2} solving x = 8\sqrt{2} hence hypotenuse is 16 hence B. How is the value of x calculated here ? perimeter = 2x + x\sqrt{2} = 16+16\sqrt{2} solving x = 8\sqrt{2} Thanks



Intern
Joined: 08 Jan 2013
Posts: 39

Hypotenuse of isosceles right triangle given perimeter [#permalink]
Show Tags
18 May 2013, 06:21
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)
Please attack my approach the question.
P = 16 + 16 * sqrt(2)
The "isosceles right triangle" must have the angles 90:45:45, and bisecting through the middle of the base at 90 degrees must give two smaller "isosceles right triangles" with sides equal to half the original hypotenuse.
Therefore, the hypotenuse is 2 * P / 4. This gives something approximating 19. E is also approximately 19. Why is the answer exactly 16?
Thanks,



Manager
Status: *Lost and found*
Joined: 25 Feb 2013
Posts: 123
Location: India
Concentration: General Management, Technology
GPA: 3.5
WE: Web Development (Computer Software)

Re: Hypotenuse of isosceles right triangle given perimeter [#permalink]
Show Tags
18 May 2013, 06:43
1
This post received KUDOS
stormbind 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)
Please attack my approach the question.
P = 16 + 16 * sqrt(2)
The "isosceles right triangle" must have the angles 90:45:45, and bisecting through the middle of the base at 90 degrees must give two smaller "isosceles right triangles" with sides equal to half the original hypotenuse.
Therefore, the hypotenuse is 2 * P / 4. This gives something approximating 19. E is also approximately 19. Why is the answer exactly 16?
Thanks, The answer is [B]. For an "isosceles right triangle", the sides can be {r, r and \(sqrt(2)*r\)}. Hence, perimeter = \(2r + sqrt(2)*r = 16 + 16 * sqrt(2)\) and r comes out to be \(16(1+sqrt(2))/(2+sqrt(2))\). Now the hypotenuse is \(sqrt(2)*r\) from the above. Hence multiplying sqrt(2) we have the value = 16. Answer! Hope the answer is accurate! Regards, Arpan
_________________
Feed me some KUDOS! *always hungry*
My Thread : Recommendation Letters



NonHuman User
Joined: 09 Sep 2013
Posts: 13757

Re: The perimeter of a right isoscles triangle is 32+32 [#permalink]
Show Tags
28 Jun 2014, 07:33
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



NonHuman User
Joined: 09 Sep 2013
Posts: 13757

Re: The perimeter of a right isoscles triangle is 32+32 [#permalink]
Show Tags
15 Jul 2015, 10:09
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Board of Directors
Joined: 17 Jul 2014
Posts: 2736
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: The perimeter of a right isoscles triangle is 32+32 [#permalink]
Show Tags
28 Nov 2015, 09:38
this is a tricky question, yet, we have to remember that right isosceles triangle has the ratio of the sides x:x:x(sqrt2)
clearly, something doesn't add up. if 32(sqrt2) is the hypotenuse, then the other sides should be 32 each, and the total would be 64+32sqrt2 ok, but what if the hypotenuse is 32? x(sqrt2) = 32 x = 32/sqrt2 or 32*sqrt2/2 = 16(sqrt2) ok, now that's something better. we have 2 x's and one hypotenuse. 2x = 32(sqrt2) and hypotenuse 32. the answer is C.



NonHuman User
Joined: 09 Sep 2013
Posts: 13757

Re: The perimeter of a right isoscles triangle is 32+32 [#permalink]
Show Tags
17 Apr 2017, 11:50
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: The perimeter of a right isoscles triangle is 32+32
[#permalink]
17 Apr 2017, 11:50






