Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 22 May 2015, 10:25

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Need Help With Math Problem: Triangles

Author Message
TAGS:
Intern
Joined: 18 Nov 2010
Posts: 10
Followers: 0

Kudos [?]: 8 [0], given: 8

Need Help With Math Problem: Triangles [#permalink]  18 Nov 2010, 07:33
00:00

Difficulty:

35% (medium)

Question Stats:

50% (02:11) correct 50% (00:29) wrong based on 10 sessions
The perimeter of a certain isoceles right triangle is 16 + (16 * Square Root (2)). What is the length of the hypotenuse.

8
4 * Square Root (2)
8 * Square Root (2)
16 * Square Root (2)

Thanks!
[Reveal] Spoiler: OA
Manager
Status: Still Struggling
Joined: 02 Nov 2010
Posts: 138
Location: India
GMAT Date: 10-15-2011
GPA: 3.71
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 30 [1] , given: 8

Re: Need Help With Math Problem: Triangles [#permalink]  18 Nov 2010, 08:34
1
KUDOS
ANS : B

lets say we have an isoceles triangle ABC with B = 90 degress and AB=BC=X and CA=Y.

now applying rt angle formula sqr Y = sqr X + sqr X = 2(sqr X).......................1
also, perimeter = 2X + Y = 16 + 16(sqrt of 2).................2

solve both the equations for Y.. in the end, we would get the equation something like:

Y + Y(sqrt 2) = 16 + 16(sqrt 2).

From this, its easily seen that Y=16.

----------------------------------------------------------------------
Consider KUDOS if my post helped you!!
_________________

Knewton Free Test 10/03 - 710 (49/37)
Princeton Free Test 10/08 - 610 (44/31)
Kaplan Test 1- 10/10 - 630
Veritas Prep- 10/11 - 630 (42/37)
MGMAT 1 - 10/12 - 680 (45/34)

Intern
Joined: 17 Aug 2009
Posts: 41
Location: United States
Concentration: Finance, Entrepreneurship
GMAT 1: 740 Q49 V42
GPA: 3.29
WE: Engineering (Consulting)
Followers: 0

Kudos [?]: 11 [1] , given: 4

Re: Need Help With Math Problem: Triangles [#permalink]  18 Nov 2010, 09:51
1
KUDOS
Since we are dealing with isosceles triangle, we know that its 2 legs are equal.

The equation for its perimeter = hypotenuse + 2*leg.

and since isosceles right triangle is a 45 45 right triangle,

$$leg = \frac{hypotenuse}{\sqrt{2}}$$

which can be simplified to $$leg = \frac{hypotenuse}{2}\sqrt{2}$$

So the perimeter equation becomes

$$perimeter = hypotenuse + 2\frac{hypotenuse}{2}\sqrt{2}$$

$$perimeter = hypotenuse + hypotenuse\sqrt{2}$$

and since we are given

$$perimeter = 16 + 16\sqrt{2}$$

therefore hypotenuse = 16.
Manager
Joined: 01 Nov 2010
Posts: 182
Location: Zürich, Switzerland
Followers: 2

Kudos [?]: 21 [0], given: 20

Re: Need Help With Math Problem: Triangles [#permalink]  18 Nov 2010, 16:02
Remember the equation for isoscelleous right triangle sides- x:x:x*sqrt(2)

Let one leg= a Thus,
2a+sqrt(2)*a= 16+16*sqrt(2)

2a+sqrt(2)*a= 16*sqrt(2)+16

thus a = 8 *sqrt(2)
And hypotenus= 16

Ans:- B
Manager
Joined: 20 Apr 2010
Posts: 240
WE 1: 4.6 years Exp IT prof
Followers: 8

Kudos [?]: 29 [0], given: 47

Re: Need Help With Math Problem: Triangles [#permalink]  22 Nov 2010, 13:45
messed up....
_________________

I will give a Fight till the End

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
- Bernard Edmonds

A person who is afraid of Failure can never succeed -- Amneet Padda

Don't Forget to give the KUDOS

Veritas Prep GMAT Instructor
Joined: 26 Jul 2010
Posts: 226
Followers: 141

Kudos [?]: 307 [0], given: 27

Re: Need Help With Math Problem: Triangles [#permalink]  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

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

Re: Need Help With Math Problem: Triangles   [#permalink] 22 Nov 2010, 18:14
Similar topics Replies Last post
Similar
Topics:
Help needed with GMAT Math Problem? 2 07 Oct 2008, 02:08
Need help in Math!! 2 31 Jul 2008, 15:44
Help with simple maths problem 4 11 Jan 2007, 07:26
Triangle Problem DS Help 5 14 Jul 2006, 16:11
Need Help - Explain Math Problem 4 04 Dec 2005, 10:01
Display posts from previous: Sort by