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.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
A certain right triangle has sides of length x, y, and z [#permalink]
17 Jan 2010, 01:39
2
This post received KUDOS
6
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
55% (hard)
Question Stats:
67% (02:16) correct
33% (01:45) wrong based on 145 sessions
A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?
A. \(y >\sqrt{2}\) B. \(\frac{\sqrt{3}}{2}<y<\sqrt{2}\) C. \(\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}\) D. \(\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}\) E. \(y<\frac{\sqrt{3}}{4}\)
Re: Possible Values of side of Right Angle Triangle [#permalink]
17 Jan 2010, 01:55
vibhaj wrote:
A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? A. y > sqr root (2) B. sqr root (3)/2 < y < sqr root (2) C. sqr root (2)/3 < y < sqr root (3)/2 D. sqr root (3)/4 < y < sqr root (2)/3 E. y < sqr root (3)/4
OA after discussion.
--Edited to denote the square roots more clearly
IMO it should be A,
area of the triangle=1 = 1/2 X*Y
or X*Y = 2
Now , also X<Y , now we by using the POE, we can see that apart from 1 option rest all doesn't satisfy this equation.
Re: Possible Values of side of Right Angle Triangle [#permalink]
17 Jan 2010, 11:30
16
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
vibhaj wrote:
Thanks Nitish.is there a better way to solve this , without POE ..?
The area of the triangle is \(\frac{xy}{2}=1\) (\(x<y<z\) means that hypotenuse is \(z\)) --> \(x=\frac{2}{y}\). As \(x<y\), then \(\frac{2}{y}<y\) --> \(2<y^2\) --> \(\sqrt{2}<y\).
Also note that max value of \(y\) is not limited at all. For example \(y\) can be \(1,000,000\) and in this case \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) --> \(x=\frac{2}{1,000,000}\).
A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? A. y > \(sqrt 2\) B. \((sqrt 3)/2\) < y <\(sqrt 2\) C. \((sqrt 2)/3\) < y < \((sqrt 3)/2\) D. \((sqrt 3)/4\) < y < \((sqrt 2)/3\) E. y < \((sqrt 3)/4\)
I'M LOOKING FOR THE ALGEBRAIC SOLUTION, WITHOUT SUPPOSING (PLUGGING) THE VALUES OF X,Y AND Z..!!
A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? A. y > \(sqrt 2\) B. \((sqrt 3)/2\) < y <\(sqrt 2\) C. \((sqrt 2)/3\) < y < \((sqrt 3)/2\) D. \((sqrt 3)/4\) < y < \((sqrt 2)/3\) E. y < \((sqrt 3)/4\)
I'M LOOKING FOR THE ALGEBRAIC SOLUTION, WITHOUT SUPPOSING (PLUGGING) THE VALUES OF X,Y AND Z..!!
Some questions .... please help [#permalink]
07 May 2010, 21:05
1.A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? A. y > sqrt.2 B. sqrt.3/2 < y < sqrt. 2 C. sqrt.2/3 < y < sqrt.3/2 D. sqrt.3/4 < y < sqrt.2/3 E. y < sqrt.3/4
Re: Some questions .... please help [#permalink]
07 May 2010, 22:42
ok question 1 doesnt seem GMATesque but i have come up with the solution ...let me know what the source is and what the OA is
1) we know that area of triangle is 2 so xy/2 = 1 so xy = 2 moreover x < y .....so think about the two graphs ...y =2/x and x=y ....there will be a point on x-axis beyond which y<x ...to find this point lets equate the two eqns
2/x = x --> x = sqrt(2)
beyond this point y<x ...so y > sqrt(2)
so answer should be A ...in all other choices the value of y is less than sqrt(2)
A certain right triangle [#permalink]
21 May 2010, 02:17
A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? A. y > sqrt.2 B. sqrt.3/2 < y < sqrt. 2 C. sqrt.2/3 < y < sqrt.3/2 D. sqrt.3/4 < y < sqrt.2/3 E. y < sqrt.3/4
Re: A certain right triangle [#permalink]
21 May 2010, 02:24
dimitri92 wrote:
A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? A. y > sqrt.2 B. sqrt.3/2 < y < sqrt. 2 C. sqrt.2/3 < y < sqrt.3/2 D. sqrt.3/4 < y < sqrt.2/3 E. y < sqrt.3/4
This has already been discussed, please follow the link
Re: Possible Values of side of Right Angle Triangle [#permalink]
10 Mar 2011, 20:03
2
This post received KUDOS
Expert's post
gmat1220 wrote:
Solved this intuitively - pls verify this reasoning. The above stem means that one angle > 45 and another angle is < 45
At x = y (hypothetically) i.e. Both angles equal. x = z sin (45) = z/ sqrt(2) y = z cos (45) = z/ sqrt(2)
1/2 x y = 1 Hence z = 2
x = sqrt(2) y = sqrt(2)
If one angle > 45 and another angle is < 45, And knowing x < y < z So 0 < x < sqrt(2) and sqrt(2) < y < Infinity
Yes, it can be solved intuitively. It is a very interesting way of thinking. There is one small thing: z may not be 2. e.g. When x = 1 and y = 2, z = sqrt(5)
Think this way: We know x < y and (1/2)xy = 1 (z is the hypotenuse) So xy = 2 If x = y, then x = y = sqrt(2) But since x < y, x is less than sqrt(2) and y is greater than sqrt(2). Makes sense. Good thinking. _________________
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...