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:
If P and Q are two points on the line 3x + 4y =-15 such that [#permalink]
21 Oct 2010, 16:22
4
This post received KUDOS
Expert's post
8
This post was BOOKMARKED
kobinaot wrote:
help pls!
2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by
a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)
Look at the the diagram below:
OC is perpendicular to AB (PQ), so it's height of AOB and POQ;
\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);
As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);
Re: Good Question ...Number system [#permalink]
22 Oct 2010, 21:08
Bunuel wrote:
kobinaot wrote:
help pls!
2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by
a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)
Look at the the diagram below:
Attachment:
graph.php.png
OC is perpendicular to AB (PQ), so it's height of AOB and POQ;
\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);
As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);
Re: Good Question ...Number system [#permalink]
23 Oct 2010, 04:35
1
This post received KUDOS
Expert's post
utin wrote:
Bunuel, what makes us think that point o is Origin in this case???
Well, we must know the coordinates of the point O to solve the question. As origin in coordinate geometry usually marked with letter O (as the center of a circle) then we can assume that O is origin in this case. Though I think on GMAT it would be explicitly stated. _________________
Re: Good Question ...Number system [#permalink]
23 Oct 2010, 20:02
Bunuel wrote:
utin wrote:
Bunuel, what makes us think that point o is Origin in this case???
Well, we must know the coordinates of the point O to solve the question. As origin in coordinate geometry usually marked with letter O (as the center of a circle) then we can assume that O is origin in this case. Though I think on GMAT it would be explicitly stated.
Re: Good Question ...Number system [#permalink]
13 Jul 2012, 03:30
Bunuel wrote:
kobinaot wrote:
help pls!
2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by
a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)
Look at the the diagram below:
Attachment:
graph.php.png
OC is perpendicular to AB (PQ), so it's height of AOB and POQ;
\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);
As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);
Re: Good Question ...Number system [#permalink]
13 Jul 2012, 03:34
Expert's post
MacFauz wrote:
Bunuel wrote:
kobinaot wrote:
help pls!
2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by
a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)
Look at the the diagram below:
Attachment:
graph.php.png
OC is perpendicular to AB (PQ), so it's height of AOB and POQ;
\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);
As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);
Re: If P and Q are two points on the line 3x + 4y =-15 such that [#permalink]
01 Mar 2014, 11:16
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. _________________
Re: Good Question ...Number system [#permalink]
05 Apr 2014, 15:28
Bunuel could you please elaborate a little bit more on the relation between similar triangles? How could you get OC/OB = OA/AB, I'm not able to see it.
Re: Good Question ...Number system [#permalink]
23 Apr 2014, 06:54
Expert's post
jlgdr wrote:
Bunuel could you please elaborate a little bit more on the relation between similar triangles? How could you get OC/OB = OA/AB, I'm not able to see it.
Re: If P and Q are two points on the line 3x + 4y =-15 such that [#permalink]
24 Apr 2014, 01:04
Expert's post
pretzel wrote:
Hi Bunnel,
Could you please explain how we can derive SQRT (OA^2 + OB^2) = 25/4?
AB is the hypotenuse of right triangle AOB, hence \(AB=hypotenuse=\sqrt{OA^2+OB^2}\), where \(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively).
Re: Good Question ...Number system [#permalink]
07 May 2014, 01:26
Bunuel wrote:
kobinaot wrote:
help pls!
2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by
a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)
Look at the the diagram below:
Attachment:
graph.php.png
OC is perpendicular to AB (PQ), so it's height of AOB and POQ;
\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);
As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);
Re: Good Question ...Number system [#permalink]
07 May 2014, 01:34
Expert's post
himanshujovi wrote:
Bunuel wrote:
kobinaot wrote:
help pls!
2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by
a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)
Look at the the diagram below:
Attachment:
graph.php.png
OC is perpendicular to AB (PQ), so it's height of AOB and POQ;
\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);
As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);
Re: If P and Q are two points on the line 3x + 4y =-15 such that [#permalink]
07 Mar 2015, 08:04
Bunuel wrote:
kobinaot wrote:
help pls!
2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by
a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)
Look at the the diagram below:
Attachment:
graph.php.png
OC is perpendicular to AB (PQ), so it's height of AOB and POQ;
\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);
As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...