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

It is currently 30 Aug 2014, 11:00

Close

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
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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

In the figure above, points P and Q lie on the circle with

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
avatar
Joined: 30 Aug 2009
Posts: 290
Location: India
Concentration: General Management
Followers: 2

Kudos [?]: 93 [0], given: 5

GMAT Tests User
Re: coordinate geometry [#permalink] New post 13 Nov 2009, 06:32
IndianGuardian wrote:
How would you solve it by the distance formula kp1811?


from the figure we can see that OP=OQ = radius of the semicircle. Also Angle POQ = 90degree

thus PQ^2 = OP^2 + OQ^2-------eqn1

using distance formula (O is origin)
OP = sqrt [(1-0)^2 + (sqrt -3 -0)^2] = sqrt [4] = 2 = OQ

also OQ = sqrt [ (s-0)^2 + (t-0)^2] = sqrt[s^2 + t^2]
OQ^2 = s^2+t^2 =4 ----eqn2

now PQ = 2sqrt 2 when we put values of OP and OQ in eqn 1

using distance formula length of PQ = sqrt [ (s+ sqrt3)^2 + (t-1)^2]
PQ^2 = (s+sqrt3)^2 + (t-1)^2 = 8--------eqn3

soving eqn 2 and 3 we get sqrt 3 s = t.Equating this in eqn 2 we get s=1
Senior Manager
Senior Manager
avatar
Joined: 05 Oct 2008
Posts: 273
Followers: 3

Kudos [?]: 42 [0], given: 22

GMAT Tests User
gmat prep line semi circle [#permalink] New post 17 Nov 2009, 09:52
I solved it in the foll manner:

X is point 1 on the y axis

POQ = 90 deg - given
PX = rt 3 - given
XO = 1 - given

Hence PO = 2 (30, 60, 90 triangle)
PO is radius. So PO = OQ = RADIUS = 2

POQ is 90 deg
90 deg angle rule - a, a, rt2 a
PO = 2
QO = 2
Hence PQ = 2 rt2

XQ = 2 rt2 - rt3
approx - 2.8 - 1.7 = 1

Last edited by study on 02 Dec 2009, 10:43, edited 3 times in total.
Senior Manager
Senior Manager
avatar
Joined: 05 Oct 2008
Posts: 273
Followers: 3

Kudos [?]: 42 [0], given: 22

GMAT Tests User
Re: coordinate geometry [#permalink] New post 17 Nov 2009, 22:51
Because the explanation by IndianGuardian seems incorrect:

Lets see how:
First draw a perpendicular from the x-axis to the point P. Lets call the point on the x-axis N. Now we have a right angle triangle \triangle NOP. We are given the co-ordinates of P as (-\sqrt{3}, 1). i.e. ON=\sqrt{3} and PN=1. Also we know angle PNO=90\textdegree. If you notice this is a 30\textdegree-60\textdegree-90\textdegreetriangle with sides 1-sqrt3-2 and angle NOP=30\textdegree.


The coordinates of point P is (-\sqrt{3}, 1) ie ON = 1 and PN = root3 - not the other way round as explained in his posting quoted in blue above.

I get PO = QO = 2 (RADIUS)
POQ is a rt angled Triangle.
Hence, isn't PN = 2 sqrt2 ?
So wouldn't s = 2 sqrt2 - sqrt3

Can someone explain how to solve this. Thanks.
Manager
Manager
avatar
Joined: 23 Apr 2010
Posts: 140
Location: Tx
Schools: NYU,UCLA,BOOTH,STANFORD
Followers: 1

Kudos [?]: 10 [0], given: 36

GMAT Tests User
Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 18 Jun 2010, 03:35
I understood bunuel's solution ok but i have a problem with the other one.

I got every step of maths but how did you get (t-1) too. However How did you decide that "t" is bigger than 1. Because it can be less then one then it should became (1-t). Can anyone explain it plz just that part??
Intern
Intern
User avatar
Joined: 27 Jul 2010
Posts: 20
Location: Bangalore
Followers: 0

Kudos [?]: 2 [0], given: 0

Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 30 Sep 2010, 22:53
DenisSh wrote:
Please help me to solve this problem.

In the figure above, points P and Q lie on the circle with center O. What is the value of s?



This is how I solved it.. This will take less than a minute...

Drop perpendiculars from P and Q. Mark the points as X and Y.

Now, PXO and QYO are right angled triangles.

We know, PO =1 and XO = sqrt(3). So PO = radius = 2.

XOP + POQ + QOY =180
so, QOY = 30,
This gives, t= sqrt (3) and s = 1.
_________________

Nothing is free.. You 've to earn it!!!

Manager
Manager
User avatar
Joined: 15 Apr 2010
Posts: 170
Followers: 4

Kudos [?]: 28 [0], given: 3

GMAT Tests User
Re: gmatprep geo [#permalink] New post 05 Oct 2010, 11:58
OP and OR are prependicular to each other. Hence slope of one must be negative inverse of the other.
Slope of OP = -1/sqrt{3}
Slope of OQ = s/r and it should be equal to sqrt{3}.
Hence s = r * sqrt{3} -----------------(1)

Since OP and OQ represent radius of the circle, their lengths must be equal.
Length of OP = sqrt{(sqrt{3})^2 + 1^2}
Length of OQ = sqrt{r^2 + s^2}

OP = OQ, Also substitue the value of s obtained in (1).
Upon solving, we'll get two values of r +1 and -1. Now since r lies in quad I r has to be +ve. Hence +1.
Current Student
User avatar
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 129
Concentration: Finance, General Management
GMAT 1: 720 Q49 V40
Followers: 7

Kudos [?]: 61 [0], given: 210

Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 04 Jan 2011, 10:14
That's exactly what I needed. I fully understand the problem now. Thanks Bunuel!
Manager
Manager
User avatar
Joined: 19 Aug 2010
Posts: 78
Followers: 2

Kudos [?]: 9 [0], given: 2

Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 13 Feb 2011, 08:40
jullysabat wrote:
BalakumaranP wrote:
DenisSh wrote:
Please help me to solve this problem.

In the figure above, points P and Q lie on the circle with center O. What is the value of s?



This is how I solved it.. This will take less than a minute...

Drop perpendiculars from P and Q. Mark the points as X and Y.

Now, PXO and QYO are right angled triangles.

We know, PO =1 and XO = sqrt(3). So PO = radius = 2.

XOP + POQ + QOY =180
so, QOY = 30,
This gives, t= sqrt (3) and s = 1.



Yeah this is the quickest way I think...
But I think the underlined portion is stated wrongly....


The method with the angles is the quickest one. However, I solved it using the slopes. PO and QO have negative reciprocal slopes. The slope of PO=\frac{-1}{\sqrt{3}}
The slope QO is the native reciprocal to PO:\sqrt{3}
This means that we have to solve the system:
\frac{t}{s}=\sqrt{3}
r^2+t^2=4

Solve for s and we have s=1
Director
Director
avatar
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 931
Followers: 11

Kudos [?]: 191 [0], given: 123

Reviews Badge
Re: GMAT prep question - Geometry [#permalink] New post 15 Apr 2011, 02:55
The equation of the line PO is y = -1/sqrt(3) x

The slope of the perpendicular line QO is sqrt(3). Hence the equation of line QO => y = sqrt(3) x

One way is finding the intersection of the circle and the line QO. We know that radius = QO = 2

The equation of the circle is x^2 + y^2 = 2^2 ----(1)
y = sqrt(3) x ----(2)

Solving 1 and 2 we get x = 1, -1 and y = sqrt(3) and -sqrt(3)

Hence x = 1 since we are looking in the first quadrant. Answer B
SVP
SVP
avatar
Joined: 16 Nov 2010
Posts: 1691
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 30

Kudos [?]: 288 [0], given: 36

GMAT Tests User Premium Member Reviews Badge
Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 15 Apr 2011, 04:27
t/s * -1/root(3) = -1

t/s = root(3)

3 + 1 = t^2 + s^2

=> 4 = 4s^2

=> s = 1 (as s is +ve)

Answer - B
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4688
Location: Pune, India
Followers: 1084

Kudos [?]: 4872 [0], given: 163

Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 17 Apr 2011, 16:38
Expert's post
ravsg wrote:
had it been a DS question, was it okay to assume O as origin ?? :)


It is mentioned in the question that it is a circle with center O. If it is obviously the center, it will be given. If there are doubts and it is not mentioned, you cannot assume it. You cannot assume that the figure will be to scale.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Senior Manager
Senior Manager
User avatar
Joined: 08 Nov 2010
Posts: 422
WE 1: Business Development
Followers: 7

Kudos [?]: 34 [0], given: 161

GMAT ToolKit User GMAT Tests User
Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 19 Apr 2011, 05:04
I solved it with triangles and not lines.
If we can c that the left triangle have two sides (squrt 3 and 1)
we can immediately know that the hypotenuse = 2 (thats the radii as well)
so we also know very easy that the angels are 30,60,90
and if we know the radii and the angels - we can also know all of that very easy on the right triangle as well - just make sure u put the 30 and 60 deg in the right place.
_________________

Get the best GMAT Prep Resources with GMAT Club Premium Membership

VP
VP
avatar
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1365
Followers: 11

Kudos [?]: 136 [0], given: 10

GMAT Tests User
Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 30 Apr 2011, 19:47
for 30 deg triangle sides are in the ratio 3^(1/2) : 1 : 2
for 60 deg triangle sides are in the ratio 1: 3^(1/2) : 2

Hence S = 1.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Manager
Manager
avatar
Joined: 25 Sep 2010
Posts: 85
Schools: HBS, LBS, Wharton, Kelloggs, Booth
Followers: 0

Kudos [?]: 2 [0], given: 15

Re: ps - value of s [#permalink] New post 19 Jul 2011, 21:56
s = 1.
According to the diagram,
we see that OP and OQ are perpendicular.
Let us not assume anything here. The figure might be misleading to some because some may think the points are symmetrical. They are not.

Since, OP is perp. to OQ,
slope of OP * slope of OQ = -1
=> (-1/sqrt3)*(t/s) = -1
=> t/s = sqrt3
=> t = s*sqrt3
Hence, point Q can be written as Q(s,s*sqrt3)

Now, since both points are on the circumference of the circle,
distance to the centre(origin here) will be the same.

Distance from (x,y) to (0,0) is sqrt(x^2 + y^2)
=>dist. from (-sqrt3,1) to (0,0) = sqrt(3+1) = 2 ----(1)
similarly dist. from (s,s*sqrt3) to (0,0) = 2s -----(2)

equating (1) and (2), we get s = 1.
Hope it helps.
Senior Manager
Senior Manager
avatar
Joined: 07 Sep 2010
Posts: 340
Followers: 3

Kudos [?]: 119 [0], given: 136

Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 31 Mar 2012, 21:01
Hi Bunuel,
Request you to provide a detailed reasoning on the below concept mentioned by you.
I am aware of the reflection concepts but it is failing over here. i.e If Point(A,B) is reflected over y axis then the reflected points become (-A,B). I applied this concept and got the answer incorrect. Could you please explain where I am going wrong.
Can't we apply the Reflection concept. If yes, then the angles forming between the lines joining from origin(0,0) to point(A,B) and origin(0,0) to point(-A,B) should be 90.

Please explain.

Thanks
H

Bunuel wrote:
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.


In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant (1, \sqrt{3});
Point P in II quadrant (-\sqrt{3}, 1);
Point T in III quadrant (-1, -\sqrt{3}) (OT perpendicular to OP);
Point R in IV quadrant (\sqrt{3},-1) (OR perpendicular to OQ).

_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Senior Manager
Senior Manager
User avatar
Joined: 23 Oct 2010
Posts: 384
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Followers: 12

Kudos [?]: 131 [0], given: 73

GMAT ToolKit User
Re: In the figure above, points P and Q lie on the circle with [#permalink] New post 31 Mar 2012, 23:58
since these two lines are perpendicular, the result of multiplication of their slopes is (-1)

((1-0)/(-sqrt3-0))*((t-0)/(s-0))=-1

t/s=sqrt3/1

s=1
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Intern
Intern
avatar
Joined: 07 May 2011
Posts: 42
GMAT 1: Q V
GMAT 2: Q V
Followers: 0

Kudos [?]: 10 [0], given: 11

Re: In the figure above, points P and Q lie on the circle with [#permalink] New post 29 May 2012, 17:14
Beautiful solution by Karishma! Turns out a strong imagination is all it takes to solve this problem. Turn the radius joined by origin and (- root3,1) 90 degrees and you will notice the values will be swapped, with their signs dependent on the quadrant into which the shift takes place.
If still not clear, you can think of any point on the x axis, say (5,0). what happens if you turn it 90 degrees upwards? It ends up on the Y-axis at (0,5). Once you see it, you can't unsee it. I am floored! Thank you Karishma! Following you from here on.
Intern
Intern
avatar
Joined: 19 Aug 2012
Posts: 18
Followers: 0

Kudos [?]: 15 [0], given: 10

In the figure above, points P and Q lie on the circle with [#permalink] New post 18 Nov 2012, 07:19
you can break the triangle into 2 individual triangles, find the height and base for the one the left i.e. h=1 and b=-root(3)

therefore b^2+h^2=r^2---->1^2+root(-3)^2=r^2=4

r=2

Use this now to find the Height on the other side, the BASE will remain the same as root(-3)
Expert Post
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Status: Preparing for the another shot...!
Joined: 03 Feb 2011
Posts: 1425
Location: India
Concentration: Finance, Marketing
GPA: 3.75
Followers: 127

Kudos [?]: 599 [0], given: 62

GMAT ToolKit User GMAT Tests User Premium Member
Re: Coordinate System and Geometry [#permalink] New post 19 Nov 2012, 22:30
Expert's post
are you sure about \sqrt{3} as the answer.
As per my calculations i got \sqrt{2} as the answer.
If I am correct, then I will surely post my answer.
_________________

Prepositional Phrases Clarified|Elimination of BEING| Absolute Phrases Clarified
Rules For Posting
www.Univ-Scholarships.com

Expert Post
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Status: Preparing for the another shot...!
Joined: 03 Feb 2011
Posts: 1425
Location: India
Concentration: Finance, Marketing
GPA: 3.75
Followers: 127

Kudos [?]: 599 [0], given: 62

GMAT ToolKit User GMAT Tests User Premium Member
Re: Coordinate System and Geometry [#permalink] New post 19 Nov 2012, 22:47
Expert's post
Step 1: Measure the radius.
That will come out to be 2.
Step 2: Observe the diagram. Since both OP and OQ are the radii, their measure will be 2 only.
Step 3: Draw a line parallel to x axis, from P to Q.
Step 4: Angle OQP and angle OPQ= 45
In the diagram attached, the angles marked X are 45 and 45 degrees.
Step 5: Draw a perpendicular from Q to x axis. The point at which it meets the X axis, mark it as A.
Step 5: The triangle OPA is a 45:45: 90 degree tiangle. This triangle has the property that the hypotenuese:side 1:side 2 will be in the ration \sqrt{2}:1:1.
Step 6: Since side 1(OA) and side 2(AQ) are equal, therefore let them be x. These will be equal to 2/\sqrt{2}.
Step 7: The length of side 1 will come out to be 2/\sqrt{2} or simply \sqrt{2}.

Please let me know what my mistake, if my solution is wrong.
Attachments

solution.png
solution.png [ 13.5 KiB | Viewed 2742 times ]


_________________

Prepositional Phrases Clarified|Elimination of BEING| Absolute Phrases Clarified
Rules For Posting
www.Univ-Scholarships.com

Re: Coordinate System and Geometry   [#permalink] 19 Nov 2012, 22:47
    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic In the figure above, points P and Q lie on the circle with Murmeltier 1 29 Jan 2012, 11:41
In the figure above, points P and Q smarinov 1 24 Dec 2008, 15:45
1 In the figure below, points P and Q lie on the circle with Skewed 15 25 Oct 2007, 10:13
In the figure, point P and Q lie on the circle with center mbunny 11 03 Sep 2007, 08:07
In the attached figure, points P and Q lie on the circle paddyboy 8 15 Jun 2006, 23:55
Display posts from previous: Sort by

In the figure above, points P and Q lie on the circle with

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page   Previous    1   2   3    Next  [ 47 posts ] 



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.