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

It is currently 26 Jun 2016, 21:31
GMAT Club Tests

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

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
VP
VP
User avatar
Joined: 18 May 2008
Posts: 1287
Followers: 14

Kudos [?]: 303 [0], given: 0

In the figure above, points P [#permalink]

Show Tags

New post 13 Nov 2008, 03:35
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Help
Attachments

semicircle.JPG
semicircle.JPG [ 49.33 KiB | Viewed 1083 times ]

Manager
Manager
avatar
Joined: 08 Aug 2008
Posts: 234
Followers: 1

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

Re: GMATPREP- semicircle [#permalink]

Show Tags

New post 13 Nov 2008, 08:01
good one really...
at 1st sight, s looks as \(sqrt3\)

radius=2
now, distance between P and Q=2\(sqrt2\)

now (s+\(sqrt3\))^2 + (t-1)^2 =8
to solve this equation i need \(sqrt3\) value for T, hence t=\(sqrt3\)

Hence s=1.
(if u look at the equation closely, it'll be clear....)
1 KUDOS received
Manager
Manager
avatar
Joined: 23 Jul 2008
Posts: 203
Followers: 1

Kudos [?]: 76 [1] , given: 0

Re: GMATPREP- semicircle [#permalink]

Show Tags

New post 13 Nov 2008, 12:06
1
This post received
KUDOS
prasun84 wrote:
good one really...
at 1st sight, s looks as \(sqrt3\)

radius=2
now, distance between P and Q=2\(sqrt2\)

now (s+\(sqrt3\))^2 + (t-1)^2 =8
to solve this equation i need \(sqrt3\) value for T, hence t=\(sqrt3\)

Hence s=1.
(if u look at the equation closely, it'll be clear....)


it s an interesting method we will have to rely on hit and trial to get the answer
My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

hence s is 1
Director
Director
avatar
Joined: 14 Aug 2007
Posts: 733
Followers: 9

Kudos [?]: 159 [0], given: 0

Re: GMATPREP- semicircle [#permalink]

Show Tags

New post 13 Nov 2008, 20:00
calculate radius of the circle
radius^2= (sqrt(3))^2 + 1^2
radius = 2 (cant be -ve)

the equation of line passing through point P is,

y = -1/sqrt(3)x [ slope = (1-0)/(-sqrt(3) - 0) and y intercept of the line is 0)

The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s

Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1

S= 1 (being in Ist quadrant)
VP
VP
User avatar
Joined: 18 May 2008
Posts: 1287
Followers: 14

Kudos [?]: 303 [0], given: 0

Re: GMATPREP- semicircle [#permalink]

Show Tags

New post 13 Nov 2008, 21:42
Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)
alpha_plus_gamma wrote:
calculate radius of the circle
radius^2= (sqrt(3))^2 + 1^2
radius = 2 (cant be -ve)

the equation of line passing through point P is,

y = -sqrt(3)x [ slope = (-sqrt(3) - 0) / (1-0) and y intercept of the line is 0)
The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s

Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1

S= 1 (being in Ist quadrant)
VP
VP
User avatar
Joined: 18 May 2008
Posts: 1287
Followers: 14

Kudos [?]: 303 [0], given: 0

Re: GMATPREP- semicircle [#permalink]

Show Tags

New post 13 Nov 2008, 21:48
Wow ! wht a shortcut. Applying angle formula never occured to me.
+1 from me
hibloom wrote:
prasun84 wrote:
good one really...
at 1st sight, s looks as \(sqrt3\)

radius=2
now, distance between P and Q=2\(sqrt2\)

now (s+\(sqrt3\))^2 + (t-1)^2 =8
to solve this equation i need \(sqrt3\) value for T, hence t=\(sqrt3\)

Hence s=1.
(if u look at the equation closely, it'll be clear....)


it s an interesting method we will have to rely on hit and trial to get the answer
My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

hence s is 1
Director
Director
avatar
Joined: 14 Aug 2007
Posts: 733
Followers: 9

Kudos [?]: 159 [0], given: 0

Re: GMATPREP- semicircle [#permalink]

Show Tags

New post 13 Nov 2008, 21:55
ritula wrote:
Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)


Yes you are correct. I have edited my post. thanks
Manager
Manager
User avatar
Joined: 21 Oct 2008
Posts: 131
Schools: Rady School of Management at UC San Diego GO TRITONS
Followers: 3

Kudos [?]: 14 [0], given: 0

Re: GMATPREP- semicircle [#permalink]

Show Tags

New post 14 Nov 2008, 12:28
There's a much faster way to do this problem. To find the coordinates of q, all you have to do is switch the x- and y- values of the coordinates of P, then multiply the new y-value (t) by -1.

To understand why, imagine the full circle. Draw any two perpendicular lines through the center. They will intersect the circle at (a, b), (b, -a), (-a, -b), and (-b, a).
Director
Director
User avatar
Joined: 27 Jun 2008
Posts: 546
WE 1: Investment Banking - 6yrs
Followers: 2

Kudos [?]: 55 [0], given: 92

Re: GMATPREP- semicircle [#permalink]

Show Tags

New post 14 Nov 2008, 12:45
Please check

7-t71146
Re: GMATPREP- semicircle   [#permalink] 14 Nov 2008, 12:45
Display posts from previous: Sort by

In the figure above, points P

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| 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®.