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

 It is currently 29 Mar 2015, 21:53

### 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

# PS # half circle

Author Message
TAGS:
SVP
Joined: 28 May 2005
Posts: 1743
Location: Dhaka
Followers: 6

Kudos [?]: 67 [0], given: 0

PS # half circle [#permalink]  02 Nov 2005, 13:26
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
Attachments

half circle.jpg [ 9.14 KiB | Viewed 800 times ]

_________________

hey ya......

Manager
Joined: 20 Mar 2005
Posts: 201
Location: Colombia, South America
Followers: 1

Kudos [?]: 6 [0], given: 0

it looks like it is sqrt(3) as the angle POQ is 90 degrees and therefore it is a reflection.

another way to think about it
what is the result of rotating a radius 90 degrees along the circumference.
in that case what is the effect in the cosine?

make sense?
Current Student
Joined: 28 Dec 2004
Posts: 3392
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

can you repost the question..not sure what we are asking for s, t?
Manager
Joined: 04 Oct 2005
Posts: 246
Followers: 1

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

I get 2sqrt2 - sqrt3... which is neither of the choices...

We know that the radii must be the same for both. We have two triangles...
one with legs of length 1 and sqrt3.... a similar triangle, thus the 3rd length, or the radius, must be 2.
The second triangle with the 90┬░ angle is then a 45-45-90. So the line connecting points P and Q should have a length of sqrt2.

Deduct the length of the x coordinate from the first point, gives you the x coordinate of the second.

:EDIT:
just realized, that this only works when the line PQ is parallel to the x axis... which must not be the case.
Intern
Joined: 22 Mar 2005
Posts: 39
Location: Houston
Followers: 0

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

Wow, this is a great one. Don't let the diagram fool you here.

First, as was previously mentioned you can see that the radius is 2.

Now calculate the distance between the two points by drawing a line between them. Obviously the distance between them is equal to the hypotenuse of the triangle with the two radii as its legs.

So distance between them = sqrt(2^2+2^2) = sqrt(8)

Now the distance formula between to points says that:
DF = (s+sqrt(3))^2+(t-1)^2 had better equal sqrt(8)^2 = 8 as well.

We need one more equation that relates s and t and we are ready to go.
That equation of course is that s^2+t^2 = 2^2 (eqn of the circle)

Now solve for t^2 = 4 - s^2 and t = sqrt(4-s^2)

Now go back to the distance formula above (DF), multipy it out and then plug in the expressions above for t^2 and t. Now solve for s and you will see that s = 1 and t = sqrt(3).

So the drawing is playing a little trick as the y axis isn't really bisecting the 90 deg angle. As a quick check if it were the other way around and s = sqrt(3) and y = 1, then the distance between the points would simply be sqrt(3) + sqrt(3) = 2sqrt(3) which does not equal sqrt(8) (or 2sqrt(2)).
SVP
Joined: 24 Sep 2005
Posts: 1895
Followers: 11

Kudos [?]: 133 [0], given: 0

first of all, we should find the function of the two lines OP and OQ
from the coordinates provided we can easily find:
OP: y=-1/sqrt3 *x
OQ: y= t/s*x
Becoz OP and OQ are perpendicular --> t/s* ( -1/sqrt3)= -1
---> t/s= sqrt3--> t= s*sqrt3-->t^2= 3*s^2
We have:
--> s^2+3*s^2= 4 ---> s=- or + 1
becoz Q belongs to the II portion of the coordination --> s=1

Last edited by laxieqv on 02 Nov 2005, 21:30, edited 1 time in total.
Current Student
Joined: 29 Jan 2005
Posts: 5244
Followers: 23

Kudos [?]: 175 [0], given: 0

S = sqrt3 by "eyeballing" and under 10 seconds

OA?
SVP
Joined: 24 Sep 2005
Posts: 1895
Followers: 11

Kudos [?]: 133 [0], given: 0

first of all, we should find the function of the two lines OP and OQ
from the coordinates provided we can easily find:
OP: y=-1/sqrt3 *x
OQ: y= t/s*x
Becoz OP and OQ are perpendicular --> t/s* ( -1/sqrt3)= -1
---> t/s= sqrt3--> t= s*sqrt3-->t^2= 3*s^2
We have:
--> s^2+3*s^2= 4 ---> s=- or + 1
becoz Q belongs to the II portion of the coordination --> s=1
SVP
Joined: 28 May 2005
Posts: 1743
Location: Dhaka
Followers: 6

Kudos [?]: 67 [0], given: 0

good job guys... OA is 1.

how long did it take you guys to do it... for me it took over 4 minutes... is it okay?
_________________

hey ya......

Similar topics Replies Last post
Similar
Topics:
1 A circle is inscribed in a half circle with a diameter of 4 23 Feb 2012, 00:53
inscribed circle in half circle 4 30 Aug 2010, 07:21
GMATPrep Question: Half Circle 1 10 Nov 2007, 17:21
PS- circle 2 28 Sep 2006, 13:10
ps:circle 2 26 May 2006, 08:38
Display posts from previous: Sort by