In the circle with center O above, chord QS is perpendicular : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 16 Jan 2017, 07:27

### 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

# In the circle with center O above, chord QS is perpendicular

Author Message
TAGS:

### Hide Tags

Intern
Joined: 25 Mar 2013
Posts: 3
Location: Italy
WE: General Management (Other)
Followers: 0

Kudos [?]: 16 [0], given: 6

In the circle with center O above, chord QS is perpendicular [#permalink]

### Show Tags

27 Mar 2013, 03:05
1
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

70% (03:38) correct 30% (03:33) wrong based on 74 sessions

### HideShow timer Statistics

In the circle with center O above, chord QS is perpendicular to radius OR. If QS = 16 and PR = 4, what is the area of the circle ?

(A) 80 π
(B) 100 π
(C) 125 π
(D) 144 π
(E) 256 π

Please see the above mentioned circle in the attachment . The OA should be (B), this problem is part of a set of GMAT PS exercises given to me for training by a friend .
[Reveal] Spoiler: OA

Attachments

gmat q 22.jpg [ 96.4 KiB | Viewed 1958 times ]

_________________

Labor omnia vincit

Intern
Joined: 22 Jan 2010
Posts: 25
Location: India
Concentration: Finance, Technology
GPA: 3.5
WE: Programming (Telecommunications)
Followers: 0

Kudos [?]: 25 [2] , given: 3

Re: In the circle with centre O above, chord QS is perpendicular [#permalink]

### Show Tags

27 Mar 2013, 03:37
2
KUDOS
In the given circle,join OS ans OQ.
Let OR = OS = OQ = r. So, OP = r-4.

OR will bisect QS,Hence, QP = PS = 8.
Now solve for r in right angle triangle OPS.

OS ^2 = OP ^2 + PS ^ 2
=> r^2 = (r-4)^2 + 8^2
=> r = 10

So area = pi * 10 ^ 2 = 100 pi.
Hence option B.
_________________

Please press +1 KUDOS if you like my post.

Intern
Joined: 28 Oct 2015
Posts: 11
Location: United States
Concentration: Entrepreneurship, Technology
GMAT 1: 710 Q47 V41
GPA: 3.16
WE: Research (Computer Software)
Followers: 0

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

Re: In the circle with center O above, chord QS is perpendicular [#permalink]

### Show Tags

07 Mar 2016, 10:34
How do you solve the quadratic efficiently?
Manager
Joined: 01 Mar 2014
Posts: 140
Schools: Tepper '18
Followers: 2

Kudos [?]: 8 [0], given: 616

Re: In the circle with center O above, chord QS is perpendicular [#permalink]

### Show Tags

26 Mar 2016, 22:20
subhendu009 wrote:
In the given circle,join OS ans OQ.
Let OR = OS = OQ = r. So, OP = r-4.

OR will bisect QS,Hence, QP = PS = 8.
Now solve for r in right angle triangle OPS.

OS ^2 = OP ^2 + PS ^ 2
=> r^2 = (r-4)^2 + 8^2
=> r = 10

So area = pi * 10 ^ 2 = 100 pi.
Hence option B.

Got the same answer. I took OR to be the bisector of QS, but i cannot remember the theorem for the same. Can you please suggest??
Intern
Joined: 16 Mar 2016
Posts: 4
Followers: 0

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

Re: In the circle with center O above, chord QS is perpendicular [#permalink]

### Show Tags

27 Mar 2016, 06:14
How do you solve the quadratic efficiently?

The r^2 cancel out, so no need to solve the quadratic. you will just get r=10
Senior Manager
Status: Exam scheduled!!
Joined: 05 Sep 2016
Posts: 411
Location: United States (WI)
Concentration: Marketing, Technology
WE: Other (Law)
Followers: 3

Kudos [?]: 13 [0], given: 234

Re: In the circle with center O above, chord QS is perpendicular [#permalink]

### Show Tags

21 Oct 2016, 19:22
The following can be used to solve this problem:

8^2 + (x-4)^2 = x^2

Manipulate --> x = 10

Area = Pi(r^2) = (10)^2 pi = 100 pi
Re: In the circle with center O above, chord QS is perpendicular   [#permalink] 21 Oct 2016, 19:22
Similar topics Replies Last post
Similar
Topics:
4 In the diagram above, S is the center of the circle. If QS = 5 and QR 9 02 Mar 2015, 07:33
7 If O is the center of the circle above, what fraction of the 11 11 Sep 2012, 03:33
150 In the figure above, point O is the center of the circle and 39 16 Jan 2011, 15:56
35 In the figure above, the radius of circle with center O is 1 20 27 Aug 2009, 11:03
31 In the figure above, the radius of the circle with center O 17 11 Sep 2008, 04:49
Display posts from previous: Sort by