BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Interesting GMATFix Problem-42

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Community Manager
Posts: 1048
Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members
Location: India
Thanked: 51 times
GMAT Score:670
by arora007 » Mon Oct 04, 2010 4:19 am
The circle with center O bisects the side AB and points A and C lie on the circle such that AC is the diameter. IF BC=4, what is the maximum possible area of triangle ABC?
A)4
B)4root3
C)8
D)8root3
E)12
https://www.skiponemeal.org/
https://twitter.com/skiponemeal
Few things are impossible to diligence & skill.Great works are performed not by strength,but by perseverance

pm me if you find junk/spam/abusive language, Lets keep our community clean!!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Senior | Next Rank: 100 Posts
Posts: 82
Joined: Sat Aug 21, 2010 8:18 am
Location: India
Thanked: 5 times

by sumit.sinha » Mon Oct 04, 2010 4:32 am
arora007 wrote:The circle with center O bisects the side AB and points A and C lie on the circle such that AC is the diameter. IF BC=4, what is the maximum possible area of triangle ABC?
A)4
B)4root3
C)8
D)8root3
E)12
IMO C
Cheers,
Sumit
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1119
Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members
Thanked: 29 times

by diebeatsthegmat » Mon Oct 04, 2010 4:55 pm
arora007 wrote:The circle with center O bisects the side AB and points A and C lie on the circle such that AC is the diameter. IF BC=4, what is the maximum possible area of triangle ABC?
A)4
B)4root3
C)8
D)8root3
E)12
i just guess the answer,,,,
is the answer C???
i dont understand the world bisect much but i consider that O is midpoint of diametter AC and OB is perpendicular AC and AB=AB then we will get the maximum =1/2 AB*BC
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members
Location: Milpitas, CA
Thanked: 447 times

by Rahul@gurome » Mon Oct 04, 2010 7:30 pm
Solution:
O is the centre of circle and AC is the diameter.
Or O lies on AC as its midpoint.
This circle bisects AB.
Let it bisect AB at P.
So P is the midpoint of AB.
Join P to O.
Consider the triangle ABC.
Using the midpoint theorem which states that the straight line joining the mid-points of two sides of a triangle is parallel to and equal to half the third side, we get that PO is half of BC.
Since BC is 4, PO is 2.
PO is also the radius of the circle.
Also AC is the diameter.
So AC is 2*2 = 4.
So area of triangle ABC is ½ * BC * AC * sin (angle ACB) = ½ * 4 * 4 * sin (angle ACB).
Now the maximum value of sin (angle ACB) is 1.
So the maximum value of the area of triangle ABC is ½ * 4 * 4 = 8.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1