Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 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
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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
GMATOnDemandStart 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

Geometry

Expert replies
by vish150783 » Tue Dec 02, 2008 10:10 am
Faced this in GMATPrep

What is the greatest possible area of a triangular region with one vortex at teh centre of teh circle or raidus 1 and the other two vertices on the circle?

A. (Sqrt 3)/4
B. 1/2
C. pi/4
D. 1
E. sqrt 2
Join the discussion
Source: — Problem Solving |

by reneerenee » Tue Dec 02, 2008 5:39 pm
i think its C. the triangle that you get when the angle is 90 is half of a 1x1 square and it's area is 1/2.

there is a similar question like this up at the gmat and gre math blog (gmatgremath.blocked) He or she or they or whoever runs that blog has really great geometric pictures to explaing the problems.

(also, i hope you meant vertex instead of vortex! :) )
Join the discussion

by ifairo » Tue Dec 02, 2008 5:57 pm
Answer "1/2" means choice B.
Join the discussion

by anayeri » Tue Dec 02, 2008 6:04 pm
Answer: B.

The maximum area of a triangle, must be a right triangle. So (1*1)/2 = 1/2.

But, why do we necessarily need a right triangle to maximize area? Well, we know for sure that our base is 1 (pitcure a clock, with the base being drawn from 3:00 to the centre of the clock). The right triangle would have the following sides: 3:00-to centre, centre to 12:00, 12:00 to 3:00.

But, if we had a triangle drawn from 10:00 to centre, centre to 3:00, and 3:00 to 10:00, then the base would be the same, but the height would be smaller than 1 (because 10:00 is lower than 12:00, seeing as 12:00 is the apex of the clock). Same if you did it the other way, instead of 10:00, say the other vertex is at 1:00. Seeing as 1:00 is lower than 12:00, we still wouldn't be maximizing the area of our triangle within the clock.

So, the largest triangle inside a circle must go from the centre to two dots that are "3 hours" apart from one another, hence a right triangle.

Hope that helps!
Join the discussion

by reneerenee » Tue Dec 02, 2008 6:05 pm
:D whoops! sorry about that! I totally meant 1/2, and 1/2 = B. you're right. I'm gonna get killed on the gmat if i can't fill in the right letters!
Join the discussion

by vish150783 » Tue Dec 02, 2008 6:51 pm
Hey thanks a lot for your post. the OA is 1/2.

I was getting confused as to what the triangular region meant. When I saw this problem, I could not really picture a triangle. I was looking at a sector.

So, I began with a theta/360 * pi r^2 formula.

I guess wording plays a role in PS as well...yiikeees..

But I guess the general concept I learnt from all your posts is that a given a constant base, the right angled triangle gives the maximum area.

Am I right?
Join the discussion

by anayeri » Tue Dec 02, 2008 7:47 pm
Vish, that's a great way to put it!

Let's hope that comes up on both our exams!
Join the discussion

by anayeri » Tue Dec 02, 2008 8:22 pm
Vish, that's a great way to put it!

Let's hope that comes up on both our exams!
Join the discussion