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 jkelk » Wed Mar 14, 2012 10:50 am
Hi

Please help solve the attached.

Thanks,
J
[Moderator Edit: Moved the post to a relevant forum - neelgandham]
You do not have the required permissions to view the files attached to this post.
Join the discussion
Source: — Problem Solving |

by zueswoods » Thu Mar 15, 2012 12:20 pm
Remember this:

"To maximize the area of a triangle given a base, make it a 90 degree isoceles, 90-45-45" -MGMAT book

Therefore with radius 1 (base =1), the height equals 1 and thus the area equalls 1/2.
Join the discussion

by karunamay » Wed Apr 11, 2012 7:31 am
HI

In this problem it is given that one vertex of the triangle is at the center of a unit circle.
From this we conclude that if you join this vertex to other two vertex somewhere at the circle, then length of these two sides of triangle will be equal to radius of circle.
Let these sides be a, b , therefore a = b =radius of circle = 1
Let angle between these two sides is C.

Then area of the triangle =1/2*a*b* sin C

=1/2*1*1* sin C = 1/2* sin C
here we see area of the triangle solely depends on angle C.
Therefor, area of the triangle will be maximum where sin C is maximum.

Maximu value that sin function takes is 1, cosequently (sin C)max =1.

Hence, required area of the triangle = 1/2*1 =1/2.

Therefore, option B is correct.
Join the discussion