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 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

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
Limited cohort · enrollment openTarget 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.

Geometry question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 26
Joined: Tue May 08, 2007 8:22 pm
Thanked: 1 times

by mendiratta » Tue May 08, 2007 8:55 pm
whwre's the problem figure?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 26
Joined: Tue May 08, 2007 8:22 pm
Thanked: 1 times

by mendiratta » Tue May 08, 2007 9:00 pm
I think it is sqrt(3)/2.
Angle subtended by diameter is always 90 degree.
So it is right angled triangle with 2 sides as 1 & 2(twice the radius).
There the third side is sqrt(3). ( 1,2,sqrt(3) ) kind of triangle.
therefore
area = (base*height)/2
area = ( sqrt(3) * 1 )/2
area = sqrt(3)/2.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 559
Joined: Tue Mar 27, 2007 1:29 amFollowed by: 2 members
Thanked: 5 times

by Cybermusings » Sat May 12, 2007 2:41 am
Remember the theorem for your exam...An angle in a semicircle is a right-angle...

Here angle ABC is a right angle...

AC = Diameter = 2
BC = 1
So AB = sqr rt. 3
Now Area = 1/2 * sqr rt 3 * 1 = sqr rt. 3/ 2

Hence Choice C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1