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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Area of a circular region

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members
Thanked: 82 times
GMAT Score:720
by maihuna » Fri Dec 25, 2009 10:21 am
Is the area of a circular region less than the area of a semi circular region?

1. The radius of the circle is less than the radius of the semi circle

2. The circumference of the circle is less than the perimeter of the semi circular region.
Charged up again to beat the beast :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 124
Joined: Tue Dec 22, 2009 8:32 am
Location: Karjat
Thanked: 3 times

by linkinpark » Sat Dec 26, 2009 1:18 am
I would go with E,
stmt 1 and 2 essentially give same info i.e. radius of circle is less than that of semi-circle

now consider r1=4 for full circle and r2=8 for semi-circle
a1=area of circle = 16pi and a2=area of semi-circle = 1/2*pi*8^2 = 32pi, in this case a1<a2 but what if r1=4 and r2=5 in that case 16pi > 1/2*pi*25 i.e. 16pi > 12.5pi
so we can't conclude
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members
Thanked: 82 times
GMAT Score:720

by maihuna » Sat Dec 26, 2009 7:12 am
linkinpark wrote:I would go with E,
stmt 1 and 2 essentially give same info i.e. radius of circle is less than that of semi-circle

now consider r1=4 for full circle and r2=8 for semi-circle
a1=area of circle = 16pi and a2=area of semi-circle = 1/2*pi*8^2 = 32pi, in this case a1<a2 but what if r1=4 and r2=5 in that case 16pi > 1/2*pi*25 i.e. 16pi > 12.5pi
so we can't conclude
Good explanation on part 1, but remember in part 2, one is circumference another is perimeter, for a semicircle perimeter means PI*r+2r if radius of wholecircle is R then given : PI*R^2 = PI*r+2r how can you conclude R>r?
Charged up again to beat the beast :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 124
Joined: Tue Dec 22, 2009 8:32 am
Location: Karjat
Thanked: 3 times

by linkinpark » Sat Dec 26, 2009 7:31 am
maihuna wrote:
linkinpark wrote:I would go with E,
stmt 1 and 2 essentially give same info i.e. radius of circle is less than that of semi-circle

now consider r1=4 for full circle and r2=8 for semi-circle
a1=area of circle = 16pi and a2=area of semi-circle = 1/2*pi*8^2 = 32pi, in this case a1<a2 but what if r1=4 and r2=5 in that case 16pi > 1/2*pi*25 i.e. 16pi > 12.5pi
so we can't conclude
Good explanation on part 1, but remember in part 2, one is circumference another is perimeter, for a semicircle perimeter means PI*r+2r if radius of wholecircle is R then given : PI*R^2 = PI*r+2r how can you conclude R>r?
in stmt2 we're told
pi * R + 2R > 2*pi*r i.e. perimeter > circumference
R*(pi+2) > 2*pi*r isn't that sufficient to indicate R > r because 2*pi is definitely > pi + 2
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members
Thanked: 82 times
GMAT Score:720

by maihuna » Sat Dec 26, 2009 9:10 am
linkinpark wrote:
maihuna wrote:
linkinpark wrote:I would go with E,
stmt 1 and 2 essentially give same info i.e. radius of circle is less than that of semi-circle

now consider r1=4 for full circle and r2=8 for semi-circle
a1=area of circle = 16pi and a2=area of semi-circle = 1/2*pi*8^2 = 32pi, in this case a1<a2 but what if r1=4 and r2=5 in that case 16pi > 1/2*pi*25 i.e. 16pi > 12.5pi
so we can't conclude
Good explanation on part 1, but remember in part 2, one is circumference another is perimeter, for a semicircle perimeter means PI*r+2r if radius of wholecircle is R then given : PI*R^2 = PI*r+2r how can you conclude R>r?
in stmt2 we're told
pi * R + 2R > 2*pi*r i.e. perimeter > circumference
R*(pi+2) > 2*pi*r isn't that sufficient to indicate R > r because 2*pi is definitely > pi + 2
R > r*(2*PI)/(PI+2)
R > 1.1r

humm yeas since R > 1.1 r so R>r, right
Charged up again to beat the beast :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1