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

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

geometry

Expert replies
Source: — Problem Solving |

by lightbulb » Thu Jul 10, 2008 10:07 pm
Area = (pi) r^2 - (pi) (r - s)^2

= (pi)(r - r + s)(r + r - s) [Since (a^2 - b^2 = (a - b)(a + b))]
= (pi)(s)(2r - s)
Join the discussion

by lordpapi63 » Thu Jul 10, 2008 10:31 pm
Using the backsolving method, I am getting B for an answer.

Am I doing something wrong?

I plugged in 8 for the radius of the medallion and 3 for the distance of the frame...and B is the only answer that matches if you subtract...
Join the discussion

by cornell » Fri Jul 11, 2008 1:55 am
From the question we can find:

Area of Metal Frame = (phi)r^2 - (phi) (r-s)^2

= (phi)r^2 - (phi) (r^2 - 2rs + s^2)
= (phi)r^2 - (phi)r^2 + 2(phi)rs - (phi)s^2
= 2(phi)rs - (phi)s^2
= (phi)s (2r - s)

Hence the answer is D (CMIIW ya)

Hope this helps...Enjoy your day
Life is about choices
Join the discussion

by ildude02 » Fri Jul 11, 2008 6:51 am
lightbulb wrote:Area = (pi) r^2 - (pi) (r - s)^2

= (pi)(r - r + s)(r + r - s) [Since (a^2 - b^2 = (a - b)(a + b))]
= (pi)(s)(2r - s)
What made you realize that "r" is the radius of the complete circle, encompassing the non shaded region and the shaded region? I thought r was radius of the the non shaded circle and s is the width of the frame(width of the shaded frame). So (r+s) was the complete circle radius. Appreciate your response.

btw, the answer is E.
Join the discussion

by dbart06 » Fri Jul 11, 2008 10:10 am
Not quite understanding why you are subtracting s from r. The total radius is r+s. Wouldn't that give the total radius? Then you would subtract non shaded area which is r & then you would get answer. My calculation:

Pi(r+s)^2-pir^2 => radius of whole cir - radius on inner cir

I get = pi s(2r+s)

Any clarification would be helpful
Join the discussion

by asigheartau » Fri Jul 11, 2008 10:53 am
The figure shows the TOP SIDE of a circular medallion made by a CIRCULAR piece of glass surrounded by a metal frame represented by the shaded region. In other words, the top side is the metal frame with width S. The back side= the cercle with radius r.


pi(r-s)^2= Area of unshaded region.
pi(r)^2= Area of the ushaded + shaded region

subtracting these two you get the result

The others did a good job explaining as well so I will not go over the calculations.

I hope it helped
Alin Sigheartau
Join the discussion

by dbart06 » Fri Jul 11, 2008 11:10 am
Not quite understanding why you are subtracting s from r. The total radius is r+s. Wouldn't that give the total radius? Then you would subtract non shaded area which is r & then you would get answer. My calculation:

Pi(r+s)^2-pir^2 => radius of whole cir - radius on inner cir

I get = pi s(2r+s)

Any clarification would be helpful
Join the discussion

by parallel_chase » Fri Jul 11, 2008 12:29 pm
ildude02 wrote:
lightbulb wrote:Area = (pi) r^2 - (pi) (r - s)^2

= (pi)(r - r + s)(r + r - s) [Since (a^2 - b^2 = (a - b)(a + b))]
= (pi)(s)(2r - s)
What made you realize that "r" is the radius of the complete circle, encompassing the non shaded region and the shaded region? I thought r was radius of the the non shaded circle and s is the width of the frame(width of the shaded frame). So (r+s) was the complete circle radius. Appreciate your response.

btw, the answer is E.
There is a lot of confusion about which answer is correct. I got the point that r is the radius of the entire medallion and and s which is actually part of the medallion.

Therefore E is the answer.

Now it would be nice if could post the OA, so there is no confusion.

Anyways I am also getting B as the answer when I plugged in numbers.

Thanks dude
Join the discussion

by dbart06 » Fri Jul 11, 2008 2:49 pm
parrallel_chase

you are correct..statement says the medallion has radius of "r" and the medallion consist of glass and a metal ring (which is "s" in thicknes)

could you show your work so I can get a better understanding.

thanks
Join the discussion