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.

rectangle in circle.

Expert replies
by goyalsau » Wed Nov 24, 2010 2:35 am
Hi! Guys This is the Last question on Circles that i posted recently,
I hope you enjoyed them Please share your views on this one, Because i don't have any idea from where to start on this one.........
Attachments
Untitled.jpg
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Join the discussion
Source: — Problem Solving |

by Rahul@gurome » Wed Nov 24, 2010 4:21 am
AC and BD will intersect at O.
Now, in ∆ADE and ∆BCD,
  • (1) angle ADE = angle BDC
    (2) angle DAE = angle BCD (= 90°)
    (3) AD = BD (opposite sides of rectangle)
Therefore ∆ADE and ∆BCD are similar.
Hence, AE/AD = BC/CD

Say, radius of the circle = r and sides of the rectangle, AB = a and AD = b (a > b)
Then, in ∆ABD, (AB)² + (AD)² = (BD)²
=> a² + b² = 4r² ............................................. (i)

Now, area of the circle = πr²
and area of the rectangle = ab

Thus, πr²/ab = π/√3
=> ab = √3r² .................................................. (ii)

Form (i) and (ii), a = √3r and b = r

Thus, AE/AD = BC/CD = b/a = 1/√3

The correct answer is A.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Join the discussion

by [email protected] » Thu Nov 10, 2011 12:58 am
Can you explain how you derived "From (i) and (ii), a = √3r and b = r "?
Join the discussion

by GMATGuruNY » Thu Nov 10, 2011 8:29 am
Image
We can plug in the answers, which represent AE:AD.
Anytime √3 appears in the problem or in the answer choices, LOOK FOR A 30-60-90 triangle.
The most likely answer choice is A, which says that AE:AD = 1:√3, implying that ∆ADE is a 30-60-90 triangle whose sides are proportioned 1:√3:2.

Image

When a rectangle is inscribed in a circle, DRAW THE DIAGONAL, since the diagonal of the rectangle = the diameter of the circle.
Thus, in the figure above, BD is both the diagonal of rectangle ABCD and the diameter of the circle.
If ∆ADE is a 30-60-90 triangle, then so is ∆BCD, since it is given that angle ADE = angle BDC.
Thus, the sides of ∆BCD are proportioned 1:√3:2.
Since the shortest side is √3, and 1:√3:2 = √3 : 3 : 2√3, we get:
BC = √3, CD = 3, and BD = 2√3.

Area of the circle:
Since BD = 2√3, r=√3.
Area = πr² = π(√3)² = 3π.

Area of rectangle ABCD:
CD*AD = 3√3.

Thus:
Circle:triangle = 3π : 3√3 = π:√3.
Success!

The correct answer is A.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion