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
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
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

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

Square EFGH

Expert replies
Source: — Data Sufficiency |

by mals24 » Sun Dec 21, 2008 10:42 am
IMO B

If we connect points E and G, line EG = AD.
B says ABCD is a square so AD = EG = 4. We can find the side of a square with the diagonal and hence the area.
Join the discussion

by Brent@GMATPrepNow » Sun Dec 21, 2008 12:05 pm
My sincerest apologies, I forgot to add the 90 degree marks
Without them, this question is CRAZY hard.
I need to pay more attention.


Image
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by Brent@GMATPrepNow » Sun Dec 21, 2008 12:56 pm
Answer: A
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by msd_2008 » Sun Dec 21, 2008 5:20 pm
I dont get this.....how can the answer be A?
Statement I tells us that Angle BEF is 60 degrees. While this is an important piece of information to derive a 30-6-90 triangle for BEF and similarly for other 3....we dont know what is the length of BF and neither BE. So how can we deduce the measure of EF?

Regards
MSD
When the going gets tough, the tough gets going.
Join the discussion

by Brent@GMATPrepNow » Sun Dec 21, 2008 6:41 pm
That's a good question, MSD.
We don't actually need to find the measures of BF and BE etc. We need only show that we could (i.e., we have sufficient data to do so) find the measures.

First, we'll show that ABCD is a square [without using (1)]

Since EFGH is a square, we'll let each side of the square be of length x
You have already deduced that all of the triangles are 30-60-90 triangles, and now we know that they all have a hypotenuse of lenght x.

This means that each of the 30-60-90 triangles has lengths x/2, sqrt(3)x/2, and x

This will show that ABCD has sides of equal length so it is a square.
Now, what is the length of each side of square ABCD?
It's x/2 + sqrt(3)x/2
But we already know the length of each side of ABCD. Since the area is 16, we know that each side is length 4.

This gives us the equation x/2 + sqrt(3)x/2 = 4

This is an equation (a linear equation) with one unknown, which we could solve if we were so inclined. Solving for x would give us the dimensions of the inner square (EFGH) so we could determine its area.

So, (1) is enough info
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by msd_2008 » Sun Dec 21, 2008 7:22 pm
Thank you very much! I got it...
When the going gets tough, the tough gets going.
Join the discussion

by ronniecoleman » Sun Dec 21, 2008 10:32 pm
IMO A
If angle angle bef is know that we can know each and every angle..

find angle bef and cfg ... use trigo to find the lenght of outer square... in terms of inner lenght of side of inner square...
solve you are done..
Admission champion, Hauz khaz
011-27565856
Join the discussion