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 Problem HELPPPPPPPPPP urgentttt NEWWWWWW

Expert replies
by arifaisal » Fri Nov 11, 2011 11:11 pm
Q1-In ABC, angle AFB=angle AEC=90. If BF=CE, which of the following must be true?

A)AB=AC
B)AE=CE
C)BF=AF
D)AB=BC
E)none

Q2-In the figure, AD||BC. AB=5 cm and AD=7 cm. If angle BCD=1/2 angle BAD, what is the length of BC in cm?
A)10
B)12
C)17
D)19
E) cannot be determined

Q3-In the figure below, D is the midpoint of AB and E is the midpoint of AC. What is the relation between the area of BFC and the area of ADFE?

A)Area BFC> area ADFE
B)Area BFC= area ADFE
C)Area BFC< area ADFE
D)area ADFE= 1.5 times Area BFC
E) either A or C

please give full explanation and check the figure for reference
Attachments
p1.JPG
for Q1
p2.JPG
for q2
p3.JPG
for q3
Join the discussion
Source: — Problem Solving |

by shankar.ashwin » Sat Nov 12, 2011 12:23 am
Pls stick to one question per post.

Question 2:


The given figure will be a trapezoid, let the angles be A,B,C and

A + B = 180 and C + D = 180

A + B = 180 and 2A + D = 180. Given AB =5 and AD = 7, we cannot determine the finite angles for the figure IMO. BC would vary depending on angles C and A. Cannot be determined IMO. Not very sure though.
Join the discussion

by Neo Anderson » Sat Nov 12, 2011 3:09 am
Question 1

Area of the trianle here can be calculated as:
1/2 * base* height = 1/2 *AB*CE = 1/2 *AC*BF
with given condition (BF=CE)
=> AB=AC
thus A
Image
Join the discussion

by Neo Anderson » Sat Nov 12, 2011 3:36 am
Question 2
Image
extend the lines BA and CD to meet at o as shown above
now angle BCD = ADO = Q
as angle BAD = 2Q => angle DAO = 180 - 2Q => angle AOD = Q (sum of angles in trianle = 180
Thus now from two isoceles triangles AOD you have y=7 and from asoceles triangle you have x= 5+7=12
hence B
Join the discussion

by Neo Anderson » Sat Nov 12, 2011 4:36 am
Question 3: IMO E; but for certain set of values (for length of the sides) the 2 area's can be equal also; :?:
Join the discussion

by arifaisal » Sat Nov 12, 2011 5:03 am
Hey neo thanks alot...really helped me a lot...but q3 is not totally clear to me
Join the discussion

by mankey » Mon Nov 14, 2011 9:44 am
For Ques3: IMO: B. Both areas will be equal. There is a property of triangles that three medians will divide triangle in 4 triangles of equal areas.

Thanks.
Join the discussion

by saketk » Mon Nov 14, 2011 11:12 am
Image


Answering Question 3 here --

This is the property of Median of a Triangle The three medians divide the triangle into 6 smaller triangles that all have the same area, even though they may have different shapes.


Please refer to the image I've attached.

The 3rd Median is dividing ADEF into 2 equal parts. i.e ADF and AEF. Also, it divided BFC into 2 equal parts.

That means ADEF combines will be EQUAL to BFC (two parts combined).

Answer should be Option B
Join the discussion