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 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
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
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.

130-point 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
by jainrahul1985 » Sat Jul 25, 2009 9:38 pm
In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area of region
ABCDE ?
A. 76
B. 84
C. 92
D. 100
E. 108
Attachments
attachment.doc
(25 KiB) Downloaded 462 times
Join the discussion
Source: — Problem Solving |

by truplayer256 » Sun Jul 26, 2009 6:49 am
Area of trapezoid ABDE= 1/2*h(base1+base2)= 1/2(8)(10)=40 square units

Area of triangle BCD= 1/2*b*h=1/2*(10)(12)=60 square units

60+40=100 square units D
Join the discussion

by hariharakarthi » Sun Jul 26, 2009 3:08 pm
@truplayer256

Can you explain how did you find the hight of the triangle BCD as 12?

Regards,
hhk.
Join the discussion

by truplayer256 » Sun Jul 26, 2009 4:40 pm
@truplayer256

Can you explain how did you find the hight of the triangle BCD as 12?

Regards,
hhk.
Sure. Drop a line across B and D to form triangle BCD. Now drop a perpendicular line in between B and D. Let's call the middle point between B and D, point M. Line segment MC is the height of the triangle and in order to find the heigh of the triangle, all we have to do is apply the pythagorean theorem. We have a right triangle with sides 5 and 13, so:

(MC)^(2)+ (5)^(2)=(13)^(2)

169-25=144

MC=sqrt(144)=12

We have to use the pythagoren theorem again to find the base of the triangle. If you look closely, you can see that the two legs of the triangle
that has the base of triangle BCD as the hypotenuse are 6 and 8, so:

6^2+8^2= (BD)^(2)

100=(BD)^(2)

BD=10

Area of triangle BCD= 1/2*base*height= 1/2*10*12=60 square units.
Join the discussion

by kaulnikhil » Mon Jul 27, 2009 2:15 am
join b and d ... u get bd as 10
use herons formulae to calculate area of triangle bdc .. which comes to 60
now u have a triangle and a rectangle remaning with areas 16 and 24 respectively
add them u get 60
Join the discussion

by maihuna » Tue Jul 28, 2009 12:04 pm
I also got 100 fig attached
Attachments
attachment (4).doc
(25 KiB) Downloaded 398 times
Charged up again to beat the beast :)
Join the discussion