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 question.....Need immediate help....

Expert replies
by The Jock » Fri Oct 22, 2010 1:26 am
Question: In triangle ABC, AD is the bisector of |A, AB=10 cm, AC=14 cm and area of triangle ABD = 140 sq cm. Find area of triangle ACD.

The solution in the document is as follows:
Any angle bisector of any angle between 2 sides of a triangle divides the Area of the triangle into the ratio of sides .
Area of any triangle is 1/2 *(Product of any 2 sides of the triangle) * (Sin of Angle between those 2 sides)

Now coming to the question at concern.
Here area of ABD => 140 = 1/2*(AB * AD) *(Sin of angle BAD) ---eqn (1)
Area of ACD = 1/2*(AC*AD) * (Sin of angle DAC) ---eqn(2)
angle DAC = angle BAD ---eqn(3) as angle A is bisected

Using eqn 1 and 2 and 3, gives 196 as area of ACD.

But I am not able to get it fully and want another explanation.
Thanks in advance.

Source: Beat the GMAT math questions collectionby Papgust.
Thanks and Regards,
Varun
https://mbayogi.wordpress.com/
Join the discussion
Source: — Problem Solving |

by Rahul@gurome » Fri Oct 22, 2010 2:06 am
Area of triangle = ½ * product of any 2 sides * sine of angle between these two sides.
Let angle BAD be x.
So angle DAC is also x.
So area of triangle ABD is ½ * AB * AD * sin x = 140.
Or ½ * 10 * AD * sin x = 140.
So AD*sin x = 28.
Now area of triangle ACD is ½ * AC * AD * sin x = ½ * 14 * AD * sin x = 7 * 28 = 196.
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 N:Dure » Sat Oct 23, 2010 7:02 am
lol!
Join the discussion

by goyalsau » Sat Oct 23, 2010 9:01 am
Rahul@gurome wrote:
Area of triangle = ½ * product of any 2 sides * sine of angle between these two sides.

.
Awesome, Awesome....................
It always been learning at BTG,
I never knew about this formula
Area of triangle =
1/2 * product of any two sides * sine of angle between them

But i was thinking it should have been product of two adjacent sides and that must share a common vertices as per this question .

Rahul please correct me if am wrong.
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Join the discussion

by Rahul@gurome » Sat Oct 23, 2010 11:06 am
goyalsau wrote: ...But i was thinking it should have been product of two adjacent sides and that must share a common vertices as per this question .

Rahul please correct me if am wrong.
Well, any two sides of a triangle always share a common vertex. Isn't it?
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 Ian Stewart » Sat Oct 23, 2010 12:37 pm
You will never need to use sines or cosines on a real GMAT question. Just ignore this one and move on to more relevant material.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Join the discussion