Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 0408:30 AM PDT
-09:30 AM PDT
For most test takers, Data Insights is the most challenging section on the GMAT, with test takers scoring several points lower on average on DI than on Quant or Verbal and completing the section with less time to spare.
Jun 0809:00 AM EDT
-11:00 AM EDT
A powerful GMAT course taught live online + 6 months of access to TTP OnDemand GMAT Masterclass included! date of event Class begins June 8, 2026- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program.. 
Jun 1111:00 AM EDT
-01:00 PM EDT
6 months of access to over 400 hours of video lessons date of event No date.
21 Jun 2019, 00:14
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
68% (02:08) correct
32%
(02:03)
wrong
based on 104
sessions
History
Date
Time
Result
Not Attempted Yet
What is the area of the triangle ABC?
(A) 15
(B) \(15\sqrt{3}\)
(C) \(\frac{5\sqrt{119}}{2}\)
(D) 32.5
(E) 36
Attachment:
Capture1 (1).JPG [ 7.33 KiB | Viewed 3470 times ]
21 Jun 2019, 00:54
Kudos
Bookmarks
Bunuel
The altitude to the base from A will form a 30-60-90 triangle with Hypo as 12. So Other altitude using 1-root3-2 will be 6root3 (as opposite 60degrees) so area is 15root3 IMO B
21 Jun 2019, 02:18
Kudos
Bookmarks
Bunuel
Draw a perpendicular from point C to meet side AB at point D
CDB is a 30-60-90 triangle with
∠C = 30 deg, ∠D = 90 deg & ∠B = 60 deg
--> BD: CD: BC = 1: \(\sqrt{3}\): 2
--> BD = k, CD = \(\sqrt{3}\)k, BC = 2k
Given BC = 2k = 5
--> k = 5/2
Altitude = \(\sqrt{3}\)k = 5/2*\(\sqrt{3}\)
Area of triangle ABC = 1/2*base*height = 1/2*AB*CD
= 1/2*12*5/2*\(\sqrt{3}\)
= 15\(\sqrt{3}\)
IMO Option B
Pls Hit Kudos if you like the solution
Attachments
33.png [ 35.79 KiB | Viewed 3699 times ]
