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.
May 2008:00 AM PDT
-08:30 AM PDT
What’s in it for you- Live Profile Evaluation Chat Session with Jenifer Turtschanow, CEO, ARINGO. Come with your details prepared and ARINGO will share insights! Pre-MBA Role/Industry, YOE, Exam Score, C/GPA, ECs Post-MBA Role/ Industry & School List.
Jun 1006:00 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..
01 Apr 2020, 08:01
Kudos
Bookmarks
C
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:
70% (01:36) correct
30%
(01:47)
wrong
based on 152
sessions
History
Date
Time
Result
Not Attempted Yet
In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is the area of the triangle?
A. √3/2
B. 1
C. √3
D. 2
E. 2√3
A. √3/2
B. 1
C. √3
D. 2
E. 2√3
Attachments
Screenshot 2020-04-01 at 8.31.03 PM.png [ 11.4 KiB | Viewed 3979 times ]
01 Apr 2020, 09:40
Kudos
Bookmarks
Kritisood
Aside: Although we can, indeed, solve this question using trigonometric ratios, the GMAT test makers don't expect students to know about trigonometric ratios. So another approach is to apply what we know about special 30-60-90 right triangles
Since we're told that AB = BC, we know that the triangle is an isosceles triangle, which means ∠BCA = 30°
So if we draw in the altitude from point B, we get two identical right triangles.
Also notice that these right triangles are both special 30-60-90 right triangles
So let's compare these triangles with our base 30-60-90 right triangle
When we do this, we see that the two red right triangles are identical to the base triangle
So, we can add the following measurements to our diagram:
At this point we're ready to find the area of the triangle.
Area of triangle = (base)(height)/2
So, the area \(= \frac{(\sqrt{3}+\sqrt{3})(1)}{2}= \frac{(2\sqrt{3})(1)}{2}=\sqrt{3}\)
Answer: C
Cheers,
Brent
General Discussion
ArunSharma12
Joined: 25 Oct 2015
Last visit: 20 Jul 2022
Posts: 512
Own Kudos:
Given Kudos: 74
Location: India
Schools: IIML-IPMX '22 (A)
GMAT 1: 650 Q48 V31
GMAT 2: 720 Q49 V38 (Online)
GPA: 4
Products:
01 Apr 2020, 09:17
Kudos
Bookmarks
Kritisood
Attachment:
Img.png [ 11.91 KiB | Viewed 3467 times ]
\(\frac{2}{sin(30)}=\frac{x}{sin(120)}\)
\(x = \frac{2*sin(120)}{sin(30)}\)
\(x = \frac{2*cos(30)}{sin(30)}\) = \(\frac{2*2*\sqrt{3}}{2}\); sin(90+x)=cos(x)
\(x = 2*\sqrt{3}\)
To find area we can use formula
\(A =\frac{1*a*b*sin(x)}{2} \); where a,b are two adjacent sides with angle x between them
\(A =\frac{1*2*2*\sqrt{3}*sin(30)}{2} \)
\(A =\sqrt{3}\)
Ans: C
