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.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
29 Jul 2019, 01:23
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
55% (02:27) correct
45%
(02:35)
wrong
based on 78
sessions
History
Date
Time
Result
Not Attempted Yet
[GMAT math practice question]
\(ABCD\) is a square and \(ADE\) is an equilateral triangle. What is the value of \(x\)?
7.29(ps).png [ 6.57 KiB | Viewed 3437 times ]
\(A. 120^o\)
\(B. 130^o\)
\(C. 140^o\)
\(D. 150^o\)
\(E. 160^o\)
\(ABCD\) is a square and \(ADE\) is an equilateral triangle. What is the value of \(x\)?
Attachment:
7.29(ps).png [ 6.57 KiB | Viewed 3437 times ]
\(A. 120^o\)
\(B. 130^o\)
\(C. 140^o\)
\(D. 150^o\)
\(E. 160^o\)
29 Jul 2019, 01:46
Kudos
Bookmarks
Please, find the attached file.
Posted from my mobile device
Posted from my mobile device
Attachments
File comment: PFA
96A066B9-5425-4FF7-87DA-38F927817F10.jpeg [ 2.61 MiB | Viewed 3371 times ]
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 24 Jul 2026
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
29 Jul 2019, 03:08
Kudos
Bookmarks
Since ADE is an equilateral triange,
\(AE=DE=AD=\)Side length of square and \(\angle DAE=\angle ADE=\angle AED=60^{\circ}\)
So, \(\angle BAE=90-\angle DAE=90-60=30^{\circ}\)
We know that \(AE=AD\) and \(AD=AB\)(side length of square) and so \(AE=AB\)
Therefore, triangle ABE is isoceles and \(\angle ABE=\angle AEB\)
So in triangle ABE, \(30+\angle ABE+\angle AEB=180\) or
\(30+2*\angle AEB=180\)
\(\angle AEB=75^{\circ}\)
Similarly, \(\angle DEC=75^{\circ}\)
Now \(\angle AEB+x+\angle DEC+\angle AED= 360\)
\(75+x+75+60=360\)
\(x=150^{\circ}\)
Answer is (D)
Hit Kudos if this helped!
Posted from my mobile device
\(AE=DE=AD=\)Side length of square and \(\angle DAE=\angle ADE=\angle AED=60^{\circ}\)
So, \(\angle BAE=90-\angle DAE=90-60=30^{\circ}\)
We know that \(AE=AD\) and \(AD=AB\)(side length of square) and so \(AE=AB\)
Therefore, triangle ABE is isoceles and \(\angle ABE=\angle AEB\)
So in triangle ABE, \(30+\angle ABE+\angle AEB=180\) or
\(30+2*\angle AEB=180\)
\(\angle AEB=75^{\circ}\)
Similarly, \(\angle DEC=75^{\circ}\)
Now \(\angle AEB+x+\angle DEC+\angle AED= 360\)
\(75+x+75+60=360\)
\(x=150^{\circ}\)
Answer is (D)
Hit Kudos if this helped!
Posted from my mobile device
