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.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
10 Jun 2020, 04:55
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
85% (01:50) correct
15%
(01:39)
wrong
based on 53
sessions
History
Date
Time
Result
Not Attempted Yet
ABCD has area equal to 28. BC is parallel to AD. BA is perpendicular to AD. If BC is 6 and AD is 8, then what is CD?
A. 2√2
B. 2√3
C. 4
D. 2√5
E. 6
Attachment:
1.jpg [ 3.06 KiB | Viewed 10883 times ]
10 Jun 2020, 05:03
Kudos
Bookmarks
2 root 5. i guess it is D
since AB=4.
area of triangle+ recatangle is 6*h+2*h/2=7h=28 so h=AB=4
2^2+4^2=cd^2
20=cd^2
cd=root 20 or root of (4*5)=== 2* root 5
since AB=4.
area of triangle+ recatangle is 6*h+2*h/2=7h=28 so h=AB=4
2^2+4^2=cd^2
20=cd^2
cd=root 20 or root of (4*5)=== 2* root 5
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,089
Own Kudos:
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
10 Jun 2020, 05:36
Kudos
Bookmarks
Extended BC to an exterior point E, Such that when E connected to D. E is perpendicular on D.
Such that
ABED is a rectangle.
Let AB = x
BE = 6 = AD
The area of rectangle = l*b = 8*x = 8x
Now, CDE is a right angle triangle.. Right angle at E
Where CE = 2. DE = x
Area if triangle = 1/2*2*x = x
Area of ABCD = Area of ABED - Area of CDE
28 = 8x - x
7x = 28
X = 4 = AB = DE
Now. Value if CD =
CD^2 = DE^2 + BC^2
Pythagoras Theoram
CD^2 = 4^2 + 2^2
CD^2 = 16+4
CD = √20
CD = 2√5
Answer is D
Posted from my mobile device
Such that
ABED is a rectangle.
Let AB = x
BE = 6 = AD
The area of rectangle = l*b = 8*x = 8x
Now, CDE is a right angle triangle.. Right angle at E
Where CE = 2. DE = x
Area if triangle = 1/2*2*x = x
Area of ABCD = Area of ABED - Area of CDE
28 = 8x - x
7x = 28
X = 4 = AB = DE
Now. Value if CD =
CD^2 = DE^2 + BC^2
Pythagoras Theoram
CD^2 = 4^2 + 2^2
CD^2 = 16+4
CD = √20
CD = 2√5
Answer is D
Posted from my mobile device
