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 1312:30 AM EDT
-01:30 AM EDT
Struggling to find the right strategies to score a 99 %ile on GMAT Focus? Riya (GMAT 715) boosted her score by 100-points in just 15 days! Discover how the right mentorship, tailored strategies, and an unwavering mindset can transform your GMAT prep.
Sep 1301:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
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
09 Oct 2013, 03:04
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
45% (02:47) correct
55%
(02:51)
wrong
based on 49
sessions
History
Date
Time
Result
Not Attempted Yet
In trapezoid ABCD, AD is parallel to BC. Also BD is perpendicular to DC. The point F is chosen on line BD so that AF is perpendicular to BD. AF is extended to meet BC at point E. If AB=41, AD=50 and BF=9, what is the area of quadrilateral FECD?
A. 900
B. 1523.5
C. 960
D. 910
E. None of these
A. 900
B. 1523.5
C. 960
D. 910
E. None of these
Updated on: 10 Oct 2013, 00:39
Originally posted by rockstar23 on 09 Oct 2013, 04:01.
Last edited by rockstar23 on 10 Oct 2013, 00:39, edited 1 time in total.
Last edited by rockstar23 on 10 Oct 2013, 00:39, edited 1 time in total.
Kudos
Bookmarks
Hi, IMO C
Refer to the figure,
trapezoid.png [ 6.71 KiB | Viewed 22753 times ]
Using Pythagoras theorem for triangle AFB, we get,
\(AF^2 = 41^2 - 9^2\)
\(AF = 40\)
Next, applying Pythagoras theorem for triangle AFD, we get,
\(DF^2 = 50^2 - 40^2\)
\(DF = 30\)
Now we have, AB = 51, AD=50, AF=40, DF=30, BF=9, EC=50(Since quad ADCE is a parallelogram)
Now area of triangle ABD =\(1/2*AF*BD\)
Also area of triangle ABD = \(1/2*BO*AD\)
\(\frac{1}{2}*AF*BD\) = \(\frac{1}{2}*BO*AD\)
\(40*39\) = \(BO*50\)
BO = 31.2
But BO is also the height of quad AECD
area of quad AECD = \(BO*EC\) = \(50*31.2\) = \(1560\)
Now,
area of quad FECD = area of quad AECD - area of triangle AFD
= \(1560 - {\frac{1}{2}*AF*DF}\)
= \(1560 - {\frac{1}{2}*40*30}\)
= \(1560 - 600\)
= \(960\)
Hence C
Hope I am correct
Refer to the figure,
Attachment:
trapezoid.png [ 6.71 KiB | Viewed 22753 times ]
Using Pythagoras theorem for triangle AFB, we get,
\(AF^2 = 41^2 - 9^2\)
\(AF = 40\)
Next, applying Pythagoras theorem for triangle AFD, we get,
\(DF^2 = 50^2 - 40^2\)
\(DF = 30\)
Now we have, AB = 51, AD=50, AF=40, DF=30, BF=9, EC=50(Since quad ADCE is a parallelogram)
Now area of triangle ABD =\(1/2*AF*BD\)
Also area of triangle ABD = \(1/2*BO*AD\)
\(\frac{1}{2}*AF*BD\) = \(\frac{1}{2}*BO*AD\)
\(40*39\) = \(BO*50\)
BO = 31.2
But BO is also the height of quad AECD
area of quad AECD = \(BO*EC\) = \(50*31.2\) = \(1560\)
Now,
area of quad FECD = area of quad AECD - area of triangle AFD
= \(1560 - {\frac{1}{2}*AF*DF}\)
= \(1560 - {\frac{1}{2}*40*30}\)
= \(1560 - 600\)
= \(960\)
Hence C
Hope I am correct
Updated on: 10 Oct 2013, 00:30
Originally posted by Chiranjeevee on 09 Oct 2013, 23:20.
Last edited by Chiranjeevee on 10 Oct 2013, 00:30, edited 2 times in total.
Last edited by Chiranjeevee on 10 Oct 2013, 00:30, edited 2 times in total.
Kudos
Bookmarks
I really wasted so much time finding why my answer is different and it came as 960. that is C.
Are u sure aboout OA?? I think its not E, rather than its C.
ADCE is parellelogram, since AD || CE (Given) and AE || DC ( since AF and CD are both perpendicular to BD).
AB = 41
BF = 9
so, AF = 40 (By pythagoras theorm)
and FD = 30
In the above figure above lets say height of the triangle ABD be h, from B to AD (This height will be same as the height of parallelogram ADCE, taking base as 50).
Now area of Triangle, ABD = (1/2)*h*AD = (1/2)*AF*(BF+FD),
(1/2)*h*50 = (1/2)*40*39
solve for h = (4/5)*39
Now since we got h, area of parellelogram, ADCE = 50 * h = 50 * (4/5)*39 = 1560.
Area of triangle AFD = (1/2)*30*40 = 600
Required area(quad DFEC) = 1560-600 = 960
Are u sure aboout OA?? I think its not E, rather than its C.
ADCE is parellelogram, since AD || CE (Given) and AE || DC ( since AF and CD are both perpendicular to BD).
AB = 41
BF = 9
so, AF = 40 (By pythagoras theorm)
and FD = 30
In the above figure above lets say height of the triangle ABD be h, from B to AD (This height will be same as the height of parallelogram ADCE, taking base as 50).
Now area of Triangle, ABD = (1/2)*h*AD = (1/2)*AF*(BF+FD),
(1/2)*h*50 = (1/2)*40*39
solve for h = (4/5)*39
Now since we got h, area of parellelogram, ADCE = 50 * h = 50 * (4/5)*39 = 1560.
Area of triangle AFD = (1/2)*30*40 = 600
Required area(quad DFEC) = 1560-600 = 960
Is there any update on the OA ? 
