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09 Oct 2013, 03:04
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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
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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.
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Hi, IMO C
Refer to the figure,
trapezoid.png [ 6.71 KiB | Viewed 22426 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 22426 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.
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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 ? 
