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In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi

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ROckHIsT
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In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink] New post   09 Oct 2013, 03:04
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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
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C
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rockstar23
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Re: In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink] New post   Updated on: 10 Oct 2013, 00:39
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Hi, IMO C

Refer to the figure,

Attachment:
trapezoid.png
trapezoid.png [ 6.71 KiB | Viewed 18992 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 :)

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.
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Re: In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink] New post   Updated on: 10 Oct 2013, 00:30
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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

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.
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Re: In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink] New post   10 Oct 2013, 00:17
Hi Chiranjeevee,

I am not able to understand this step
Chiranjeevee wrote:
Now area of parellogram, ADCE = (1/2)*h*AD = (1/2)*AF*(BF+FD),
(1/2)*h*50 = (1/2)*40*39
solve for h = (4/5)*39


Could you pls explain it :)
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Re: In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink] New post   10 Oct 2013, 00:27
rockstar23 wrote:
Hi Chiranjeevee,

I am not able to understand this step
Chiranjeevee wrote:
Now area of parellogram, ADCE = (1/2)*h*AD = (1/2)*AF*(BF+FD),
(1/2)*h*50 = (1/2)*40*39
solve for h = (4/5)*39


Could you pls explain it :)



Sorry, did a typo error, already changed thanks :-D
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Re: In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink] New post   10 Oct 2013, 01:34
Chiranjeevee wrote:
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



I think your answer is seems to be correct. Yet I would like to confirm it by an expert. Will notify you asap. :)

Regards
Rohit
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Re: In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink] New post   25 Nov 2013, 04:33
rgyanani wrote:
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


OA was not provided. Kudos to Bunuel (ADMIN) who helped me post my first question properly. :-D


I too got 960 :( Is there any update on the OA ?
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Re: In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink] New post   12 Dec 2013, 15:25
Put me down for C as well.

We have triangle ABF which is a right triangle (it shares a straight line with another right triangle) and we have two measurements for it's sides which means we can find the third side via Pythagorean theorem:

9^2 + b^2 = 41^2
81 + b^2 = 1681
b^2 = 1600
b = 40

Now we have another triangle ADF which is also right because it shares a vertical angle with right triangle BFE. We also have two sides of this triangle so we can find the third by the Pythagorean theorem:

40^2 + b^2 = 50^2
1600 + b^2 = 2500
b^2 = 900
b = 30

Now look at triangle FEB and FAD. They both share a right angle and two parallel lines. These triangles are similar. We can set up a ratio knowing two sides of FAD and one side and one unknown side of FEB to find that side:

(30/40) : (9/x)
30x = 360
x = 12

We can use another set of similar triangles to find the base for triangle BDC. Note that line FA is parallel to DC and that AD is parallel to CB. Also, both triangles have a right angle at points D and F. If you were to rotate FAD onto BDC (as done with FAD and FEB) you would see that their angle measures are the same and all of their corresponding sides parallel.

(30/50) : (39/x)
30x = 1950
x = 65

We need to find the measure for line DC to get the other base of FEDC. Triangle FEB is similar to DCB because they both share a right angle (at f and d respectively) and because they share two sets of parallel lines.

(9/12) : (39/x)
3/4 : 39/x
3x = 156
x=52

(We actually do not need to find the base of BC)

Now we have the two bases of FEDC and the height:

A = h * (b1 + b2) / 2
A = 30 * (64) / 2
A = 30 * 32
A = 960

C
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Re: In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink] New post   01 Sep 2021, 00:54
960

Official answer is the correct one?

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Re: In trapezoid ABCD, AD is parallel to BC. Also BD is perpendi [#permalink]
01 Sep 2021, 00:54
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