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If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the

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If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the [#permalink]

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New post 28 Aug 2010, 10:21
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If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? (Figure attached)

A) 12
B) 18
C) 24
D) 30
E) 48

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-be-cd-and-bc-ab-3-ae-4-and-cd-10-what-is-127060.html
[Reveal] Spoiler: OA
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Re: Trapezium area [#permalink]

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New post 28 Aug 2010, 11:10
jananijayakumar wrote:
If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? (Figure attached)
A) 12
B) 18
C) 24
D) 30
E) 48
Attachment:
Untitled.jpg


I guess it's BE||CD (BE is parallel to CD).

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Hope it's clear.
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Re: Trapezium area [#permalink]

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New post 28 Aug 2010, 12:30
Bunuel - In this case it turned out to be a rt angled triangle so the area calc was easy.. Lets say the bigger triangle were 5 7 9 in sides and the smaller was correspondingly half in size inside of it.. In that case, what would be the easiest way to calculate the trapezium area.. I guess does GMAT require us to remember the formula of the area of a triangle (not the .5 base height) but the one with the sides? Is there an easy way to calculate the area where the sides are given for any triangle?
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Re: Trapezium area [#permalink]

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New post 28 Aug 2010, 12:48
mainhoon wrote:
Bunuel - In this case it turned out to be a rt angled triangle so the area calc was easy.. Lets say the bigger triangle were 5 7 9 in sides and the smaller was correspondingly half in size inside of it.. In that case, what would be the easiest way to calculate the trapezium area.. I guess does GMAT require us to remember the formula of the area of a triangle (not the .5 base height) but the one with the sides? Is there an easy way to calculate the area where the sides are given for any triangle?


If it were not right triangle we could calculate the area of triangles with Heron's Formula (as we still would knew the lengths of all sides), though I've never seen a GMAT question requiring it.
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Re: Trapezium area [#permalink]

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New post 13 Sep 2010, 07:18
Bunuel wrote:
jananijayakumar wrote:
If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? (Figure attached)
A) 12
B) 18
C) 24
D) 30
E) 48
Attachment:
Untitled.jpg


I guess it's BE||CD (BE is parallel to CD).

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Hope it's clear.


Bunuel... If you say that
BC = AB = 3, AE = 4
then how can BE||CD (BE is parallel to CD) true. I am not able to visualize such a figure!!! :(
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Re: Trapezium area [#permalink]

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New post 13 Sep 2010, 08:42
utin wrote:
Bunuel wrote:
jananijayakumar wrote:
If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? (Figure attached)
A) 12
B) 18
C) 24
D) 30
E) 48
Attachment:
Untitled.jpg


I guess it's BE||CD (BE is parallel to CD).

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: \(\frac{AB}{AC}=\frac{BE}{CD}\) --> \(\frac{3}{6}=\frac{BE}{10}\) --> \(BE=5\) and \(AD=2AE=8\).

So in triangle ABE sides are \(AB=3\), \(AE=4\) and \(BE=5\): we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6-8-10 right angle triangle ACD.

Now, the \(area_{BEDC}=area_{ACD}-area_{ABE}\) --> \(area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18\).

Answer: B.

Hope it's clear.


Bunuel... If you say that
BC = AB = 3, AE = 4
then how can BE||CD (BE is parallel to CD) true. I am not able to visualize such a figure!!! :(


Why not? Consider triangle CAD, where CA=6 and AD=8. Now draw midsegment BE of a triangle (a line segment joining the midpoints of two sides of a triangle). Properties of a midsegment:

• The midsegment is always parallel to the third side of the triangle.
• The midsegment is always half the length of the third side.

Check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Trapezium area [#permalink]

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New post 18 Sep 2010, 08:11
Thanks again Bunuel!!! I was thinking of one triangle in the attachment only...
didn't generalize...
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Trapezoid In Triangle [#permalink]

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New post 29 Nov 2010, 14:17
If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?
a) 12
b) 18
c) 24
d) 30
e) 48

Source: Manhattan CAT Exam

Solution to follow.
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Screen shot 2010-11-29 at 4.17.27 PM.jpg
Screen shot 2010-11-29 at 4.17.27 PM.jpg [ 17.06 KiB | Viewed 4524 times ]


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Trapezoid In Triangle   [#permalink] 29 Nov 2010, 14:17
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