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

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Manager
Joined: 28 Jun 2009
Posts: 51
Location: Moscow Russia
If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the  [#permalink]

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03 Aug 2009, 23:09
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10
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Question Stats:

77% (01:03) correct 23% (02:29) wrong based on 308 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-be-cd-and-bc-ab-3-ae-4-and-cd-10-what-is-127060.html
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q.doc [68.5 KiB]

Intern
Joined: 01 Aug 2009
Posts: 28
Location: Australia
Re: Geo Q from ManhGmat. How to  [#permalink]

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04 Aug 2009, 05:59
4
1

Solution:
Since BE||CD, it implies that the triangle ABE and ACD are similar figures.

For triangle ABE,
AB=3
AE=4
BE =?

For triangle ACD,
AC=AB+BC = 3+3 = 6
CD =10

Since ABE and ACD are similar figures and AC = 2 AB (6=2*3), we can deduce the unknown sides, as the sides of ABE and ACD will follow the ratio of 1:2

For triangle ABE,
AB=3
AE=4
BE =(1/2)CD = (1/2)10 = 5

For triangle ACD,
AC=AB+BC = 3+3 = 6
AD = 2AE = 2*4 = 8
CD=10

Now that we know all sides, Area(BCDE) = Area(ACD) – Area(ABC)

Area of a triangle = sqrt(s(s-a)(s-b)(s-c)), where s=(a+b+c)/2 and abc are lengths of the sides.

Area(ACD) = sqrt(12*6*2*4) = 24
Area(ABC) = sqrt(6*3*2*1) = 6

Area(BCDE) = 24-6 = 18
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Director
Joined: 13 Nov 2003
Posts: 775
Location: BULGARIA
Re: Geo Q from ManhGmat. How to  [#permalink]

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04 Aug 2009, 06:36
5
3
Hi,

since BE is middle sector, parallel to the CD, E divides AD in two pieces of 4.
Triangle CDA is a right triangle with right angle at A , since sides are AC=6, CD=10 and AD=8.

Think that the solution is obvious. Area ACD=24, Area ABE=6 area of trapezoid is 18

regards
Intern
Joined: 06 Oct 2012
Posts: 5
Re: Geo Q from ManhGmat. How to  [#permalink]

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12 May 2013, 21:34
1
scarish wrote:

Solution:
Since BE||CD, it implies that the triangle ABE and ACD are similar figures.

For triangle ABE,
AB=3
AE=4
BE =?

For triangle ACD,
AC=AB+BC = 3+3 = 6
CD =10

Since ABE and ACD are similar figures and AC = 2 AB (6=2*3), we can deduce the unknown sides, as the sides of ABE and ACD will follow the ratio of 1:2

For triangle ABE,
AB=3
AE=4
BE =(1/2)CD = (1/2)10 = 5

For triangle ACD,
AC=AB+BC = 3+3 = 6
AD = 2AE = 2*4 = 8
CD=10

Now that we know all sides, Area(BCDE) = Area(ACD) – Area(ABC)

Area of a triangle = sqrt(s(s-a)(s-b)(s-c)), where s=(a+b+c)/2 and abc are lengths of the sides.

Area(ACD) = sqrt(12*6*2*4) = 24
Area(ABC) = sqrt(6*3*2*1) = 6

Area(BCDE) = 24-6 = 18

please post a diagram of the aforesaid figure, it will help a lot, I simply can't just visualize it.
Intern
Joined: 05 Mar 2013
Posts: 44
Location: India
Concentration: Entrepreneurship, Marketing
GMAT Date: 06-05-2013
GPA: 3.2
Re: If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the  [#permalink]

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13 May 2013, 01:35
1
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

Be is parallel to cd so triangles ABE and ACD are similar. Ratio of their areas is square of the ratio of their similar sides.

Both the triangles are also similar so we can get ED as 4 which will indicate that ACD is a right angled triangle. so we can get the area as 24.and using ratio of areas statement stated above we can get the area of ABE as 6. So are of trapezoid is 24-6 = 18
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Joined: 02 Sep 2009
Posts: 49430
Re: If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the  [#permalink]

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13 May 2013, 03:39
8
6

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

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 $$area_{BEDC}=area_{ACD}-area_{ABE}$$ --> $$area_{BEDC}=\frac{6*8}{2}-\frac{3*4}{2}=18$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-be-cd-and-bc-ab-3-ae-4-and-cd-10-what-is-127060.html

Also discussed here: trapezium-area-99966.html
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Re: If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the  [#permalink]

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16 Jul 2018, 08:21
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