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

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If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is [#permalink] New post 29 Aug 2007, 11:46
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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
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VP
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Re: PS triangle [#permalink] New post 29 Aug 2007, 12:08
Piter wrote:
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


Very nice PS.

look, use similarities of traingle rule.
AB/AC=AE/AD
3/6=4/(4+ED)
ED=4

AC=6, AD=8, CD=10 these are attributes of a right traingle
now we know that ACD is a right traingle thus ABE is a right traingle as well as the agne A is right and BE is parallel to CD, also 3 and 4 leads ton BE = 5

now find the Area of ACD and subtranct the Area of ABE
Sacd=6*8/2=24
Sabe=3*4/2=6

Sbedc=24-6=18

B
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 [#permalink] New post 29 Aug 2007, 12:36
Where the point A came from? Is it a standard GMAT format or something?
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 [#permalink] New post 29 Aug 2007, 12:42
antihero wrote:
Where the point A came from? Is it a standard GMAT format or something?


Point A? is it angle A u mean?

it is a right angle if you are about Angle A.
use similarities rule
then u will come up with the idea of right traingle.

GMAT never gives a problem if it takes longer than 2 min to solve it.
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 [#permalink] New post 29 Aug 2007, 13:18
Ravshonbek wrote:
antihero wrote:
Where the point A came from? Is it a standard GMAT format or something?


Point A? is it angle A u mean?

it is a right angle if you are about Angle A.
use similarities rule
then u will come up with the idea of right traingle.

GMAT never gives a problem if it takes longer than 2 min to solve it.


Oh lol, i didnt see the image, so i couldnt understand where point A i coming from. Now i see it.
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 [#permalink] New post 29 Aug 2007, 18:03
The Most Common of all right triangles tested in GMAT = 3:4:5 ratio.

Once we can catch that..this problem becomes easy.

Agree with B
  [#permalink] 29 Aug 2007, 18:03
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