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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] New post 28 Sep 2006, 06:27
If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?

Please show your steps and rules used.
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 [#permalink] New post 28 Sep 2006, 08:23
There are two similar triangles in there Tri ABE and Tri ACD

The sides are then proportional.. which means AB/AC = AE/AD = BE /CD


3/6 = x/ 10. so X is 5 (3-4-5 triangle => right triangle)

Area of the trapezoid = Area of the larger triangle - smaller triangle

= 1/2*6 * 8 - 1/2* 4*3
= 24 - 6 = 18


So Ans. B
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 [#permalink] New post 28 Sep 2006, 08:48
gk3.14 wrote:
There are two similar triangles in there Tri ABE and Tri ACD

The sides are then proportional.. which means AB/AC = AE/AD = BE /CD


3/6 = x/ 10. so X is 5 (3-4-5 triangle => right triangle)

Area of the trapezoid = Area of the larger triangle - smaller triangle

= 1/2*6 * 8 - 1/2* 4*3
= 24 - 6 = 18


So Ans. B


great explanation. thank you
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 [#permalink] New post 29 Sep 2006, 00:28
Clearly both ACD and ABE are right angled triangles.

Area of ACD = 1/2 x 6 x 8 = 24
Area of ABE = 1/2 x 3 x 4 = 6

So area of Trapezium = 24-6 = 18.

Since BE || CD and C bisects AB, obviously D bisects AE
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Last edited by cicerone on 25 Sep 2008, 00:12, edited 1 time in total.
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 [#permalink] New post 29 Sep 2006, 06:50
cicerone wrote:
Clearly both ACD and ABE are right angled triangles.

Area of ACD = 1/2 x 6 x 8 = 24
Area of ABE = 1/2 x 3 x 4 = 6

So area of Trapezium = 24-6 = 18.

Since BE || CD and C bisects AB, obviously D bisects AE


Nice catch Cicerone. That made the problem a whole lot easier. :-D

AE must equal ED because BE is parallel to CD.
  [#permalink] 29 Sep 2006, 06:50
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