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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

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

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artuurss 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

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

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