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

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

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05 Feb 2012, 03:42

Since triangles ABE and ACD are similar we get BE = 5 and ED = 4 To determine the height --> 3^2 = h^2+a^2 and 4^2 = h^2+b^2 where a+b+5 = 10 so we get a+b= 5 putting this value in the above equation we get b-a = 7/5 From this value we determine the value of h =12/5 since area of trapezium = 1/2*ht*(sum of parallel sides) = 1/2*12/5*15 = 18

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

Since BE is II to CD AND B is mid point of AC therefore BE is Midsegment which means that BE = 1/2 (CD) = 5 and AE = ED = 4, Observing triangle ACD : its a 6,8,10 Right Triangle & therefore area is 1/2(b/h) = 24 similarly traingle ABE : its a 3,4,5 triangle & therfore area is 1/2 (b/h) = 6

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

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27 May 2012, 19:42

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Quote:

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.

Bunuel, any time we see a triangle with sides 3,4,5, can we assume its a right triangle?

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

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28 May 2012, 00:20

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pgmat wrote:

Quote:

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.

Bunuel, any time we see a triangle with sides 3,4,5, can we assume its a right triangle?

Yes, any triangle whose sides are in the ratio 3:4:5 is a right triangle. Such triangles that have their sides in the ratio of whole numbers are called Pythagorean Triples. There are an infinite number of them, and this is just the smallest. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.

A Pythagorean triple consists of three positive integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\). Such a triple is commonly written \((a, b, c)\), and a well-known example is \((3, 4, 5)\). If \((a, b, c)\) is a Pythagorean triple, then so is \((ka, kb, kc)\) for any positive integer \(k\).

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

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31 May 2012, 07:20

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Bunuel wrote:

pgmat wrote:

Quote:

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.

Bunuel, any time we see a triangle with sides 3,4,5, can we assume its a right triangle?

Yes, any triangle whose sides are in the ratio 3:4:5 is a right triangle. Such triangles that have their sides in the ratio of whole numbers are called Pythagorean Triples. There are an infinite number of them, and this is just the smallest. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.

A Pythagorean triple consists of three positive integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\). Such a triple is commonly written \((a, b, c)\), and a well-known example is \((3, 4, 5)\). If \((a, b, c)\) is a Pythagorean triple, then so is \((ka, kb, kc)\) for any positive integer \(k\).

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

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31 May 2012, 10:00

Bunuel wrote:

manulath wrote:

But MGMAT says that if you see the 3-4-5 as the side ratio, it does not implies it is a rt angled triangle???

I doubt that. But if it does then it's a mistake.

That's because converse of the Pythagorean theorem is also true.

For any triangle with sides a, b, c, if a^2 + b^2 = c^2, then the angle between a and b measures 90°.

Since 3^2+4^2=5^2 then any triangle whose sides are in the ratio 3:4:5 is a right triangle.

Hope it's clear.

Thanks for the prompt clarification. I believe, what you said should be true. As in this problem, it took me about 4 min to solve without the Pythagorean triplet. On using the triplet, the time required was substantially less. The problem is required to be solved within 2 min. I shall go with the triplet

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

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13 Jun 2012, 08:34

enigma123 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

I am struggling to solve this. What's the concept? I can think of area of similar triangles as what Bunuel has said previously.

Hi,

Area of a triangle is directly proportional to square of a side. or area (ABC) = \(BE^2k\) (where k, is constant of proportionality) & area (ACD) = \(CD^2k\) & k = AB/AC=BE/CD=1/2 Thus, area (trap BCDE) = area (ACD) - area (ABC) =100k-64k=36k =36(1/2) =18

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

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02 Sep 2012, 11:48

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The problem I am having with this question is, why are we assuming that ADC is a triangle at all? It looks like a triangle, but AB and AE also look like they are equivalent lengths, which they are not. So would couldn't BC and ED skew off into different directions while still maintaining the parallel nature of BE and CD? Because I didn't know why we should make these assumptions, I guessed. I thought we were always supposed to assume that the picture is not drawn to scale.

If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? 12 18 24 30 48

How to solve this

BE || CD and B is the middle of AC, implies that E is the middle of AD, so AD = 8. Since \(AC^2+AD^2=CD^2 \, \, (6^2+8^2=10^2)\) we can deduce that ACD is right angled triangle and its area is 6 * 8 / 2 = 24. The area of the small triangle ABE is 1/4 of the area of the large triangle ACD. Triangle ABE is similar to triangle ACD, similarity ratio being 1:2. The corresponding areas are in relation 1:4.

Therefore, the area of the trapezoid BEDC is 3/4 of the area of the large triangle i.e. (3/4) * 24 = 18.

Answer B. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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03 Sep 2012, 04:38

Expert's post

dandarth1 wrote:

The problem I am having with this question is, why are we assuming that ADC is a triangle at all? It looks like a triangle, but AB and AE also look like they are equivalent lengths, which they are not. So would couldn't BC and ED skew off into different directions while still maintaining the parallel nature of BE and CD? Because I didn't know why we should make these assumptions, I guessed. I thought we were always supposed to assume that the picture is not drawn to scale.

Merging similar topics. Please refer to the solutions above.

As for your doubt, OG13, page150: Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

OG13, page 272: A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2). Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight. You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees. All figures lie in a plane unless otherwise indicated.

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

Straightforward: Where did you see in the question that A, B and C are aligned? The height can take any value from 0 to 3 from my understanding. Could you please clarify ? Thank you in advance.

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

Straightforward: Where did you see in the question that A, B and C are aligned? The height can take any value from 0 to 3 from my understanding. Could you please clarify ? Thank you in advance.

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

Straightforward: Where did you see in the question that A, B and C are aligned? The height can take any value from 0 to 3 from my understanding. Could you please clarify ? Thank you in advance.

Bunuel, It doesn t help. When i have read that thread, i GUESSED that there should have been a picture, which is not the case in the application. I have reported to add the picture if it is THE element missing for my understanding of the question. As is -without picture-, i maintain than we can only find a range.

Bunuel, It doesn t help. When i have read that thread, i GUESSED that there should have been a picture, which is not the case in the application. I have reported to add the picture if it is THE element missing for my understanding of the question. As is -without picture-, i maintain than we can only find a range.

The following may be a rather dumb question...please bear with me.

So, I've always known that a 3-4-5 triangle was a "special" right triangle but the explanation for the problem above essentially said that if a triangle has sides in the proportion 3-4-5, it is automatically a right triangle. Can someone either confirm or deny this?

The following may be a rather dumb question...please bear with me.

So, I've always known that a 3-4-5 triangle was a "special" right triangle but the explanation for the problem above essentially said that if a triangle has sides in the proportion 3-4-5, it is automatically a right triangle. Can someone either confirm or deny this?

Again, apologies in advance for the brain fart...

IMO yes...every triangle whose side are in 3:4:5 ration is a right angle triangle because (3x)^2+(4x)^2=(5x)^2.

Coming to the solution of the problem:

Given that BE||CD and B is midpoint point of AC => B divides AC in 1:1 ratio => E divides AD in 1:1 ration too. hence AE = 4

According to basic proportionality theorem BE = 1/2 CD i.e 5.

Now we have two right angle triangles ABC (3-4-5) and ACD (6-8-10)

Area of trapezium is area of ACD-area of ABC i.e 1/2 * 6*8 - 1/2 * 3*4 = 24-6 =18

Basic proportionality theorem: When a line parallel to third side divides the remaining two sides in m:n ration the lenght of the parallel line is m/ (m+n) third side of triangle here AB= 1/ (1+1) CD

The converse is also true A line drawn through one point on one side of a triangle and parallel to second side will cut the third side at a point which will divide the third side in the same ratio as the first point divided the first side

The following may be a rather dumb question...please bear with me.

So, I've always known that a 3-4-5 triangle was a "special" right triangle but the explanation for the problem above essentially said that if a triangle has sides in the proportion 3-4-5, it is automatically a right triangle. Can someone either confirm or deny this?

Again, apologies in advance for the brain fart...

IMO yes...every triangle whose side are in 3:4:5 ration is a right angle triangle because (3x)^2+(4x)^2=(5x)^2.

Coming to the solution of the problem:

Given that BE||CD and B is midpoint point of AC => B divides AC in 1:1 ratio => E divides AD in 1:1 ration too. hence AE = 4

According to basic proportionality theorem BE = 1/2 CD i.e 5.

Now we have two right angle triangles ABC (3-4-5) and ACD (6-8-10)

Area of trapezium is area of ACD-area of ABC i.e 1/2 * 6*8 - 1/2 * 3*4 = 24-6 =18

Basic proportionality theorem: When a line parallel to third side divides of triangle the remaining two sides in m:n ration the lenght of the line line is m/ (m+n) third side of triangle here AB= 1/ (1+1) CD

The converse is also true A line drawn through one point on one side of a triangle and parallel to second side will cut the third side at a point which will divide the third side in the same ratio as the first point divided the first side

Thanks for the thorough response, however, I'm just focusing on the first part you said (in bold): "every triangle whose side are in 3:4:5 ration is a right angle triangle because (3x)^2+(4x)^2=(5x)^2".

However, you can only use the Pythagorean Theorem IF AND ONLY IF the triangle is a right triangle. The more I think about this problem, the more it's like the saying: "which came first, the chicken or the egg"...

gmatclubot

Re: If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the
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18 Nov 2013, 18:50

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