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If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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04 Feb 2012, 18:48
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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 Attachment:
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If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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04 Feb 2012, 18:54
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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Hope it helps. Attachment:
Trapezoid.GIF [ 1.77 KiB  Viewed 84253 times ]
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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#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 ba = 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



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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05 Feb 2012, 22:32
Bunuel wrote: Attachment: Trapezoid.GIF 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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Also discussed here: trapeziumarea99966.htmlHope it helps. This is how I approached: 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 Therefore area of trapezoid is = 24 6 = 18. Answer is B



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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27 May 2012, 19:42
Quote: So in triangle ABE sides are AB=3, AE=4 and BE=5: we have 345 right triangle ABE (with right angle CAD, as hypotenuse is BE=5) and 6810 right angle triangle ACD. Bunuel, any time we see a triangle with sides 3,4,5, can we assume its a right triangle?



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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28 May 2012, 00:20
pgmat wrote: Quote: So in triangle ABE sides are AB=3, AE=4 and BE=5: we have 345 right triangle ABE (with right angle CAD, as hypotenuse is BE=5) and 6810 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 wellknown example is \((3, 4, 5)\). If \((a, b, c)\) is a Pythagorean triple, then so is \((ka, kb, kc)\) for any positive integer \(k\). For more on this check Triangles chapter of Math Book: mathtriangles87197.htmlHope it helps.
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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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31 May 2012, 07:20
Bunuel wrote: pgmat wrote: Quote: So in triangle ABE sides are AB=3, AE=4 and BE=5: we have 345 right triangle ABE (with right angle CAD, as hypotenuse is BE=5) and 6810 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 wellknown example is \((3, 4, 5)\). If \((a, b, c)\) is a Pythagorean triple, then so is \((ka, kb, kc)\) for any positive integer \(k\). For more on this check Triangles chapter of Math Book: mathtriangles87197.htmlHope it helps. But MGMAT says that if you see the 345 as the side ratio, it does not implies it is a rt angled triangle???



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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31 May 2012, 07:28



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#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) =100k64k=36k =36(1/2) =18 Answer is (B) Regards,



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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02 Sep 2012, 11:48
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.



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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02 Sep 2012, 13:37
artuurss wrote: 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.
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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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03 Sep 2012, 04:38
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. Hope it helps.
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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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07 Jan 2014, 06:30
Bunuel wrote: Attachment: Trapezoid.GIF 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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Also discussed here: trapeziumarea99966.htmlHope it helps. I got that how it is a right triangle.. I have another question in my mind.. that is... If BE and CD are parallel, so does that mean BE bisecting AC?? thats y we have assumed that if AB is 3 then BC is 3 too?
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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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07 Jan 2014, 06:37
sanjoo wrote: Bunuel 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 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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Also discussed here: trapeziumarea99966.htmlHope it helps. I got that how it is a right triangle.. I have another question in my mind.. that is... If BE and CD are parallel, so does that mean BE bisecting AC?? thats y we have assumed that if AB is 3 then BC is 3 too? No, BE  CD does not imply that BE bisects AC. The fact that BC = AB = 3 is given in the stem.
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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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18 May 2014, 17:52
Bunuel 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 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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Also discussed here: trapeziumarea99966.htmlHope it helps. Hi Bunuel, It's quite clear from your explanation that once we apply the properties of similar triangles, ED has to equal 4. That being said, one thing that stumped was whether ED should have been 3 or 4? If lines BE and CD are parallel, don't they have to have a uniform distance between the two? If so, why doesn't BC = ED? EDIT: What is the main indicator that these are parallel? Is it the fact that they both have the same angle BAE or is it because they have parallel lines? Would having just one vs. the other suffice in identifying that these are parallel?



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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18 May 2014, 19:54
russ9 wrote: Bunuel wrote: Attachment: The attachment Trapezoid.GIF is no longer available 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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Hope it helps. Hi Bunuel, It's quite clear from your explanation that once we apply the properties of similar triangles, ED has to equal 4. That being said, one thing that stumped was whether ED should have been 3 or 4? If lines BE and CD are parallel, don't they have to have a uniform distance between the two? If so, why doesn't BC = ED? EDIT: What is the main indicator that these are parallel? Is it the fact that they both have the same angle BAE or is it because they have parallel lines? Would having just one vs. the other suffice in identifying that these are parallel? I think you are getting confused because of the figure. The figure shown is not drawn to scale, and it shows AE = AD which is not the case. It is not necessary that distance between two parallel line is always the same. Think of a line drawn perpendicular to both the parallel lines and then a line making 30 degrees with these parallel lines. The lengths would be different.



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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19 May 2014, 02:14
russ9 wrote: Bunuel 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 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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Also discussed here: trapeziumarea99966.htmlHope it helps. Hi Bunuel, It's quite clear from your explanation that once we apply the properties of similar triangles, ED has to equal 4. That being said, one thing that stumped was whether ED should have been 3 or 4? If lines BE and CD are parallel, don't they have to have a uniform distance between the two? If so, why doesn't BC = ED? EDIT: What is the main indicator that these are parallel? Is it the fact that they both have the same angle BAE or is it because they have parallel lines? Would having just one vs. the other suffice in identifying that these are parallel? If two lines are parallel, then they are constant distance apart (definition of parallel lines). But the distance between parallel lines is perpendicular from one to another, thus neither BC nor ED is the distances between BE and CD, hence the difference. As for the other question: we know that BE is parallel to CD, because we are directly told about it. BE  CD, means BE is parallel to CD ( implies parallel). Hope it's clear.
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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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19 May 2014, 02:15
mittalg wrote: russ9 wrote: Bunuel wrote: Attachment: Trapezoid.GIF 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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Hope it helps. Hi Bunuel, It's quite clear from your explanation that once we apply the properties of similar triangles, ED has to equal 4. That being said, one thing that stumped was whether ED should have been 3 or 4? If lines BE and CD are parallel, don't they have to have a uniform distance between the two? If so, why doesn't BC = ED? EDIT: What is the main indicator that these are parallel? Is it the fact that they both have the same angle BAE or is it because they have parallel lines? Would having just one vs. the other suffice in identifying that these are parallel? I think you are getting confused because of the figure. The figure shown is not drawn to scale, and it shows AE = AD which is not the case. It is not necessary that distance between two parallel line is always the same. Think of a line drawn perpendicular to both the parallel lines and then a line making 30 degrees with these parallel lines. The lengths would be different. The red part is not correct: if two lines are parallel, then they are constant distance apart (definition of parallel lines).
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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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19 May 2014, 02:35
russ9 wrote: Bunuel wrote: Attachment: Trapezoid.GIF 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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Hope it helps. Hi Bunuel, It's quite clear from your explanation that once we apply the properties of similar triangles, ED has to equal 4. That being said, one thing that stumped was whether ED should have been 3 or 4? If lines BE and CD are parallel, don't they have to have a uniform distance between the two? If so, why doesn't BC = ED? EDIT: What is the main indicator that these are parallel? Is it the fact that they both have the same angle BAE or is it because they have parallel lines? Would having just one vs. the other suffice in identifying that these are parallel? I think you are getting confused because of the figure. The figure shown is not drawn to scale, and it shows AE = AD which is not the case. It is not necessary that distance between two parallel line is always the same. Think of a line drawn perpendicular to both the parallel lines and then a line making 30 degrees with these parallel lines. The lengths would be different.[/quote] The red part is not correct: if two lines are parallel, then they are constant distance apart (definition of parallel lines).[/quote] Yes, you are right. What I was trying to say was the lengths of the various lines drawn between there two parallel lines might not be the same. The distance is the length of perpendicular drawn between two parallel lines which is always the same. My bad!!! Thanks for pointing it out.



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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of [#permalink]
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19 May 2014, 18:31
Bunuel wrote: russ9 wrote: Bunuel 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 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 345 right triangle ABE (with right angle CAD, as hypotenuse is \(BE=5\)) and 6810 right angle triangle ACD. Now, the \(area_{BEDC}=area_{ACD}area_{ABE}\) > \(area_{BEDC}=\frac{6*8}{2}\frac{3*4}{2}=18\). Answer: B. Also discussed here: trapeziumarea99966.htmlHope it helps. Hi Bunuel, It's quite clear from your explanation that once we apply the properties of similar triangles, ED has to equal 4. That being said, one thing that stumped was whether ED should have been 3 or 4? If lines BE and CD are parallel, don't they have to have a uniform distance between the two? If so, why doesn't BC = ED? EDIT: What is the main indicator that these are parallel? Is it the fact that they both have the same angle BAE or is it because they have parallel lines? Would having just one vs. the other suffice in identifying that these are parallel?If two lines are parallel, then they are constant distance apart (definition of parallel lines). But the distance between parallel lines is perpendicular from one to another, thus neither BC nor ED is the distances between BE and CD, hence the difference. As for the other question: we know that BE is parallel to CD, because we are directly told about it. BE  CD, means BE is parallel to CD ( implies parallel). Hope it's clear. My apologies  I meant to ask "what is the main indicator that these triangles are similar triangles" as opposed to mistakenly asking "what is the main indicator that these are parallel lines"? How do you know that these are similar triangles? is it because we have two parallel lines or is it because one triangle is inscribed in the other? If one was inscribed in the other but we weren't told that the bases were parallel  we wouldn't be able to conclude that they were similar triangles. Is that correct? Thanks and sorry for the confusion.




Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of
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