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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, 17:48
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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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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: http://gmatclub.com/forum/trapeziumarea99966.htmlHope it helps. Attachment:
Trapezoid.GIF [ 1.77 KiB  Viewed 72579 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, 02:42
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, 21:32
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, 18:42
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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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27 May 2012, 23:20
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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, 06:20
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Bunuel wrote: pgmat wrote: Quote: So in triangle ABE sides are AB=3, AE=4 and BE=5: we have 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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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, 09:00
Bunuel wrote: manulath wrote: But MGMAT says that if you see the 345 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



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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, 07:34
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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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, 12:37
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, 03:38
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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18 Nov 2013, 14:00
If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? The following may be a rather dumb question...please bear with me. So, I've always known that a 345 triangle was a "special" right triangle but the explanation for the problem above essentially said that if a triangle has sides in the proportion 345, it is automatically a right triangle. Can someone either confirm or deny this? Again, apologies in advance for the brain fart...



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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 Nov 2013, 17:47
NvrEvrGvUp wrote: If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? The following may be a rather dumb question...please bear with me. So, I've always known that a 345 triangle was a "special" right triangle but the explanation for the problem above essentially said that if a triangle has sides in the proportion 345, 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 BECD 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 (345) and ACD (6810) Area of trapezium is area of ACDarea of ABC i.e 1/2 * 6*8  1/2 * 3*4 = 246 =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



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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 Nov 2013, 17:50
adityapagadala wrote: NvrEvrGvUp wrote: If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? The following may be a rather dumb question...please bear with me. So, I've always known that a 345 triangle was a "special" right triangle but the explanation for the problem above essentially said that if a triangle has sides in the proportion 345, 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 BECD 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 (345) and ACD (6810) Area of trapezium is area of ACDarea of ABC i.e 1/2 * 6*8  1/2 * 3*4 = 246 =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"...



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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 Nov 2013, 18:12
i think u accept that we can definitely construct atleast one right angled triangle if sides are in 3x 4x 5x  say ABC (as per pythogorous theorem). Now the question is  is the converse of pythorogous theorem is true of not?
I think it is true. Let say the converse of theorem is false i.e there should be a non right angle triangle (we know nothing about angles) say DEF whose sides are 3x 4x 5x.
But if we see both ABC and DEF are congruent triangles (SSS property). Hence their corresponding angles must also be same. As ABC is right angle DEF must also be right angle
Correct me if im wrong some where..!!



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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 Nov 2013, 21:21
NvrEvrGvUp wrote: If BE CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC? The following may be a rather dumb question...please bear with me. So, I've always known that a 345 triangle was a "special" right triangle but the explanation for the problem above essentially said that if a triangle has sides in the proportion 345, it is automatically a right triangle. Can someone either confirm or deny this? Again, apologies in advance for the brain fart... Funny one. Anyways, think of it in a different way : Lets say you have a stick of 3 units, and a stick of 4 units. You place a common end of each stick at say a point B.Keep this fixed. Now, you can move around the other arms of both the sticks freely in any way you want.However, there will be only a particular angle(unique if i may) at B, for which , the 2 other ends of these sticks would have a distance of 5 units between them. Now, you are confused about whether angle B is a right angle or not.Say it is not.Then, as previously proved, angle B has a value which is not 90 degrees and is unique. But, we already know that a 345 is a perfectly legitimate triangle with angle B as 90 degrees.Thus, B can have only one value which is , 90 degrees. Hope this didn't confuse you even more.
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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, 05:30
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, 05:37
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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