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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.



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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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20 May 2014, 00:04
russ9 wrote: Bunuel wrote: russ9 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. 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. Similar Triangles are triangles in which the three angles are identical. It is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º. So, for our question, triangles ABE and ACD are similar because their three angles are identical: 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, which means that they are similar.
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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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21 May 2014, 11:40
A line parallel to one side of a triangle cuts the other two sides in same ratio.As BE  CD it will cut AC and AD in same ratio i.e. AB/BC = AE/ED 3/3 = 4/ED => ED =4 Now we have , AC = 3+3 = 6 AD = 4 +4 = 8 and CD = 10 , we have a triangle with sides 6,8 and 10 which is a Right Triangle Smaller triangle would also be a right triangle with sides 3,4,5 Area of trapezoid = Area of Bigger Triangle  Smaller triangle = 1/2 (8*6 4*3) = 1/2 * 36 = 18 Hence 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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06 Jul 2014, 10:04
cyberjadugar wrote: 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 Attachment: Trapezoid.GIF 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, Having trouble with this one.... Why is the second term in the final equation,100k64k=36k, "64k"? Through the explanation it should be "25k"... is this a mistake?



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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 Jul 2014, 22:46
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. HI Bunuel, Can we directly use here prop BE  CD . so can we say BE = 1/2(CD) = 1/2*10 = 5 Thanks



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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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01 Aug 2014, 00:13
PathFinder007 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. Also discussed here: trapeziumarea99966.htmlHope it helps. HI Bunuel, Can we directly use here prop BE  CD . so can we say BE = 1/2(CD) = 1/2*10 = 5 Thanks BC=AB=3,meaning B is the mid point of AB and since BE CD means E is the mid point of AD and hence by Mid point theorem ..BE=5
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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 Sep 2014, 18:35
How can we determine the triangle heights? I've read through the above and haven't been able to grasp it. I'm sure its very simple, so thanks in advance, and excuse my oversight.



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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 Sep 2014, 21:58
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JackSparr0w wrote: How can we determine the triangle heights? I've read through the above and haven't been able to grasp it. I'm sure its very simple, so thanks in advance, and excuse my oversight. No problem!! Attachment:
Untitled.jpg [ 24.18 KiB  Viewed 1978 times ]
Notice that Triangle BAE is right angled at A and Angle A=90.. So BA is perpendicular to AE...so Area of ABE will be 1/2*3*4=6 Area of Bigger Triangle CAD (right angled at A) =1/2*6*8=24 So area of Trapezium is Area of CAD Area of BAE= 246=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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06 Jun 2016, 08:27
enigma123 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 Consider the triangle ABE, The two sides are 3 and 4. Hence the third side would be 5 (3  4  5 right triangle) Area = 1/2 * 3 * 4 = 6 In triangle ABC, The sides of this triangle are twice the sides of triangle ABE, hence area would be 2^2 times. Area of ABC = 4*6 = 24 Area od the trapezoid = 24  6 = 18 Correct Option: B



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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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15 Oct 2017, 02:40
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: http://gmatclub.com/forum/trapeziumarea99966.htmlHope it helps. Hi Bunuel, Could you help to explain the area? Why we could use the side to calculate the area of triangle? While formula for area of triangle = (1/2)*(Base)*(Height)
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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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15 Oct 2017, 02:52
hazelnut 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: http://gmatclub.com/forum/trapeziumarea99966.htmlHope it helps. Hi Bunuel, Could you help to explain the area? Why we could use the side to calculate the area of triangle? While formula for area of triangle = (1/2)*(Base)*(Height)ABE and ACD are right triangles: area = 1/2*(leg1)(leg2), which is the same as (1/2)*(Base)*(Height), because legs in a right triangle are perpendicular to each other (so we can consider one of them to be base and another to be the height).
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Re: If BE  CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of
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15 Oct 2017, 02:52



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